Answer:
a) Mean = 1.6, standard deviation = 1.21
b) 18.87% probability that zero online retail orders will turn out to be fraudulent.
c) 32.82% probability that one online retail order will turn out to be fraudulent.
d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The mean of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]p = 0.08, n = 20[/tex]
a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?
Mean
[tex]E(X) = np = 20*0.08 = 1.6[/tex]
Standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21[/tex]
b. What is the probability that zero online retail orders will turn out to be fraudulent?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887[/tex]
18.87% probability that zero online retail orders will turn out to be fraudulent.
c. What is the probability that one online retail order will turn out to be fraudulent?
This is P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282[/tex]
32.82% probability that one online retail order will turn out to be fraudulent.
d. What is the probability that two or more online retail orders will turn out to be fraudulent?
Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 1) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X \leq 1)[/tex]
In which
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
From itens b and c
[tex]P(X \leq 1) = 0.1887 + 0.3282 = 0.5169[/tex]
[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831[/tex]
48.31% probability that two or more online retail orders will turn out to be fraudulent.
The probability is an illustration of a binomial distribution.
The mean and the standard deviationThe given parameters are:
n = 20
p = 0.08
The mean is calculated as:
[tex]\bar x = np[/tex]
So, we have:
[tex]\bar x = 20 * 0.08[/tex]
[tex]\bar x = 1.6[/tex]
The standard deviation is calculated as:
[tex]\sigma = \sqrt{\bar x * (1 - p)[/tex]
This gives
[tex]\sigma = \sqrt{1.6 * (1 - 0.08)[/tex]
[tex]\sigma = 1.21[/tex]
Hence, the mean is 1.6 and the standard deviation is 1.21
The probability that zero online retail orders will turn out to be fraudulentThis is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(0) = ^{20}C_0 * 0.08^0 * (1 - 0.08)^{20 - 0}[/tex]
[tex]P(0) =0.1887[/tex]
The probability that zero online retail orders will turn out to be fraudulent is 0.1887
The probability that one online retail order will turn out to be fraudulentThis is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(1) = ^{20}C_1 * 0.08^1 * (1 - 0.08)^{20 - 1}[/tex]
[tex]P(1) =0.3281[/tex]
The probability that one online retail orders will turn out to be fraudulent is 0.3281
The probability that two or more online retail orders will turn out to be fraudulentThis is calculated as:
[tex]P(x\ge 2) = 1 - P(0) - P(1)[/tex]
So, we have:
[tex]P(x\ge 2) = 1 - 0.1887 - 0.3281[/tex]
[tex]P(x\ge 2) = 0.4832[/tex]
The probability that two or more online retail orders will turn out to be fraudulent is 0.4832
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A researcher claims that the proportion of people who are right-handed is greater than 70%. To test this claim, a random sample of 600 people is taken and its determined that 424 people are right-handed.
The following is the setup for this hypothesis test:
{H0:p=0.70
Ha:p>0.70
Find the test statistic for this hypothesis test for a proportion.
The test statistic for the given hypothesis test for a proportion is calculated to be 0.356.
Given that:
The claim is that the proportion of people who are right-handed is greater than 70%.
So, p > 0.7 is the alternative hypothesis.
From the setup of the hypothesis test, the test is a one-tailed test since the alternative hypothesis has a sign ">".
Sample size, n = 600
Number of people right-handed = 424
So, the observed sample proportion is:
[tex]\hat{\text{p}}=\frac{424}{600}[/tex]
[tex]=0.7067[/tex]
The standard deviation is:
[tex]\sigma=\sqrt{\frac{p(1-p)}{n} }[/tex]
[tex]=\sqrt{ \frac{0.7(1-0.7)}{600} }[/tex]
[tex]=0.01871[/tex]
So, the test statistic is:
[tex]\text{z}=\frac{\hat{\text{p}}-\text{p}}{\sigma}[/tex]
[tex]=\frac{0.7067-0.7}{0.01871}[/tex]
[tex]=0.356[/tex]
Hence, the test statistic is 0.356.
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Find the zeros of each function
f(x)=(x-3)(x+5)
Answer:
The zeros of this function are 3 and -5.
Step-by-step explanation:
One note: The question says "Find the zeros of EACH function", but there seems to be only one function in the question. I hope I answered it well.
To complete the function provided, you need to get both terms inside both parenthesis to equal zero. So, if its (x-3)(x+5), the X's need to equal the opposite of the number that is being added or subtracted.
For the first parenthesis, (x-3), the number is -3, so the zero of that side is positive 3. (3-3 = 0)
For the second parenthesis, (x+5), the number is positive 5, so the zero of that side is -5. (-5+5 = 0)
Each of the ODEs below is second order in y, with y1 as a solution. Reduce the ODE from being second order in y to being first order in ????, with ???? being the only response variable appearing in the ODE. Combine like terms. Show your work.
