Answer:
The new mean of processed transactions is 8
Step-by-step explanation:
The teller averages 5 transactions every 15 minutes.
Taking this to a single transaction basis gives; The teller averages one transaction every 3 minutes.
So, when the time changes to 25 minutes, there is the need to find the number of 3-minutes obtainable from a 25-minute interval
New mean of processed transaction = [tex]\frac{25 minutes}{3 minutes}[/tex] = 8.333
The new mean of transaction every 25 minutes is about 8 transactions
Answer: 8.33/ 8 transactions on the average
Step-by-step explanation:
Since the teller is capable of processing five (5) different transactions on the average every fifteen minutes (15).
This means that he is capable of processing a transaction in three minutes:
15minutes/3transactions
= 3 minutes per transaction.
Judging by this speed ( 3 minutes per transaction); we can deduce the number of transactions the teller is capable of processing on the average in 25 minutes.
Since 3 minutes --------- 1 transaction, in 25 minutes the teller will process :
(25/3) × (1/1)
= 8.333 transactions
approximately 8 transactions.
Find the magnitude of the torque exerted by F on the bolt at P if |vector PQ| = 6 in. and |F| = 16 lb.
Final answer:
To calculate the magnitude of the torque exerted on a bolt, use the formula τ = rFsinθ with the given values: lever arm length of 6 inches, force of 16 lbs, and an assumed angle of 90 degrees. The calculated torque is 96 in.lb
Explanation:
The magnitude of torque (τ) can be calculated using the formula τ = rFsinθ, where r is the distance from the pivot to the point where the force is applied, F is the force applied, and θ is the angle between the force and the arm of the lever—in this case, assuming the force is perpendicular (90 degrees) to the lever arm for maximum torque.
Given that |vector PQ| = 6 in. as the lever arm (r) and |F| = 16 lb. as the force applied, and assuming the angle θ is 90 degrees (since it's not provided but necessary for calculating maximum torque):
τ = rFsinθ = (6 in.)(16 lb.)(sin90°) = 96 in.lb. because sin90° = 1.
Therefore, the magnitude of the torque exerted by the force on the bolt at point P is 96 in.lb.
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the basis of body mass index (BMI), expressed as weight/height2. An adult is considered overweight if the BMI is 25 or more but less than 30. An obese adult will have a BMI of 30 or greater. According to a January 2012 article in the Journal of the American Medical Association, 33.1% of the adult population in the United States is overweight and 35.7% is obese. Use this information to answer the following questions.A. What is the probability that a randomly selected adult is either overweight or obese? (Round your answer to 3 decimal places.)B. What is the probability that a randomly selected adult is neither overweight nor obese? (Round your answer to 3 decimal places.)C. Are the events "overweight" and "obese" exhaustive?D. Are the events "overweight" and "obese" mutually exclusive?
Answer:
(A) The probability that a randomly selected adult is either overweight or obese is 0.688.
(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.
(C) The events "overweight" and "obese" exhaustive.
(D) The events "overweight" and "obese" mutually exclusive.
Step-by-step explanation:
Denote the events as follows:
X = a person is overweight
Y = a person is obese.
The information provided is:
A person is overweight if they have BMI 25 or more but below 30.
A person is obese if they have BMI 30 or more.
P (X) = 0.331
P (Y) = 0.357
(A)
The events of a person being overweight or obese cannot occur together.
Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.
So, P (X ∩ Y) = 0.
Compute the probability that a randomly selected adult is either overweight or obese as follows:
[tex]P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688[/tex]
Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.
(B)
Commute the probability that a randomly selected adult is neither overweight nor obese as follows:
[tex]P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312[/tex]
Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.
(C)
If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.
For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.
In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.
Thus, the events "overweight" and "obese" exhaustive.
(D)
Mutually exclusive events are those events that cannot occur at the same time.
The events of a person being overweight and obese are mutually exclusive.
An English teacher needs to pick 10 books to put on her reading list for the next school year, and she needs to plan the order in which they should be read. She has narrowed down her choices to 4 novels, 6 plays, 8 poetry books, and 4 nonfiction books. Step 1 of 2: If she wants to include no more than 3 poetry books, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place. Answer Tables II Keypa a x10
Calculating the number of different reading schedules an English teacher can create involves combining and permuting selections of novels, plays, poetry (up to 3), and nonfiction books from a list.
It entails calculating the combinations of books and then the permutations for the order of reading.
Detailed mathematical operations lead to the solution, expressed in scientific notation.
Explanation:The task is to determine the number of different reading schedules possible if an English teacher selects 10 books out of a potential 22 books (4 novels, 6 plays, 8 poetry books with a restriction of choosing no more than 3, and 4 nonfiction books) to include on her reading list for the next school year, planning the order in which they should be read.
Understanding the problem involves calculating combinations and permutations.
There are two steps to solve this problem:
First, calculate the total possible combinations of selecting 10 books when up to 3 can be poetry books. This includes considering combinations of all other types of books as well.Second, calculate the permutations of these combinations to determine the order in which the books can be read.Consider the constraint on the poetry books:
Selecting 0 to 3 poetry books out of 8, we have Σ from i=0 to 3 of C(8, i); where C(n, k) is the number of combinations of n items taken k at a time.For the remaining books (4 novels, 6 plays, 4 nonfiction), we must select enough to total 10 books including the poetry books selected. This means selecting from 7 to 10 books out of 14, as we can select 0, 1, 2, or 3 poetry books respectively.The next step combines these selections and calculates the permutations of each combination to arrange them in order.
Due to the complexity and lengthiness of full calculations, and presentational limitations, detailed computations for each step are not displayed here.
However, using combinations and permutations formulas, one can calculate the total number of different reading schedules and express this number in scientific notation rounding to the hundredths place as requested.
g Two players each put one dollar into a pot. They decide to throw a pair of dice alternately. The first one who throws the two dice so that the sum of the two faces is five (5) wins the pot. How much should the player who starts add to the pot to make the game a ‘fair game’? Give your answer as a fraction in reduced form (i.e. A/B where A and B do not share any common factors). Note: A ‘fair game’ would be one where the expected payment to each player is equal
Answer:
The first player should add 1/8 more dollar to the pot
Step-by-step explanation:
The probability that the first player wins on the first turn is equivalent to the probability of getting 5 in the dices. Out of the 36 possible outcomes for a dice, only 4 are favourable in the event of obtaining 5 in the sum:
- first dice 1, second dice 4
- first dice 2, second dice 3
- first dice 3, second dice 2
- first dice 4, second dice 1
Thus, the probability that the first player wins on the first turn is 4/36 = 1/9.
