Which of the following values cannot be​ probabilities?
Select all the values that cannot be probabilities.

1​, -0.51​, 5/3​, √ 2 ​, 3/5​, 0.03​, 0​, 1.37

Answers

Answer 1

Answer:

-0.51, 5/3, sqrt (2), and 1.37

Step-by-step explanation:

As we know probability must be between 0 and 1. It cannot be greater than 1 and cannot be less than 0 or negative

1,  can be the probability as it lies between 0 and 1

-0.51,  cannot be the value of probability as it is negative

5/3 = 1.66 which is greater than 1, so cannot be the value of probability

[tex]\sqrt{2}[/tex] = 1.41 which is greater than 1, so cannot be the value of probability

3/5 = 0.6 can be the probability as it lies between 0 and 1

0.03, can be the probability as it lies between 0 and 1

1.37 which is greater than 1, so cannot be the value of probability


Related Questions

A map uses the scale 1.5 cm = 25 mi. Two cities are 190 miles apart. How far apart are the cities on the map?

Please answer quickly and I will give brainiest to whoever is correct fastest.

Answers

Use simple unitary method:

∵ 1.5 cm on map = 25 miles in reality

∴ x cm on map = 190 miles in reality

[tex]\frac{1.5}{x} = \frac{25}{190}\\ x = 11.4 cm[/tex]

Thus, the two cities are 11.4 cm apart on the map

Final answer:

To determine the map distance between two cities that are 190 miles apart using the scale 1.5 cm = 25 miles, divide the actual distance by the scale ratio (16.67 miles/cm), resulting in 11.4 cm.

Explanation:

For the map with scale 1.5 cm = 25 miles, we first calculate how many miles one centimeter represents by dividing the miles by the centimeters in the scale:

25 miles / 1.5 cm = 16.67 miles/cm

Next, we find the map distance for 190 miles by dividing the actual distance by the distance one centimeter represents:

190 miles / 16.67 miles/cm = 11.4 cm

So, the two cities are 11.4 cm apart on the map.

A certain vibrating system satisfies the equation u'' + γu' + u = 0. Find the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.

Answers

Final answer:

In damped harmonic motion, we calculate damping coefficient γ by comparing the periods of damped and undamped motion. For the given situation where the quasi-period is 90% greater than the undamped period, the damping coefficient is approximately 0.7416.

Explanation:

The subject of this question involves Damped Harmonic Motion, a concept in Physics, related to vibrations and waves. The equation given, u'' + γu' + u = 0, describes the motion where γ denotes the damping coefficient. Here, we have to calculate this damping coefficient when the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.

To solve this, we must use the relationship between damped and undamped periods. The quasi-period T' of a damped harmonic motion relates to the undamped period T as: T' = T/(sqrt(1 - (γ/2)^2)). Now, given that T' = 1.9T, we can but these two equations together:

1.9 = 1/(sqrt(1 - (γ/2)^2))

Solving this for γ, we get γ ≈ 0.7416. Hence, the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is approximately 0.7416.

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Final answer:

The value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the undamped motion is the one that satisfies γ=2*ω*0.9, where ω is the natural frequency of oscillation.

Explanation:

The given equation is for a damped harmonic oscillator, a physical system that oscillates under both a restoring force and a damping force proportional to the velocity of the system. The damping coefficient γ determines the behavior of the system and in this case, we need to find the value of γ such that the quasi period of the damped motion is 90% greater than the period of the undamped motion.

The period of the undamped motion, T₀, is calculated by the formula T₀=2π/sqrt(ω), where ω is the natural frequency of oscillation. The quasi period of the damped motion, Td, is increased by a factor of 1+η (in this case, 1.9 as the increase is 90%) and calculated by the formula Td=T₀(1+η) = T₀*1.9.

The damping ratio η is determined by the damping coefficient γ as η=γ/2ω. Therefore, by combining these expressions and rearranging the terms, we extract γ from these formulas as γ=2ω*η => γ=2*ω*(0.9). Thus, the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is the one which satisfies γ=2*ω*0.9.

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The graph shows the relationship between the number of months different students practiced baseball and the number of games they won:

The title of the graph is Baseball Games. On x axis, the label is Number of Months of Practice. On y axis, the label is Number of Games Won. The scale on the y axis is from 0 to 22 at increments of 2, and the scale on the x axis is from 0 to 12 at increments of 2. The points plotted on the graph are the ordered pairs 0, 1 and 1, 3 and 2, 5 and 3, 9 and 4, 10 and 5, 12 and 6, 13 and 7, 14 and 8,17 and 9, 18 and 10,20. A straight line is drawn joining the ordered pairs 0, 1.8 and 2, 5.6 and 4, 9.2 and 6, 13 and 8, 16.5 and 10, 20.5.