Note that if the first player doesnt win in the first turn, then the second player will be at the same position the first player was at the start of the game. If we call P the probability that the first player wins, then, the second player will have a probability of P of winning after 'surviving' on the first turn.
As a result, the probability that the second player wins is P multiplied by 8/9 (the probability that the first player doenst win in the first turn). In short
Probability that the first player wins = P
Probability that the second player wins = 8/9 * P
Since the sum should be 1 (because both events are complementary from each other), then P + P*8/9 = 1, thus
17/9 P = 1
P = 9/17
Lets call E the extra amount of money player A should include into the pot (in dollard). We have that both players should have expected gain equal to 0, and we have that
- First player gains 1 if he wins (with probability 9/17)
- First player gains -1-E if the loses (with probability 8/17)
Therefore
9/17*1+ 8/17 (-1-E) = 0
9/17 = 8/17 + E 8/17
E = (1/17) / (8/17) = 1/8
The first player should add 1/8 more dollar to the pot.
HELP ME PLS ASAP!!!!!!!!!!!
Answer:
[tex]\sigma^2=73.96 \ kg^2[/tex]
Step-by-step explanation:
Standard Deviation and Variance
If we have a data set of measured values the variance is defined as the average of the squared differences that each value has from the mean. The formula to calculate the variance is:
[tex]\displaystyle \sigma^2=\frac{\sum(x_i-\mu)^2}{n}[/tex]
Where [tex]\mu[/tex] is the mean of the measured values xi (i running from 1 to n), and n is the total number of values.
[tex]\displaystyle \mu=\frac{\sum x_i}{n}[/tex]
The standard deviation is known by the symbol [tex]\sigma[/tex] and is the square root of the variance. We know the standar deviation of the weight in kg of a group of teenagers to be 8.6 kg. Thus, the variance is
[tex]\sigma^2=8.6^2=73.96 \ kg^2[/tex]
[tex]\boxed{\sigma^2=73.96 \ kg^2}[/tex]
Consider a production process that produces batteries. A quality engineer has taken 20 samples each containing 100 batteries. The total number of defective batteries observed over the 20 samples is 200.
Construct a 95% confidence interval of the proportion of defectives.
Another sample of 100 was taken and 15 defectives batteries were found. What is your conclusion?
Answer:
The 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).
It is better to take a larger sample to derive conclusion about the true parameter value.
Step-by-step explanation:
The (1 - α) % confidence interval for proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
Given:
n = 2000
X = 200
The sample proportion is:
[tex]\hat p=\frac{X}{n}=\frac{200}{2000}=0.10[/tex]
The critical value of z for 95% confidence interval is:
[tex]z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Compute the 95% confidence interval as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.10\pm1.96\times\sqrt{\frac{0.10(1-0.10)}{2000}}\\=0.10\pm0.0034\\=(0.0966, 0.1034)[/tex]
Thus, the 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).
Now if in a sample of 100 batteries there are 15 defectives, the the 95% confidence interval for this sample is:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.15\pm1.96\times\sqrt{\frac{0.15(1-0.15)}{100}}\\=0.15\pm0.0706\\=(0.0794, 0.2206)[/tex]
It can be observed that as the sample size was decreased the width of the confidence interval was increased.
Thus, it can be concluded that it is better to take a larger sample to derive conclusion about the true parameter value.
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A particle is moved along the x-axis by a force that measures 9/(6 x) pounds at a point x feet from the origin. Find the work W done in moving the particle from the origin to a distance of 12 ft.
Answer:
5.0 ft-lbf
Step-by-step explanation:
The force is
[tex]F = \dfrac{9}{6^x}[/tex]
This force is not a constant force. For a non-constant force, the work done, W, is
[tex]W = \int\limits^{x_2}_{x_1} {F(x)} \, dx[/tex]
with [tex]x_1[/tex] and [tex]x_2[/tex] the initial and final displacements respectively.
From the question, [tex]x_1 =0[/tex] and [tex]x_2 = 12[/tex].
Then
[tex]W = \int\limits^{12}_0 {\dfrac{9}{6^x}} \, dx[/tex]
Evaluating the indefinite integral,
[tex]\int\limits \dfrac{9}{6^x} \, dx =9 \int\limits\!\left(\frac{1}{6}\right)^x \, dx[/tex]
From the rules of integration,
[tex]\int\limits a^x\, dx = \dfrac{a^x}{\ln a}[/tex]
[tex]9 \int\limits \left(\frac{1}{6}\right)^x \, dx = 9\times\dfrac{(1/6)^x}{\ln(1/6)} = -5.0229\left(\dfrac{1}{6}\right)^x[/tex]
Returning the limits,
[tex]\left.-5.0229\left(\dfrac{1}{6}\right)^x\right|^{12}_0 = -5.0229(0.1667^{12} - 0.1667^0) = 5.0229 \approx 5.0 \text{ ft-lbf}[/tex]
A researcher wants to investigate if the use of e-cigarettes differs across three racial/ethnic groups. He surveys 100 adults from each racial/ethnic group. What statistical test should be used
Options: A. Chi squared Statistics
B. ANOVA
C. Independent samples t-test.
D. z-test of a population proportion.
Answer:A. Chi squared Statistics
Step-by-step explanation: A Chi Squared Statistics is a Statistical technique or test measure that determines how an expectation compares to actual observation. Chi squared Statistics data is expected to have the following features
Such as the data must be RAW, RANDOM, MUTUALLY EXCLUSIVE, OBTAINED FROM INDEPENDENT VARIABLES, AND LARGE SAMPLES WHICH WILL BE ENOUGH.
the survey of one hundred adults from each ethnic/racial groups means the data possess all the characters of a Chi Squared Statistics or test measure.
A company says its premium mixture of nuts contains 13% Brazil nuts, 19% cashews, 17% almonds, and 8% hazelnuts, and the rest are peanuts. You buy a large can and separate the various kinds of nuts. Upon weighing them, you find there are 111 grams of Brazil nuts, 183 grams of cashews, 209 grams of almonds, 79 grams of hazelnuts, and 437 grams of peanuts. You wonder whether your mix is significantly different from what the company advertises.
a) Explain why the chi-square goodness-of-fit test is not an appropriate way to find out.
b) What might you do instead of weighing the nuts in order to use a
X2
test?
Final answer:
The chi-square goodness-of-fit test is inappropriate for the nut mixture scenario due to the data being continuous (weights) rather than categorical counts. Counting the nuts instead of weighing them could make the test applicable.
Explanation:
The question concerns whether a chi-square goodness-of-fit test is appropriate for analyzing if the mix of nuts purchased differs significantly from what the company advertises.
a) The chi-square goodness-of-fit test is not suitable in this scenario because it requires categorical data representing frequencies or counts of categories, whereas the data provided are weights of different categories of nuts. This test is designed for count data, not continuous data like weights.
b) Instead of weighing the nuts, one could count the number of individual nuts in each category. This would convert the data into a suitable form for a chi-square test since you would be working with counts of items (nuts in each category) rather than their weights, aligning with the test's requirements for categorical frequency data.
A hip joint replacement part is being stress-tested in a laboratory. The probability of successfully completing the test is 0.830. 7 randomly and independently chosen parts are tested. What is the probability that exactly two of the 7 parts successfully complete the test
Answer:
0.00205
Step-by-step explanation:
This is binomial distribution problem.
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 7
x = Number of successes required = 2
p = probability of success = 0.830
q = probability of failure = 1 - 0.83 = 0.17
P(X =2) = ⁷C₂ 0.83² 0.17⁷⁻²
P(X =2) = ⁷C₂ 0.83² 0.17⁵ = 0.00205
Benedict Company leased equipment to Mark Inc. on January 1, 2017. The lease is for an eight-year period, expiring December 31, 2024. The first of eight equal annual payments of $600,000 was made on January 1, 2017. Benedict had purchased the equipment on December 29, 2016, for $3,200,000. The lease is appropriately accounted for as a sales-type lease by Benedict. Assume that at January 1, 2017, the present value of all rental payments over the lease term discounted at a 10% interest rate was $3,520,000.
Answer:
From the calculation the interest income in 2018 = 10% of $2,612,000= $261,200
Step-by-step explanation:
The complete question states:
What amount of interest income should Benedict record in 2018 (the second year of the lease period) as a result of the lease?
To answer the question, we look at the given information
The present value given = $3,520,000
The Annual Instalmental payments = $600,000
The Period is 8 years and teh given interest rate at 10%
Based on the information we prepare the following schedule
Year 0 = Instalment + interest = Principal
$600,000 + 0 (interest) = $600,000
Balance = Present value- Principal
Balance = $3520,000 - $600,000 = $2,920,000
Year 1 (2017) = Instalment + interest = Principal
$600,000 + $292,000 (10% of $2,920,000) = $308,000
Balance = Present value- Principal
Balance = $2,920,000- $308,000= $2,612,000
Year 2 (2018) = Instalment + interest = Principal
$600,000 + $261,200(10% of $2,612,000) = $338,000
Balance = Present value- Principal
Balance = $2,612,000- $338,000= $2,350,800
From the calculation the interest income in 2018 = 10% of $2,612,000= $261,200
Tim retired during the current year at age 58. He purchased an annuity from American National Life Company for $40,000. The annuity pays Tim $500 per month for life. Compute Tim�s annual exclusion (Points : 3) a. $1,500.20
b. $1,200.40c. $3,000.20
d. $1,544.40
Answer:
A
Step-by-step explanation:
For ANOVA, the test statistic is called an ____ test statistic (also called a ____-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance).
The test statistic for ANOVA is called an F test statistic (or F-ratio), which is calculated by dividing the variance between the samples by the variance within the samples.
Explanation:The test statistic for ANOVA is called an F test statistic (also called an F-ratio). It is calculated by dividing the variance between the samples (variation due to treatment) by the variance within the samples (variation due to error or chance).
The F statistic follows an F distribution with (number of groups - 1) as the numerator degrees of freedom and (number of observations - number of groups) as the denominator degrees of freedom.
For ANOVA (Analysis of Variance), the test statistic is called an F test statistic (also called an F-ratio), which is the variance between samples (a.k.a., variation due to treatment) divided by the variance within samples (a.k.a., variation due to error or chance).
Fill in blank (2): between, Fill in blank (3): within
Analysis of Variance (ANOVA)
ANOVA is a statistical method used to compare the means of three or more samples to see if at least one of them is significantly different from the others. It does this by analyzing the variances within the data.
F-Test Statistic
The test statistic used in ANOVA is called the F-test statistic or the F-ratio. This F-ratio helps to determine whether the variances between the sample means are significantly larger than the variances within the samples. The F-ratio is calculated as follows:
[tex]\[ F = \frac{\text{variance between samples}}{\text{variance within samples}} \][/tex]
Variance Between Samples
The variance between samples (also known as between-group variance or treatment variance) measures the variability among the sample means. This variability reflects how much the group means differ from the overall mean. If the group means are very different from each other, the between-group variance will be large. This part of the variance is often attributed to the effect of the different treatments or conditions being compared.
- Blank (2): The term used here is between.
Variance Within Samples
The variance within samples (also known as within-group variance or error variance) measures the variability within each of the groups. This variability reflects how much the individual data points within each group differ from their respective group mean. This part of the variance is usually attributed to random error or chance.
- Blank (3): The term used here is within.
In the context of ANOVA:
- The F-test statistic (F-ratio) is used to compare the variance between samples to the variance within samples.
- The variance between samples represents the variation due to treatment.
- The variance within samples represents the variation due to error or chance.
The complete question is
For ANOVA, the test statistic is called an_ test statistic (also called an_-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance _(3)_samples (a.k.a., variation due to error or chance) The first two blanks are completed with the letter this author uses for the ANOVA test statistic. What is this letter? Fill in blank (2): Fill in blank (3)
The Bureau of Labor Statistics (BLS) collects data regarding occupational employment and wage estimates in the United States. The stem-and-leaf plot below represents the annual wage estimates, in thousands of dollars, for 15 select architecture and engineering occupations in May 2014. 6 0 5 7 5 10 0 1 4 5 8 4 5 8 Leaf Unit = $ 1 000 Identify all of the true statements regarding the given data.a.The median annual wage estimate was $85,000 per year for these 15 architecture and engineering occupations. b.The modal class is 100 to 110 thousand dollars, with five occupations having salaries in that class. c.The shape of the distribution is skewed to the right. d.The modal annual wage estimate was $92,000 per year for these 15 architecture and engineering occupations.e.The annual wage of $108,000 was the largest wage estimate for architecture and engineering occupations in May 2014.