Part A: What is the approximate y-intercept of the line of best fit and what does it represent? (5 points)

Part B: Write the equation for the line of best fit in slope-intercept form and use it to predict the number of games that could be won after 13 months of practice. Show your work and include the points used to calculate the slope.

Answers

Answer:

A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.

B) slope: (5.6-1.8)/(2-0) = 1.9

y = 1.9x + 1.8

(Line of best fit)

x = 13,

y = 1.9(13) + 1.8 = 26.5

Predicted no. of games won after 13 months of practice is 26.5

Final answer:

The y-intercept, representing initial games won and the equation for line of best fit predicting future games won, are determined from the graph data.

Explanation:

Part A:

The y-intercept of the line of best fit is approximately 1.8.It represents the initial number of games won when the number of months of practice is zero.

Part B:

The equation for the line of best fit in slope-intercept form is y = 1.4x + 1.8.

To predict the number of games won after 13 months of practice, substitute x = 13 into the equation:

y = 1.4(13) + 1.8 = 19.5

So, the predicted number of games that could be won after 13 months of practice is 19.5.

What length is the shortest path from A to G in the graph below?

Answers

Answer:

graph a

Step-by-step explanation:

The required shortest length to go from A to G is given as 8. Option C is correct.

What is simplification?

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
The shortest distance from the figure can be given as,
= AE + EG
= 1 + 7
= 8

Thus, the required shortest length to go from A to G is given as 8. Option C is correct.

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Once a customer fills the car with gas at one station, that customer cannot then go fill the same car with gas at another station right away. Are the outcomes E1, E2, E3, and E4 mutually exclusive? Explain.

Answers

Answer:

yes :- P ( Ei ∩ Ej ) = 0

Step-by-step explanation:

These are mutually exclusive events

that is, a customer can only go to a gas station to fill his car with gas par time

P(E1) = 1/4

P(E2) = 1/4

P(E3) = 1/4

P(E4) = 1/4

summation of the four probabilities give 1

Final answer:

The outcomes E1, E2, E3, and E4 are mutually exclusive due to the physical limitation of the car's gas tank being already full, which prevents multiple fill-ups in immediate succession.

Explanation:

The concept of mutually exclusive outcomes in the context of choosing a gas station to fill up a car is referring to the idea that a customer cannot fill their car with gas at one station and then immediately go and fill it at another station. This is because once the car's tank is full, there is no capacity to add more fuel until it has been used.

Therefore, the outcomes E1, E2, E3, and E4, if defined as filling up at station 1, station 2, station 3, and station 4 respectively, are indeed mutually exclusive. When a customer chooses one station, the other stations are no longer an option for fueling during that particular instance, as the car's tank can only be filled once until some gas is used.

In economics, understanding market dynamics like price elasticity and strategies employed by gas stations in an oligopoly to maintain profits can influence consumer behavior and competition. These concepts help in analyzing why a customer might not be able to take advantage of another gas station's prices immediately after filling their tank.

A presidential candidate's aide estimates that, among all college students, the proportion who intend to vote in the upcoming election is at least . If out of a random sample of college students expressed an intent to vote, can we reject the aide's estimate at the level of significance?

Answers

Answer:

[tex]z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24[/tex]  

[tex]p_v =P(z<-2.24)=0.0125[/tex]  

If we compare  the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

Step-by-step explanation:

Assuming the following question: A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60% . If 127 out of a random sample of 240 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 level of significance?

Data given and notation

n=240 represent the random sample taken

X=127 represent the college students expressed an intent to vote

[tex]\hat p=\frac{127}{240}=0.529[/tex] estimated proportion of college students expressed an intent to vote

[tex]p_o=0.6[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that at least 60% of students are intented to vote .:  

Null hypothesis:[tex]p \geq 0.6[/tex]  

Alternative hypothesis:[tex]p < 0.6[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<-2.24)=0.0125[/tex]  

If we compare  the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

derived the MOM and MLE for an exponential distribution with parameter ????. Conduct a Bootstrap simulation to compare the estimation of λ with sample sizes of n = 10, n = 100, and n = 500. Choose true value λ = 0.2 and use B = 1000. Calculate and compare the mean and standard error for each set of simulations to each other as well as their theoretical values.