The true statements are:
a. The median annual wage estimate was $85,000 per year for these 15 architecture and engineering occupations.c. The shape of the distribution is skewed to the right.e. The annual wage of $108,000 was the largest wage estimate for architecture and engineering occupations in May 2014.Let's analyze each statement based on the provided stem-and-leaf plot:
Given stem-and-leaf plot:
Stem | Leaf
0 | 0 1 4 5 5 8
1 | 0 0 4 5 8
a. The median annual wage estimate was $85,000 per year for these 15 architecture and engineering occupations.
The median is the middle value when the data is arranged in ascending order.
The middle value is the 8th value, which is 85 (thousands of dollars).
This statement is true.
b. The modal class is 100 to 110 thousand dollars, with five occupations having salaries in that class.
The modal class is the class with the highest frequency.
In this case, the class 100 to 110 thousand dollars has three frequencies. This statement is false.
c. The shape of the distribution is skewed to the right.
Skewed to the right means that the distribution's tail is on the right side. Looking at the stem-and-leaf plot, we can see that the data has more values on the left side and tails off to the right.
This statement is true.
d. The modal annual wage estimate was $92,000 per year for these 15 architecture and engineering occupations.
The mode is the most frequently occurring value.
The mode in this data set is not explicitly shown, but we can see that there is no class with a significantly higher frequency.
This statement is false.
e. The annual wage of $108,000 was the largest wage estimate for architecture and engineering occupations in May 2014.
The largest value in the data set is 108 (thousands of dollars), which corresponds to an annual wage of $108,000.
This statement is true.
So, the true statements are:
a. The median annual wage estimate was $85,000 per year for these 15 architecture and engineering occupations.
c. The shape of the distribution is skewed to the right.
e. The annual wage of $108,000 was the largest wage estimate for architecture and engineering occupations in May 2014.
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Without the visual plot, it's impossible to confirm any statements with certainty. However, a theoretical ordering of the given data suggests that the median annual wage isn't $85,000 but closer to $84,000, the claimed modal class has fewer than five salaries, and the shape does seem to be skewed right. The highest wage seems to be $158,000 not $108,000.
Explanation:This question involves understanding and interpreting a stem-and-leaf plot. Unfortunately, without being able to view the plot, it's impossible to conclusively confirm any of the statements provided.
However, we can do a generalized analysis based on the given data. Typically, a stem-and-leaf plot represents numerical data in an order, which makes it easier to see certain statistics like median, mode and range, and also the distribution shape. If the statement that the stem represents 'tens' and leaf represents 'ones' is true, then the data arranged in ascending order would be: 50, 57, 60, 75, 75, 80, 84, 85, 100, 101, 104, 105, 108, 145, 158 (all in thousands).
From these, it's clear: a) The median annual wage is not $85,000, it's actually $84,000. b) There aren't five salaries in the 100-110k range, there are only four. So the modal class isn't $100,000 to $110,000. c) The distribution appears to be skewed to the right, given that higher wages are less common. d) The modal wage estimate is unclear without knowing how many times each wage appears in the original data. e) The highest wage is $158,000 (not $108,00), assuming that the stem-and-leaf plot does not have repeating values.
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Consider the experimental situation described below. Ability to grow in shade may help pines in the dry forests of Arizona resist drought. How well do these pines grow in shade? Investigators planted pine seedlings in a greenhouse in either full light or light reduced to 5% of normal by shade cloth. At the end of the study, they dried the young trees and weighed them. Identify the experimental unit(s) or subject(s)?
a) shade cloth
b) pine tree seedlings
c) drought resistance
d) greenhouses
e) rainy seasons
Final answer:
The experimental unit in the given scenario is the pine tree seedlings which are used to test the hypothesis regarding their growth in shade and adaptation to drought conditions.
Explanation:
The experimental unit or subject in the described experiment is b) pine tree seedlings. These seedlings are what the investigators manipulate (by altering light conditions) and measure (by weighing after drying) to test the hypothesis regarding the growth of pines in shade as an adaptation to resist drought.
This experiment tests how pine seedlings grow in different lighting conditions to characterize features like growth rate and drought resistance, informed by observations such as acclimatization where the structure of leaves change when transitioning from sun to shade or vice versa to achieve photosynthetic efficiency.
A professor, transferred from Toronto to New York, needs to sell his house in Toronto quickly. Someone has offered to buy his house for $220,000, but the offer expires at the end of the week. The professor does not currently have a better offer but can afford to leave the house on the market for another month. From conversations with his realtor, the professor believes the price he will get by leaving the house on the market for another month is uniformly distributed between $210,000 and $235,000.
(a) If he leaves the house on the market for another month, what is the probability that he will get at least $225,000 for the house?
(b) If he leaves it on the market for another month, what is the probability he will get less than $217,000?
(c) What is the expected value and standard deviation of the house price if it is left in the market?
Answer:
(a) = 40%
(b) = 28%
(c) Expected value = $222,500
Standard deviation = $7,216.88
Step-by-step explanation:
This is a normal distribution with a = 210,000 and b =235,000
(a) The probability that he will get at least $225,000 for the house is:
[tex]P(X\geq 225,000) =1 -\frac{225,000-a}{b-a} =1-\frac{225,000-210,000}{235,000-210,000} \\P(X\geq 225,000) =0.4= 40\%[/tex]
(b)The probability he will get less than $217,000 is:
[tex]P(X\leq 217,000) =\frac{217,000-a}{b-a} =\frac{217,000-210,000}{235,000-210,000} \\P(X\leq 217,000) =0.28= 28\%[/tex]
(c) The expected value (E) and the standard deviation (S) are:
[tex]E=\frac{a+b}{2}=\frac{210,000+235,000}{2}\\ E=\$222,500\\S=\frac{b-a}{\sqrt{12}}=\frac{235,000-210,000}{\sqrt{12}}\\S=\$7,216.88[/tex]
The probabilities of the house selling for at least $225,000 and less than $217,000 are 40% and 28% respectively. The expected selling price if left on the market is $222,500 and the standard deviation is around $7,211.1.
Explanation:This question is about the calculation of probabilities and expected values related to the selling price of a house. Let's solve this step by step:
We want to calculate the probability that the price of the house will be at least $225,000. The price range is uniformly distributed between $210,000 and $235,000. Therefore, to find the probability of the price being at least $225,000, we subtract the lower limit of this range ($225,000) from the upper limit ($235,000), and divide that by the entire possible price range ($235,000 - $210,000). That is $(235,000-225,000)/(235,000-210,000) = 0.4$ or 40%.
Next, to calculate the probability that the price of the house will be less than $217,000, we subtract the lower limit of the price range ($210,000) from our desired limit ($217,000), and divide that by the full price range ($235,000 - $210,000). That is $(217,000-210,000)/(235,000-210,000) = 0.28$ or 28%.
For part (c), the expected value of a uniform distribution is the midpoint of the range, or $(210,000 + 235,000)/2 = $222,500$. The standard deviation of a uniform distribution is the square root of ((upper limit - lower limit)^2/12), or sqrt[(235,000 - 210,000)^2/12] = $7,211.1$ approximately.
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Samantha's pie stand has 17 average visitors per day for the past 6 days. How many people must visit her stand on the seventh day to maintain her quota of 20 visitors per day?
Step-by-step explanation:
Below is an attachment containing the solution.
7. Calculate the distance Derek rows
his boat if he rows a mile each day
for 11 days.
In the 2009 General Social Survey, respondents were asked if they favored or opposed death penalty for people convicted of murder. The 95% confidence interval for the population proportion who were in favor (say, p) was (0.65, 0.69). For the above data, the 99% confidence interval for the true population proportion of respondents who were opposed to the death penalty would be narrower than the one your derived above
Answer:
The calculated 99% confidence interval is wider than the 95% confidence interval.
Step-by-step explanation:
We are given the following in the question:
95% confidence interval for the population proportion
(0.65, 0.69)
Let [tex]\hat{p}[/tex] be the sample proportion
Confidence interval:
[tex]p \pm z_{stat}(\text{Standard error})[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Let x be the standard error, then, we can write
[tex]\hat{p} - 1.96x = 0.65\\\hat{p}+1.96x = 0.69[/tex]
Solving the two equations, we get,
[tex]2\hat{p} = 0.65 + 0.69\\\\\hat{p} = \dfrac{1.34}{2} = 0.67\\\\x = \dfrac{0.69 - 0.67}{1.96} \approx 0.01[/tex]
99% Confidence interval:
[tex]p \pm z_{stat}(\text{Standard error})[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.01} = 2.58[/tex]
Putting values, we get,
[tex]0.67 \pm 2.58(0.01)\\=0.67 \pm 0.0258\\=(0.6442,0.6958)[/tex]
Thus, the calculated 99% confidence interval is wider than the 95% confidence interval .
A commuter has to cross a train track each day on the way to work. The probability of having to wait for a train is .2. If she has to wait for a train, her commute takes 25 minutes; otherwise, it takes 20 minutes. What is her expected commute time?
Answer:
21 minutes
Step-by-step explanation:
Her expected commute time is given by the probability of having to wait for the train (0.2) multiplied by the commute time in this scenario (25 min), added to the probability of not having to wait for the train (1 - 0.2) multiplied by the commute time in this scenario (20 min). The expected commute time is:
[tex]E(X) = 0.2*25+(1-0.2)*20\\E(X) = 21\ minutes[/tex]
Her expected commute time is 21 minutes.
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (Enter your answer using interval notation.) ty'' + 7y = t, y(1) = 1, y'(1) = 7
Answer:
[tex] y'' + \frac{7}{t} y = 1[/tex]
For this case we can use the theorem of Existence and uniqueness that says:
Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:
[tex] y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o[/tex]
has unique solution defined for all t in [a,b]
If we apply this to our equation we have that p(t) =0 and [tex] q(t) = \frac{7}{t}[/tex] and [tex] g(t) =1[/tex]
We see that [tex] q(t)[/tex] is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by [tex] (0, \infty)[/tex]
And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.
Step-by-step explanation:
For this case we have the following differential equation given:
[tex] t y'' + 7y = t[/tex]
With the conditions y(1)= 1 and y'(1) = 7
The frist step on this case is divide both sides of the differential equation by t and we got:
[tex] y'' + \frac{7}{t} y = 1[/tex]
For this case we can use the theorem of Existence and uniqueness that says:
Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:
[tex] y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o[/tex]
has unique solution defined for all t in [a,b]
If we apply this to our equation we have that p(t) =0 and [tex] q(t) = \frac{7}{t}[/tex] and [tex] g(t) =1[/tex]
We see that [tex] q(t)[/tex] is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by [tex] (0, \infty)[/tex]
And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.
Answer:
The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is (0,∞)
Step-by-step explanation:
Given the differential equation:
ty'' + 7y = t .................................(1)
Together with the initial conditions:
y(1) = 1, y'(1) = 7
We want to determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution.
First, let us have the differential equation (1) in the form:
y'' + p(t)y' + q(t)y = r(t) ..................(2)
We do that by dividing (1) by t
So that
y''+ (7/t)y = 1 ....................................(3)
Comparing (3) with (2)
p(t) = 0
q(t) = 7/t
r(t) = 1
For t = 0, p(t) and r(t) are continuous, but q(t) = 7/0, which is undefined. Zero is certainly out of the required points.
In fact (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous. But t = 1, which is contained in the initial conditions is found in (0,∞), and that makes it the correct interval.
So the largest interval containing 1 on which p(t), q(t) and r(t) are defined and continuous is (0,∞)
he one‑sample t statistic from a sample of n = 23 observations for the two‑sided test of H 0 : μ = 15 versus H α : μ > 15 has the value t = 2.24 . Based on this information: 0.01 < P ‑ value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct. P ‑ value > 0.1 . we would reject the null hypothesis at α = 0.025 . 0.01 < P ‑ value < 0.025 .