Answers

Answer:

rm(list=ls(all=TRUE))

set.seed(12345)

N=c(10,100,500)

Rate=0.2

B=1000

MN=SE=rep()

for(i in 1:length(N))

{

n=N[i]

X=rexp(n,rate=Rate)

EST=1/mean(X)

ESTh=rep()

for(j in 1:B)

{

Xh=rexp(n,rate=EST)

ESTh[j]=1/mean(Xh)

}

MN[i]=mean(ESTh)

SE[i]=sd(ESTh)

}

cbind(N,Rate,MN,SE)

For a data set of weights​ (pounds) and highway fuel consumption amounts​ (mpg) of eight types of​ automobile, the linear correlation coefficient is found and the​ P-value is 0.044. Write a statement that interprets the​ P-value and includes a conclusion about linear correlation.

The​ P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is [WHAT PERCENT] which is [LOW OR HIGH] so there [IS OR IS NOT] sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.

​(Type an integer or a decimal. Do not​ round.)

Answers

Answer:

The​ P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is 4.4 which is LOW so there IS sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.

Step-by-step explanation:

Hello!

Remember:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

Let's say that the significance level of this correlation test is α: 0.05

If the p-value is the probability of obtaining

you can express it as a percentage: 4.4%Is a very low probability. The decision rule using the p-value is:

p-value < α ⇒ Reject the null hypothesis

p-value ≥ α ⇒ Do not reject the null hypothesis.

The p-value is less than the significance level, the decision is to reject the null hypothesis.

In a linear correlation analysis the statement "there is no linear correlation between the two variables" is always in the null hypothesis, so if you reject it, you can conclude that there is a linear correlation between the variables.

I hope it helps!

Final answer:

In statistical analysis, a P-value of 0.044 indicates there is a 4.4% chance of obtaining a linear correlation as extreme as the observed correlation coefficient. A P-value under 0.05 provides enough evidence to reject the null hypothesis of no correlation, therefore suggesting a significant correlation. Hence, there is sufficient evidence of a linear correlation between car weight and highway fuel consumption.

Explanation:

The P-value of 0.044 in this context represents the likelihood of obtaining a linear correlation coefficient for the data points in your dataset that is as extreme as, or more extreme than, the one you calculated, assuming there is no linear relationship between the two variables (weight and highway fuel consumption in automobiles). This probability is 4.4% - rather low. A common threshold for significance in many fields is 0.05, or 5%. If your P-value is below this threshold, we reject the null hypothesis that there is no correlation and conclude there may be a correlation. Therefore, as the P-value is 0.044, which is below the 0.05 threshold, there is sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.

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A survey found that​ women's heights are normally distributed with mean 63.4 in and standard deviation 2.4 in. A branch of the military requires​ women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?

Answers

Correct Answer: 98.78

A. 98. 78

A) Part 2:  ​No, because only a small percentage of women are not allowed to join this branch of the military because of their height.

B. 57.8 ,    68.3

Final answer:

Approximately 98.8% of women meet the current military height requirement of 58-80 inches. If the military change to exclude the shortest 1% and tallest 2%, the height requirements change to approximately 58.11 to 68.38 inches.

Explanation:

To answer part (a) of your question, we must utilize the notion of a Z-score. A Z-score standardizes or normalizes a data point in terms of how many standard deviations away it is from the mean. A standard deviation of 2.4 and a mean of 63.4 inches would be the reference points. Using the formula ((X - Mean) / Std Deviation) we calculate the Z-score which gives us the percentage.

For 58 inches, we get a Z-score of -2.25, and using a Z-score table, this corresponds approximately to 1.2%. For 80 inches, the Z-score is 6.92, which is practically at the extreme end of the curve, so we can take this as 100%. Hence, almost 98.8% of women fall within the height requirements of 58 inches to 80 inches.

Coming to part (b) we need to find the height that represents the shortest and tallest 1% and 2% respectively. Using the Z-score table, the Z-score for 1% is -2.33 and for 2%, it's 2.05. Hence, the height cut-off for the shortest 1% will be Mean - 2.33*Std Deviation = 58.11 inches approx, and the tallest 2% will be Mean + 2.05*Std Deviation = 68.38 inches approx. This means, if the military changes their requirements to exclude the shortest 1% and tallest 2%, their new height requirements would approximately be from 58.11 inches to 68.38 inches.

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Given an acceleration​ vector, initial velocity left angle u 0 comma v 0 right angle​, and initial position left angle x 0 comma y 0 right angle​, find the velocity and position vectors for tgreater than or equals0. a​(t)equalsleft angle 0 comma 12 right angle​, left angle u 0 comma v 0 right angleequalsleft angle 0 comma 6 right angle​, left angle x 0 comma y 0 right angleequalsleft angle 6 comma negative 1 right angle

Answers

Answer:

For the velocity vector, we have

V(t) = dR

dt = (1, 2t, 3t

2

).

For the acceleration vector, we get

A =

dV

dt = (0, 2, 6t).

The velocity vector at t = 1 is

V(1) = (1, 2, 3).

The speed at t = 1 is

kV(1)k =

p

1

2 + 22 + 32 =

14.

Data collected at Toronto Pearson International Airport suggest that an exponential distribution with mean value 2,725 hours is a good model for rainfall duration. What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

Answers

Answer:

[tex] P(X >2) [/tex]

And we can calculate this with the complement rule like this:

[tex] P(X>2) = 1-P(X<2)[/tex]

And using the cdf we got:

[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

[tex]P(X=x)=\lambda e^{-\lambda x}, x>0[/tex]

And 0 for other case. Let X the random variable of interest:

[tex]X \sim Exp(\lambda=\frac{1}{2.725})[/tex]

Solution to the problem

We want to calculate this probability:

[tex] P(X >2) [/tex]

And we can calculate this with the complement rule like this:

[tex] P(X>2) = 1-P(X<2)[/tex]

And using the cdf we got:

[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]

The lengths of the sides of a rectangle are consecutive prime numbers. The area of the rectangle is represented by a three-digit number composed only of the two smallest prime digits and it will not change if we reverse it . What is the perimeter of the rectangle?
WILL MARK BRAINLIEST!

Answers

Answer:

Perimeter = 72

Step-by-step explanation:

Mathematics Puzzle

The area of a rectangle is the product of the base and the height

A=b.h

We know the base and the height are two consecutive prime numbers, not much of useful information so far.

We also know the area is composed only of the two smallest prime digits. Those digits are 2 and 3. If the number is reversed and it's not changed, then we only have two possible values for the area: 232 and 323.

We only need to find two consecutive prime numbers which product is one of the above. The number 232 has no prime factors: 232 = 8*29.

The number 323 is the product of 17 and 19, two consecutive prime numbers, thus the dimensions of the rectangle are 17 and 19.

The perimeter of that rectangle is 2*17+2*19= 72

Answer:

72 units

Step-by-step explanation:

72 units

A sorority has 38 members, 28 of whom are full members and 10 are pledges. Two persons are selected at random from the membership list of the sorority. Find the requested probabilities. (Enter the probabilities as fractions.)

Answers

Final answer:

The student's question pertains to calculating the probability of selecting two individuals from a sorority consisting of full members and pledges. The calculation involves using combinatorial formulas to determine the likelihood of different types of selections.

Explanation:

The student is asking about finding the probability of selecting two persons at random from a group consisting of full members and pledges in a sorority. With 38 members in total, 28 full members and 10 pledges, we need to calculate the probability of different pairs of members being selected. This is a combinatorial probability question.

In such problems, if there is no specific requirement about who needs to be picked (like how many full members or pledges), the total number of ways to select two members from the group without regard to order is calculated using the combination formula C(n, k) = n! / [k!(n-k)!], where 'n' is the total number of items, and 'k' is the number of items to choose.

For instance, the probability of selecting two full members can be calculated by finding the number of ways to choose two full members from the 28 available, divided by the number of ways to choose any two members from the entire group of 38.

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a population standard deviation of 15. What is the probability we could select a sample of 50 adults and find that the mean of this sample is between 98 and 103? Show your solution.

A.) 0.3264

B.) 0.9428

C.) 0.4702

D.) 0.7471

E.) 0.6531

Answers

Answer:

D.) 0.7471

Step-by-step explanation:

Mean=μ=100

Standard deviation=σ=15

We know that IQ score are normally distributed with mean 100 and standard deviation 15.

n=50

According to central limit theorem, if the population is normally distributed with mean μ and standard deviation σ then the distribution of sample taken from this population will be normally distributed with mean μxbar and standard deviation σxbar=σ/√n.

Mean of sampling distribution=μxbar=μ=100.

Standard deviation of sampling distribution=σxbar=σ/√n=15/√50=2.1213.

We are interested in finding the probability of sample mean between 98 and 103.

P(98<xbar<103)=?

Z-score associated with 98

Z-score=(xbar-μxbar)/σxbar

Z-score=(98-100)/2.1213

Z-score=-2/2.1213

Z-score=-0.94

Z-score associated with 103

Z-score=(xbar-μxbar)/σxbar

Z-score=(103-100)/2.1213

Z-score=3/2.1213

Z-score=1.41

P(98<xbar<103)=P(-0.94<Z<1.41)

P(98<xbar<103)=P(-0.94<Z<0)+P(0<Z<1.41)

P(98<xbar<103)=0.3264+0.4207

P(98<xbar<103)=0.7471

Thus, the probability that the sample mean is between 98 and 103 is 0.7471.