Answer:
[tex] t = 2.24[/tex]
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=23-1=22[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]
And for this case we can conclude that:
[tex] 0.01 < p_v < 0.025[/tex]
And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X[/tex] represent the mean height for the sample
[tex]s[/tex] represent the sample standard deviation
[tex]n=23[/tex] sample size
[tex]\mu_o =15[/tex] represent the value that we want to test
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 15, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 15[/tex]
Alternative hypothesis:[tex]\mu > 15[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
For this case the statistic is given:
[tex] t = 2.24[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=23-1=22[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]
And for this case we can conclude that:
[tex] 0.01 < p_v < 0.025[/tex]
And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]
With a one-sample t-test (t = 2.24, n = 23) for H0: μ = 15 vs. Hα: μ > 15, P-value (0.01 < P < 0.025) supports rejecting the null at α = 0.025. The claim "P-value > 0.1" is incorrect.
Based on the given information, we have a one-sample t statistic from a sample of n = 23 observations for the two-sided test of H0: μ = 15 versus Hα: μ > 15, with a t value of 2.24.
To determine the P-value, we compare the t value to a t-distribution with n-1 degrees of freedom. In this case, since we have n = 23 observations, we would compare the t value of 2.24 to the t-distribution with 22 degrees of freedom.
The P-value is the probability of observing a t value as extreme as 2.24 or more extreme, assuming the null hypothesis is true.
Now, let's evaluate the given statements:
1. "0.01 < P-value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct."
Since the P-value is between 0.01 and 0.025, it falls within the critical region for α = 0.025. This means that the P-value is smaller than α, so we would reject the null hypothesis at α = 0.025. Therefore, the statement is correct.
2. "P-value > 0.1. We would reject the null hypothesis at α = 0.025."
Since the given P-value is not greater than 0.1, this statement is incorrect. If the P-value is larger than the significance level α (in this case, 0.025), we would fail to reject the null hypothesis. In other words, we do not have enough evidence to conclude that the population mean is greater than 15.
Based on the above analysis, the correct statement is:
- "0.01 < P-value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct."
It is important to note that the decision to reject or fail to reject the null hypothesis depends on the chosen significance level α and the P-value. The P-value measures the strength of the evidence against the null hypothesis, while the significance level determines the threshold for rejecting the null hypothesis.
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Suppose that we examine the relationship between high school GPA and college GPA. We collect data from students at a local college and find that there is a strong, positive, linear association between the variables. The linear regression predicted college GPA = 1.07 + 0.62 * high school GPA. The standard error of the regression, se, was 0.374. What does this value of the standard error of the regression tell us?
Answer:
The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).
Step-by-step explanation:
The linear regression line for College GPA based on High school GPA is:
College GPA = 1.07 + 0.62 High-school GPA
It is provided that the standard error of the regression line is,
[tex]s_{e}=0.374[/tex]
The standard error of a regression line is the average distance between the predicted value and the regression equation.
It is the square root of the average squared deviations.
It is also known as the standard error of estimate.
The standard error of 0.374 implies that:
The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).
The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA. A smaller standard error indicates a better fit of the model to the data. In this case, the small standard error suggests that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.
Explanation:The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA based on the high school GPA. In this case, the standard error of the regression is 0.374. This means that, on average, the actual college GPA for a student deviates from the predicted college GPA by approximately 0.374.
This value gives us an idea of the accuracy of the linear regression model in predicting college GPA based on high school GPA. A smaller standard error indicates a better fit of the model to the data, implying that the predicted college GPA is closer to the actual college GPA. Conversely, a larger standard error suggests that the model's predictions are less accurate.
In this case, the standard error of the regression is relatively small (0.374), which indicates that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.
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Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in order to treat peptic or digestive disorders. The distribution of plasma aluminum levels is known to be approximately normal; however its mean u and standard deviation o are not known. The mean aluminum level for the sample of n = 10 infants is found to be X = 37.20 ug/l and the sample standard deviation is s = 7.13 ug/1. Furthermore, the mean plasma aluminum level for the population of infants not receiving antacids is known to be only 4.13 ug/1.(a) Formulate the null hypothesis and complementary alternative hypothesis, for a two-sided test of whether the mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.(b) Construct a 95% confidence interval for the true mean plasma aluminum level of the population of infants receiving antacids.(c) Calculate the p-value of this sample (as best as possible), at the a=.05 significance level.(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the alternative hypothesis, at the a = .05 significance level? Interpret your conclusion: What exactly has been demonstrated, based on the empirical evidence?(e) With the knowledge that significantly elevated plasma aluminum levels are toxic to human beings, reformulate the null hypothesis and complementary alternative hypothesis, for the appropriate one-sided test of the mean plasma aluminum levels. With the same sample data as above, how does the new p-value compare with that found in part (c), and what is the resulting conclusion and interpretation?
Answer:
a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids.
b. (32.1, 42.3)
c. p-value < .00001
d. The null hypothesis is rejected at the α=0.05 significance level
e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.
p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.
Step-by-step explanation:
a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.
b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:
(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })
Substituting in the values of X bar and s, the interval becomes:
(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })
or (32.1, 42.3)
c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:
t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.
Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.
d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly changes the plasma aluminum levels of children.
e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.
Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.
To test whether the mean plasma aluminum level of infants on antacids differs from those not on antacids, a null hypothesis (that the means are equal) is established alongside an alternative. A confidence interval and p-value are calculated to assess this hypothesis, and based on these results, a decision is made to reject or not reject the null hypothesis.
Explanation:Hypotheses Formulation and Test Statistics
To analyze the plasma aluminum levels in infants receiving antacids compared to those not receiving them, one would perform a hypothesis test. The steps include formulating a null hypothesis (H0) and an alternative hypothesis (Ha), calculating the test statistic, finding the p-value, and making a decision regarding H0 based on the p-value and the confidence interval.
Null Hypothesis (H0): The mean plasma aluminum level of infants receiving antacids is equal to the mean level of those not receiving antacids (H0: μ = 4.13 µg/L).
Alternative Hypothesis (Ha): The mean plasma aluminum level of infants receiving antacids is not equal to the mean level of those not receiving antacids (Ha: μ ≠ 4.13 µg/L).
To construct a 95% confidence interval, we use the sample mean (× = 37.20 µg/L), sample standard deviation (s = 7.13 µg/L), the sample size (n = 10), and the t-distribution since the population variance is unknown. The confidence interval provides a range of values within which the true mean is likely to lie.
For the p-value, we compare it against the alpha level α=0.05. If the p-value is less than α, we reject H0; otherwise, we do not reject H0. The p-value indicates the likelihood of obtaining a sample mean at least as extreme as the one observed if H0 were true.