A group of researchers conducted a study to determine whether the final grade in an honors section of introductory psychology was related to a student’s performance on a test of math ability administered for college entrance. The researchers looked at the test scores of 200 students (n= 200) and found a correlation of r= .45 between math ability scores and final course grade. The proportion of the variability seen in final grade performance that can be predicted by math ability scores is ____.

Answers

Answer:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

And for this case [tex] r =0.45[/tex]

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]0.45^2 =0.2025[/tex], so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.

Step-by-step explanation:

For this case we asume that we fit a linear model:

[tex] y = mx+b[/tex]

Where y represent the final grade and x the math ability scores

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]  

Where:  

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]  

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]  

[tex]\bar x= \frac{\sum x_i}{n}[/tex]  

[tex]\bar y= \frac{\sum y_i}{n}[/tex]  

And we can find the intercept using this:  

[tex]b=\bar y -m \bar x[/tex]  

The correlation coeffcient is given by:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

And for this case [tex] r =0.45[/tex]

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]0.45^2 =0.2025[/tex], so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.

On a test with a population mean of 75 and standard deviation equal to 16, if the scores are normally distributed, what percentage of scores fall between 70 and 80?

Answers

Answer:

Percentage of scores that fall between 70 and 80 = 24.34%

Step-by-step explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N([tex]\mu,\sigma^{2}[/tex])

The z probability is given by;

           Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)    where, [tex]\mu[/tex] = 75  and  [tex]\sigma[/tex] = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{80-75}{16}[/tex] ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{70-75}{16}[/tex] ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

                                              = 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%

Derive the validity of universal form of part(a) of the elimination rule from the validity of universal instantiation and the valid argument called elimination in Section 2.3.

Answers

Answer:

Step-by-step explanation:

Derive the validity of universal form of part(a) of the elimination rule from the validity of universal instantiation and the valid argument called elimination in Section 2.3.

P(x)∨Q(x)

~Q(x)

∵ P(x)

Universal Instantiation has the following argument form

∀ x ∈ D, P (x)

P(a) for a particular a∈D

Universal Elimination Rule:

∀x, P(x)

∵~ P(a)

Here is a particular value.

P(a) For a particular  a∈D

Since the universal elimination is same as universal instantiation.

Therefore, Universal elimination is valid when universal instantiation and elimination rule are  valid

19) If an average of 12 customers are served per hour, what is the probability that the next customer will arrive in 3 minutes or less? Note: λ = 12/60

Answers

Answer:

The probability that the next customer will arrive in 3 minutes or less is 0.45.

Step-by-step explanation:

Let N (t) be a Poisson process with arrival rate λ. If X is the time of the next arrival then,

[tex]P(X>t)=e^{-\lambda t}[/tex]

Given:

[tex]\lambda=\frac{12}{60}[/tex]

t = 3 minutes

Compute the probability that the next customer will arrive in 3 minutes or less as follows:

P (X ≤ 3) = 1 - P (X > 3)

              [tex]=1-e^{-\frac{12}{60}\times3}\\=1-e^{-0.6}\\=1-0.55\\=0.45[/tex]

Thus, the probability that the next customer will arrive in 3 minutes or less is 0.45.

The probability that the next customer will arrive in 3 minutes or less is; 0.4512

This is a Poisson distribution problem with the formula;

P(X > t) = e^(-λt)

Where;

λ is arrival rate

t is arrival time

We are given;

λ = 12/60

t = 3 minutes

We want to find the probability that the next customer will arrive in 3 minutes or less. This is expressed as;

P (X ≤ 3) = 1 - (P(X > 3))

Thus;

P (X ≤ 3) = 1 - e^((12/60) × 3)

P (X ≤ 3) = 1 - 0.5488

P (X ≤ 3) = 0.4512

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What is the age distribution of adult shoplifters (21 years of age or older) in supermarkets? The following is based on information taken from the National Retail Federation. A random sample of 895 incidents of shoplifting gave the following age distribution. Estimate the mean age, sample variance, and sample standard deviation for the shoplifters. For the class 41 and over, use 45.5 as the class midpoint. (Enter your answers to one decimal place.)

Age range (years) 21-30 31-40 41 and over
Number of shoplifters 280 368 247

Answers

Final answer:

We can estimate the mean, variance, and standard deviation by using class midpoints and the number of observations. First, find the midpoint of each age group. Then, to find the mean, multiply each midpoint by the number of shoplifters in that age group, sum those products, and divide by the total number of observations. To find the variance, subtract the mean from each midpoint, square the result, multiply by the number of shoplifters in that group, sum those products, and divide by the total number of observations minus one. The standard deviation is the square root of the variance.

Explanation:

To estimate the mean age, sample variance, and sample standard deviation for the shoplifters from a random sample of 895 incidents we need to follow a few steps. Here's how to do it:

Calculate the midpoint of each class: For the age ranges 21-30 and 31-40, the midpoints are 25.5 and 35.5 respectively. The problem already provides 45.5 as the class midpoint for the range '41 and over'.Calculate the estimated mean (µ) by multiplying the midpoint of each class by the number of observations in that class, summing these values, and dividing by the total number of observations.Calculate the estimated variance (σ²) by subtracting the estimated mean from each class midpoint, squaring the result, multiplying by the number of observations in that class, summing these values, and dividing by the total number of observations - 1.Finally, calculate the standard deviation (σ) by taking the square root of the estimated variance.

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According to a recent​ study, 23% of U.S. mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure. A random sample of twelve mortgages was selected. What is the probability that greater than 5 of these mortgages are delinquent?

Answers

Answer:

P ( X > 5) = 0.0374

Step-by-step explanation:

Given:

n = 12

p = 0.23

Using Binomial distribution formula,

X ~ Binomial ( n = 12, p = 0.23)

[tex]=\frac{n!}{(n-x)! x!}. p^{x} q^{n-x}[/tex]

Substitute for n = 12, p = 0.23, q = 1-0.23 for  x = 6,7,8,9,10,11 and 12

P (X > 5)  = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)

P ( X > 5 ) = 0.0285 + 0.007299 + 0.00136 + 0.000181 + 0.0000162 + 1E-6 + 1E-6

P ( X > 5) = 0.0374

These data can be approximated quite well by a N(3.4, 3.1) model. Economists become alarmed when productivity decreases. According to the normal model what is the probability that the percent change in worker output per hour from the previous quarter is more than 0.6 standard deviations below the mean? .0228 Incorrect: Your answer is incorrect. Question 3. What is the probability that the percent change in worker output from the previous quarter is between -1.715 and 7.12? Use the normal model mentioned at the beginning of question 2.

Answers

Answer:

First part

[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]

And for this case we can use the z score formula given by:

[tex] z = \frac{x- \mu}{\sigma}[/tex]

And using this formula we got:

[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]

And we can use the normal standard table or excel and we got:

[tex]P(Z<-0.6) = 0.274[/tex]

Second part

For the other part of the question we want to find the following probability:

[tex] P(-1.715 <X< 7.12)[/tex]

And using the score we got:

[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]

And we can find this probability with this difference:

[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the data of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(3.4,3.1)[/tex]  

Where [tex]\mu=3.4[/tex] and [tex]\sigma=3.1[/tex]

First part

And for this case we want this probability:

[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]

And for this case we can use the z score formula given by:

[tex] z = \frac{x- \mu}{\sigma}[/tex]

And using this formula we got:

[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]

And we can use the normal standard table or excel and we got:

[tex]P(Z<-0.6) = 0.274[/tex]

Second part

For the other part of the question we want to find the following probability:

[tex] P(-1.715 <X< 7.12)[/tex]

And using the score we got:

[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]

And we can find this probability with this difference:

[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/tex]

Which expressions is equivalent to 7/10-2/10?

Answers

Answer:

5/10 & 1/2: anything that simplifies to 1/2 or .5

Final answer:

The expression 7/10-2/10 simplifies to 5/10, which can be further reduced to 1/2 by dividing both the numerator and denominator by their greatest common divisor, 5.

Explanation:

The expression 7/10-2/10 involves the subtraction of two fractions with the same denominator. When subtracting fractions with the same denominator, you simply subtract the numerators and keep the denominator the same. Thus, the calculation would be:

Numerator: 7 - 2 = 5

Denominator: 10 (remains the same)

Therefore, the expression 7/10-2/10 is equivalent to 5/10. This fraction can be further simplified by dividing the numerator and the denominator by their greatest common divisor, which in this case is 5. This gives us:

(5 ÷ 5) / (10 ÷ 5) = 1/2

So, 7/10-2/10 simplifies to 1/2, which is the equivalent fraction.

Suppose X is a normal distribution with N(210, 32). Find the following: a. P( X < 230) b. P(180 < X < 245) c. P( X >190) d. Find c such that P( X < c) = 0.0344 e. Find c such that P( X > c) = 0.7486

Answers

Using the normal distribution, it is found that:

a) P(X < 230) = 0.734.

b) P(180 < X < 245) = 0.6885.

c) P( X >190) = 0.734.

d) X = 151.76.

e) X = 188.56.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of [tex]\mu = 210[/tex].The standard deviation is of [tex]\sigma = 32[/tex].

Item a:

This probability is the p-value of Z when X = 230, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{230 - 210}{32}[/tex]

[tex]Z = 0.625[/tex]

[tex]Z = 0.625[/tex] has a p-value of 0.734.