If the confidence interval does not include the population mean of children not receiving antacids and the p-value is less than α, we reject H0 in favor of Ha. If a one-sided test is appropriate (for example, if we only want to test if the mean aluminum level is higher in the treated group), Ha would be reformulated accordingly (Ha: μ > 4.13 µg/L), potentially resulting in a different decision from the two-sided test.
Speedy Oil provides a single-server automobile oil change and lubrication service. Customers provide an arrival rate of 2.5 cars per hour. The service rate is 5 cars per hour. Assume that arrivals follow a Poisson probability distribution and that service times follow an exponential probability distribution.
If required, round your answer to the nearest whole number.
(a) What is the average number of cars in the system?
(b) What is the average time that a car waits for the oil and lubrication service to begin?
(c) What is the average time a car spends in the system?
(d) What is the probability that an arrival has to wait for service?
Answer:
(a) Average number of cars in the system is 1
(b) Average time a car waits is 12 minutes
(c) Average time a car spends in the system is 2 minutes
(d) Probability that an arrival has to wait for service is 0.08.
Step-by-step explanation:
We are given the following
Arrival Rate, A = 2.5
Service Rate B = 5
(a) Average Number of Cars in the System is determined by dividing the Arrival Rate A by the difference between the Service Rate B, and Arrival Rate A.
Average number of cars = A/(B - A)
= 2.5/(5 - 2.5)
= 2.5/2.5 = 1
There is an average of 1 car.
(b) Average time a car waits = A/B(B - A)
= 2.5/5(5 - 2.5)
= 2.5/(5 × 2.5)
= 2.5/12.5
= 1/5
= 0.20 hours
Which is 12 minutes
(c) Average time a car spends in the system is the ratio of the average time a car waits to the service rate.
Average time = 0.2/5
= 0.04 hours
= 2.4 minutes
Which is approximately 2 minutes.
(d) Probability that an arrival has to wait for service is the ratio of the average time a car waits to rate of arrivals.
Probability = 0.2/2.5
= 0.08
The average number of cars in the system can be found using Little's Law. The average time a car waits for the oil and lubrication service to begin is half of the average time in the system. The probability that an arrival has to wait for service is obtained by dividing the arrival rate by the service rate.
Explanation:(a) To find the average number of cars in the system, we can use Little's Law. Little's Law states that the average number of cars in the system (L) equals the arrival rate (λ) multiplied by the average time a car spends in the system (W). In this case, λ = 2.5 cars per hour and the service rate (μ) = 5 cars per hour.
So, L = λ * W. Rearranging the formula, W = L / λ. Substituting the values, W = (2.5 / 5) = 0.5 hours or 30 minutes.
(b) The average time that a car waits for the oil and lubrication service to begin is equal to half of the average time a car spends in the system, which is 30 minutes.
(c) The average time a car spends in the system is equal to the average waiting time plus the average service time. The average service time (1/μ) in this case is 1/5 hour or 12 minutes. Therefore, the average time a car spends in the system is 30 minutes + 12 minutes = 42 minutes.
(d) To find the probability that an arrival has to wait for service, we can use the formula P(wait) = λ / μ = 2.5 / 5 = 0.5 or 50%
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More defective components: A lot of 1060 components contains 229 that are defective. Two components are drawn at random and tested. Let A be the event that the first component drawn is defective, and let B be the event that the second component drawn is defective. Write your answer as a fraction or a decimal, rounded to four decimal places.?Explain.
Answer:
The probability of event A is 0.2160.
The probability of event B is 0.2153.
Step-by-step explanation:
Assume that the random variable X is defined as the number of defective components in a lot.
It is provided that of the 1060 component 229 are defective.
The probability of selecting a defective component is:
[tex]P(X)=\frac{229}{1060}=0.2160[/tex]
The proportion of defective components in a lot of 1060 is 0.2160.
It is provided that two components are selected to be tested.
Assuming the selection were without replacement.
A = the first component drawn is defective
B = the second component drawn is defective
Compute the probability of event A:The probability of selecting a defective component from the entire lot
of 1060 component is 0.2160.
Thus, the probability of event A is 0.2160.
Compute the probability of event B:According to event A, the first component selected was defective.
So now there are 228 defective components among 1059
components.
[tex]P(B)=\frac{228}{1059}= 0.2153[/tex]
Thus, the probability of event B is 0.2153.
Both the probabilities are almost same.
This implies that the probability of selecting a defective component from the entire population of these components is approximately 0.2160.
Answer:
Required Probability = 0.0467
Step-by-step explanation:
We are given that a lot of 1060 components contains 229 that are defective.
Two components are drawn at random and tested.
A = event that the first component drawn is defective
B = event that the second component drawn is defective
So,P(first component drawn is defective,A)=(No. of defective component)÷
(Total components)
P(A) = 229/1060
Similarly, P(second component drawn is defective,A) = 229/1060
Therefore, P(both component drawn defective) = [tex]\frac{229}{1060} * \frac{229}{1060}[/tex] = 0.0467 .
Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is 1000 ft2. What capacity (gal/min) pump would you rent to (a) keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, and (b) reduce the water accumulation in your basement at a rate of 3 in./hr even while the backup problem exists
Answer: a) 10.35gal/min
b) 41.52gal/min
Step-by-step explanation:
Let us first determine the volume of the container.
Find the attached file for solution
To keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, you would need to rent a pump with a capacity of 21.19 gallons per minute. To reduce the water accumulation in your basement at a rate of 3 inches per hour, you would need to rent a pump with a capacity of 63.57 gallons per minute.
Explanation:To keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, you would need a pump that can remove water at the same rate it is flooding. In this case, the rate of flooding is 1 inch of depth per hour, which is equivalent to 2.54 cm/hour.
To convert this to gallons per minute, we need to use the fact that 1 gallon is approximately 3.78541 liters and 1 minute is equal to 60 seconds. So, the conversion factor is: 1000 cm² * (2.54 cm/hour) * (1 gallon / 3.78541 liters) * (60 minutes / 1 hour) = 21.18974 gallons per minute.
To reduce the water accumulation in your basement at a rate of 3 inches per hour, you would need a pump that can remove water at that rate. Following the same conversion factor as before, we get: 1000 cm² * (7.62 cm / hour) * (1 gallon / 3.78541 liters) * (60 minutes / 1 hour) = 63.56922 gallons per minute.