Hence:

P(X < 230) = 0.734.

Item b:

This probability is the p-value of Z when X = 245 subtracted by the p-value of Z when X = 180, hence:

X = 245

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{245 - 210}{32}[/tex]

[tex]Z = 1.09[/tex]

[tex]Z = 1.09[/tex] has a p-value of 0.8621.

X = 180

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{180 - 210}{32}[/tex]

[tex]Z = -0.94[/tex]

[tex]Z = -0.94[/tex] has a p-value of 0.1736.

0.8621 - 0.1736 = 0.6885.

Then:

P(180 < X < 245) = 0.6885.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 190, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{190 - 210}{32}[/tex]

[tex]Z = -0.625[/tex]

[tex]Z = -0.625[/tex] has a p-value of 0.266.

1 - 0.266 = 0.734.

Hence:

P( X >190) = 0.734.

Item d:

This is X = c when Z has a p-value of 0.0344, hence X when Z = -1.82.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.82 = \frac{X - 210}{32}[/tex]

[tex]X - 210 = -1.82(32)[/tex]

[tex]X = 151.76[/tex]

Item e:

This is X when Z has a p-value of 1 - 0.7486 = 0.2514, hence X when Z = -0.67.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.67 = \frac{X - 210}{32}[/tex]

[tex]X - 210 = -0.67(32)[/tex]

[tex]X = 188.56[/tex]

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Final answer:

The question involves calculating probabilities and specific values (c) for a given normal distribution with mean 210 and standard deviation 32. The probabilities for certain ranges and tail ends are computed using the normal cumulative distribution function and its inverse.

Explanation:

The question pertains to finding probabilities and specific values associated with a normal distribution X which is denoted as N(210, 32), meaning it has a mean (μ) of 210 and a standard deviation (σ) of 32.

P(X < 230) can be found using a z-score calculation or a normal cumulative distribution function. Since this is a left-tail probability, we're interested in the area under the curve to the left of X=230.P(180 < X < 245) is the probability that X falls between these two values. We look at the area under the normal curve between these two points.P(X > 190) represents the right-tail probability, meaning the area under the curve to the right of X=190.To find c such that P(X < c) = 0.0344, we'd use the inverse of the normal cumulative distribution function, often denoted as the quantile or the percentile function.Finding the value c such that P(X > c) = 0.7486 again involves the use of the inverse normal function, but this time looking at the right-tail probability.

what is f(x)=8x^2+4x written in vertex form

Answers

Answer:

f(x) = 8 (x + ¼)² − ½

Step-by-step explanation:

f(x) = 8x² + 4x

Divide both sides by 8.

1/8 f(x) = x² + 1/2 x

Take half of the second coefficient, square it, then add to both sides.

(½ / 2)² = (1/4)² = 1/16

1/8 f(x) + 1/16 = x² + 1/2 x + 1/6

Factor the perfect square.

1/8 f(x) + 1/16 = (x + 1/4)²

Multiply both sides by 8.

f(x) + 1/2 = 8 (x + 1/4)²

Subtract 1/2 from both sides.

f(x) = 8 (x + 1/4)² − 1/2

Suppose that the waiting time for an elevator at a local shopping mall is uniformly distributed from 0 to 90 seconds.
What is the probability that a customer waits for more than 60 seconds?

Answers

Answer:

1/3

Step-by-step explanation:

60-90 is 30 numbers, right? So it is 30/90, or 1/3

slove this problem find the value of x.​

Answers

Answer:

x = 6 units.

Step-by-step explanation:

By Geometric mean property:

[tex]x = \sqrt{3 \times 12} = \sqrt{36} = 6 \\ \hspace{20 pt} \huge \orange{ \boxed{ \therefore \: x = 6}}[/tex]

Hence, x = 6 units.

Mrs. Porcelli's classroom bulletin board is 2 % feet long. Ms. Smith's bulletin board is 3
| times as long as Mrs. Porcelli's. How long is Ms. Smith's bulletin board.

Answers

Answer: 0.06 ft long

Step-by-step explanation:

Porcelli's board = 2% feet long; to convert 2% into fraction, divide by 100= 2/100 = 0.02 ft

Smith's board, from the question is 3 times porcelli's board = 3 x 0.02= 0.06 ft

I hope this helps.

14 logs are shipped to a saw mill. The decision must be made on whether they will be classified as Prime, Good, Acceptable, or Not Acceptable. The usual counts are 3 Prime, 5 Good, and 4 Acceptable. In how many ways can these logs be classified so as to match the usual counts

Answers

Answer:

2522520

Step-by-step explanation:

Number of ways logs can be classified = 14C5* 9C4*5C3

= 2002*126*10

= 2522520

Number of ways to select 5 good = 14C5, out of remaining 9, number of ways to select acceptable log = 9C4, out of remaining 5, number of ways to select prime log = 5C3 and remaking two unacceptable in 2C2 ways

Final answer:

To find how many ways 14 logs can be classified into categories to match the usual counts, use the multinomial coefficient formula based on the given category counts, resulting in a single calculations. to be 54.6.

Explanation:

The question asks in how many ways 14 logs can be classified into four categories (Prime, Good, Acceptable, Not Acceptable) if we know the usual counts for three of these categories are 3 Prime, 5 Good, 4 Acceptable, and the rest are Not Acceptable. This is a combinatorial problem that can be solved using combinations.

First, we distribute the logs into the Prime, Good, and Acceptable categories as given. This leaves us with 14 - (3 + 5 + 4) = 2 logs to be classified as Not Acceptable. Hence, all of the logs are accounted for with the specified counts.

The number of ways to classify the logs can thus be calculated as the number of ways to choose 3 out of 14 for Prime, then 5 out of the remaining 11 for Good, then 4 out of the remaining 6 for Acceptable. The remaining 2 are automatically classified as Not Acceptable. However, since we're simply fulfilling given counts, and the categories are distinct without overlap, we actually approach this as a partition of 14 objects into parts of fixed sizes, which is a straightforward calculation given by the multinomial coefficient:

The formula for the calculation is: 14! / (3! × 5! × 4! × 2!)

Which simplifies to the total number of ways these logs can be classified according to the specified counts

= 54.6

Delta Airlines quotes a flight time of 2 hours for its flights from Cincinnati to Tampa, meaning that an on-time flight would arrive in 2 hours. Suppose we believe that actual flight times are uniformly distributed between 1 hour 50min minutes and 135 minutes.a. Show the graph of the probability density function for flight time.b. What is the probability that the flight will be no more than 5 minutes late?c. What is the probability that the flight will be more than 10 minutes late?d. What is the expected flight time?

Answers

Answer:

a) Attached

b) P=0.60

c) P=0.80

d) The expected flight time is E(t)=122.5

Step-by-step explanation:

The distribution is uniform between 1 hour and 50 minutes (110 min) and 135 min.

The height of the probability function will be:

[tex]h=\frac{1}{Max-Min}=\frac{1}{135-110} =\frac{1}{25}[/tex]

Then the probability distribution can be defined as:

[tex]f(t)=\frac{1}{25}=0.04 \,\,\,\,\\\\t\in[110,135][/tex]

b) No more than 5 minutes late means the flight time is 125 or less.

The probability of having a flight time of 125 or less is P=0.60:

[tex]F(T<t)=0.04(t-min)\\\\F(T<125)=0.04*(125-110)=0.04*15=0.60[/tex]

c) More than 10 minutes late means 130 minutes or more

The probability of having a flight time of 130 or more is P=0.80:

[tex]F(T>t)=1-0.04(t-110)\\\\F(T>130)=1-0.04*(130-110)=1-0.04*20=1-0.8=0.2[/tex]

d) The expected flight time is E(t)=122.5

[tex]E(t)=\frac{1}{2}(max+min)= \frac{1}{2}(135+110)=\frac{1}{2}*245=122.5[/tex]

According to the United States Health and Human Services, the mean height for Americans is 1.757 meters for men and 1.618 meters for women. The standard deviation is 0.074 meters for men's height and 0.069 meters for women's height. Michelle’s height is 1.758 meters. What is her z-score?

Answers

Answer:

Her z-score is 2.03.

Step-by-step explanation:

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Michelle’s height is 1.758 meters. What is her z-score?

Michelle's is a woman.

The average height of women is [tex]\mu = 1.618[/tex] and the standard deviation is [tex]\sigma = 0.069[/tex]

This is Z when X = 1.758. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1.758 - 1.618}{0.069}[/tex]

[tex]Z = 2.03[/tex]

Her z-score is 2.03.

Final answer:

Michelle's z-score is approximately 2.03, which means her height is 2.03 standard deviations above the mean height for American women.

Explanation:

To calculate Michelle’s z-score for her height, we use the formula for the z-score:

Z = (X - μ) / σ

Where Z is the z-score, X is the value (Michelle's height), μ is the mean, and σ is the standard deviation. Since Michelle is a woman, we will use the mean and standard deviation for women. The mean height for American women, μ, is 1.618 meters, and the standard deviation, σ, is 0.069 meters.

Plugging in the values we get:

Z = (1.758 - 1.618) / 0.069

Z = 0.14 / 0.069

Z ≈ 2.029

Michelle's height is approximately 2.03 standard deviations above the mean height for American women.

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