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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. It is estimated that 3.7% of the general population will live past their 90th birthday. In a graduating class of 745 high school seniors, find the following probabilities. (Round your answers to four decimal places.)
a. 15 or more will live beyond their 90th birthday
b. 30 or more will live beyond their 90th birthday
c. between 25 and 35 will live beyond their 90th birthday
d. more than 40 will live beyond their 90th birthday
Answer:
a) Bi [P ( X >=15 ) ] ≈ 0.9944
b) Bi [P ( X >=30 ) ] ≈ 0.3182
c) Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623
d) Bi [P ( X >40 ) ] ≈ 0.0046
Step-by-step explanation:
Given:
- Total sample size n = 745
- The probability of success p = 0.037
- The probability of failure q = 0.963
Find:
a. 15 or more will live beyond their 90th birthday
b. 30 or more will live beyond their 90th birthday
c. between 25 and 35 will live beyond their 90th birthday
d. more than 40 will live beyond their 90th birthday
Solution:
- The condition for normal approximation to binomial distribution:
n*p = 745*0.037 = 27.565 > 5
n*q = 745*0.963 = 717.435 > 5
Normal Approximation is valid.
a) P ( X >= 15 ) ?
- Apply continuity correction for normal approximation:
Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]
- Then the parameters u mean and σ standard deviation for normal distribution are:
u = n*p = 27.565
σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522
- The random variable has approximated normal distribution as follows:
X~N ( 27.565 , 5.1522^2 )
- Now compute the Z - value for the corrected limit:
N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )
N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )
- Now use the Z-score table to evaluate the probability:
P ( Z >= -2.5358 ) = 0.9944
N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944
Hence,
Bi [P ( X >=15 ) ] ≈ 0.9944
b) P ( X >= 30 ) ?
- Apply continuity correction for normal approximation:
Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]
- Now compute the Z - value for the corrected limit:
N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )
N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )
- Now use the Z-score table to evaluate the probability:
P ( Z >= 0.37556 ) = 0.3182
N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182
Hence,
Bi [P ( X >=30 ) ] ≈ 0.3182
c) P ( 25=< X =< 35 ) ?
- Apply continuity correction for normal approximation:
Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]
- Now compute the Z - value for the corrected limit:
N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )
N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )
- Now use the Z-score table to evaluate the probability:
P ( -0.59489 =<Z =< 1.54011 ) = 0.6623
N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623
Hence,
Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623
d) P ( X > 40 ) ?
- Apply continuity correction for normal approximation:
Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]
- Now compute the Z - value for the corrected limit:
N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )
N [ P ( X > 41 ) ] = P ( Z > 2.60762 )
- Now use the Z-score table to evaluate the probability:
P ( Z > 2.60762 ) = 0.0046
N [ P ( X > 41 ) ] = P ( Z > 2.60762 ) = 0.0046
Hence,
Bi [P ( X >40 ) ] ≈ 0.0046
Using the normal approximation to binomial is appropriate for the given problem. To find probabilities, we calculate the mean and standard deviation of the binomial, then calculate the z-scores and use the normal distribution to estimate the required probabilities for each case.
Explanation:To determine if it is appropriate to use the normal approximation to the binomial distribution, we must confirm that both np and n(1-p) are greater than 5, where n is the number of trials and p is the probability of success. For a class of 745 students and probability of living past 90 being 3.7%, we have:
np = 745 * 0.037 = 27.565n(1-p) = 745 * (1 - 0.037) = 717.435Since both values are greater than 5, we can proceed with the normal approximation.
Calculations using Normal Approximation:
a. To find the probability of 15 or more students living past 90, we calculate the mean (μ) and standard deviation (σ) for the binomial distribution:
μ = np = 27.565σ = √(np(1-p)) = √(27.565*0.963) ≈ 5.15Next, we find the z-score for 14.5 (since we need more than 15, we use the continuity correction factor of 0.5) and use the standard normal distribution to estimate the probability:
P(X ≥ 15) = 1 - P(X < 15) = 1 - P(Z < (14.5 - μ) / σ)
b. - d. The procedure is similar for parts b, c, and d: calculate the z-scores for the respective values using the binomial mean and standard deviation, and then use the normal distribution to find the probabilities. Based on these calculations, we can provide the required estimates.
The yield in pounds from a day's production is normally distributed with a mean of 1500 pounds and standard deviation of 100 pounds. Assume that the yields on different days are independent random variables. Round the answers to 3 significant digits. (a) What is the probability that the production yield exceeds 1500 pounds on each of five days next week
It appears that the question is incomplete but the answer to the part given is as below.
Answer:
(a) 0.0313
Step-by-step explanation:
In a normal distribution curve, the mean divides the curve into two equal parts. Hence, the probability of being lesser or higher than the mean is 1/2. This is the probability for a single day.
From the question, the probability of a day is independent of that of another day. For 5 days, the probability is
[tex](\frac{1}{2})^5 = 0.5^5 = 0.03125 = 0.0313[/tex] to 3 significant digits.
Assume that the Poisson distribution applies and that the mean number of hurricanes in a certain area is 6.9 per year. a. Find the probability that, in a year, there will be 4 hurricanes. b. In a 45-year period, how many years are expected to have 4 hurricanes? c. How does the result from part (b) compare to a recent period of 45 years in which 4 years had 4 hurricanes? Does the Poisson distribution work well here? a. The probability is nothing. (Round to three decimal places as needed.)
Answer:
a) 9.52% probability that, in a year, there will be 4 hurricanes.
b) 4.284 years are expected to have 4 hurricanes.
c) The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
6.9 per year.
This means that [tex]\mu = 6.9[/tex]
a. Find the probability that, in a year, there will be 4 hurricanes.
This is P(X = 4).
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 4) = \frac{e^{-6.9}*(6.9)^{4}}{(4)!}[/tex]
[tex]P(X = 4) = 0.0952[/tex]
9.52% probability that, in a year, there will be 4 hurricanes.
b. In a 45-year period, how many years are expected to have 4 hurricanes?
For each year, the probability is 0.0952.
Multiplying by 45
45*0.0952 = 4.284.
4.284 years are expected to have 4 hurricanes.
c. How does the result from part (b) compare to a recent period of 45 years in which 4 years had 4 hurricanes? Does the Poisson distribution work well here?
The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.