Answer:
Step-by-step explanation:
The detailed steps and appropriate workings is as shown in the attached file.
The question asks about reducing a second-order ODE to first-order given a known solution, which is achieved using the method of reduction of order to find a new function v(y) leading to a first-order ODE.
To reduce a second-order ordinary differential equation (ODE) to a first-order ODE, given that y1 is a solution, you can apply the method of reduction of order. This involves introducing a new function v(y) such that y can be expressed as a product of the known solution y1 and this new function v(y). Essentially, you substitute y = y1*v into the original second-order ODE and differentiate as necessary to obtain an equation in terms of v and its derivatives only, effectively transforming the equation into first-order.
The steps include (i) expressing y in terms of y1 and v, (ii) differentiating this expression to find the derivatives of y, and (iii) substituting all of this into the equation that y1 obeys. The resultant first-order equation will only involve the function v and its first derivative, v'. By solving this simplified equation, you can find v and thus the second solution to the original second-order ODE.
Use mathematical induction to prove that if L is a linear transformation from V to W, then L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) g
Answer:
The proof is shown in the explanation below.
Step-by-step explanation:
Analysis:
The proof by induction focuses on n. In this case, let n = 1, and [tex]L^{1}[/tex] will be a linear operator since [tex]L^{1} = L[/tex]
The exercise will show that [tex]L^{n}[/tex] is a linear operator on V and that [tex]L^{n+1}[/tex] is also a linear operator on V. This, follows that:
[tex]L^{n+1} (av) = L(L^{m}(v_{1}+v_{2})\\ = L(L^{m} (v_{1} + L^{m}v_{2})\\ = L(L^{m}v_{1} + L(L^{m}v_{2})\\ = L^{m+1}(v_{1}) + L^{m+1}(v_{2})[/tex]
Answer/Step-by-step explanation:
For the mathematical induction,
We show that the equation
L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) is true for
L = 1,
Assume it is true for L = n and show that it is true for L = n + 1.
If L = 1, the equation become
(α1v1 + α2v2 +· · ·+αnvn)= α1(v1) + α2 (v2)+· · ·+αn (vn). Therefore, the Right Hand side(RHS) = Left Hand side(LHS)
When L = n, we assume the following is true
(α1nv1 + α2nv2 +· · ·+αnvn)= α1n(v1) + α2n (v2)+· · ·+αn (vn)
Then, when L = n + 1,
n +1 (α1v1 + α2v2 +· · ·+αvn)= α1(n +1) (v1) + α2(n +1) (v2)+· · ·+αn(n + 1)(vn).
Open the bracket,
n(α1v1 + α2v2 +· · ·+αvn) + α1v1 + α2v2 +· · ·+αnvn = α1n (v1) + α2v2 +· · ·+αvn ) + α1(v1) + α2v2+· · ·+αn(vn)
Since we assume the the equation is true for L = n, and eliminating some terms, then
L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn)
a rectangular box with two sqaure opposite ends is to hold 8000 cubic inches. find the dimensions of the cheapest box if tge recangular sides cost 15 times more as much per square inche as the top, bottom, and square ends
Answer:
the dimensions of the box that minimizes the cost are 5 in x 40 in x 40 in
Step-by-step explanation:
since the box has a volume V
V= x*y*z = b=8000 in³
since y=z (square face)
V= x*y² = b=8000 in³
and the cost function is
cost = cost of the square faces * area of square faces + cost of top and bottom * top and bottom areas + cost of the rectangular sides * area of the rectangular sides
C = a* 2*y² + a* 2*x*y + 15*a* 2*x*y = 2*a* y² + 32*a*x*y
to find the optimum we can use Lagrange multipliers , then we have 3 simultaneous equations:
x*y*z = b
Cx - λ*Vx = 0 → 32*a*y - λ*y² = 0 → y*( 32*a-λ*y) = 0 → y=32*a/λ
Cy - λ*Vy = 0 → (4*a*y + 32*a*x) - λ*2*x*y = 0
4*a*32/λ + 32*a*x - λ*2*x*32*a/λ = 0
128*a² /λ + 32*a*x - 64*a*x = 0
32*a*x = 128*a² /λ
x = 4*a/λ
x*y² = b
4*a/λ * (32*a/λ)² = b
(a/λ)³ *4096 = 8000 m³
(a/λ) = ∛ ( 8000 m³/4096 ) = 5/4 in
then
x = 4*a/λ = 4*5/4 in = 5 in
y=32*a/λ = 32*5/4 in = 40 in
then the box has dimensions 5 in x 40 in x 40 in
A soda filling machine is supposed to fill cans of soda with 12 fluid ounces. Suppose that the fills are actually normally distributed with a mean of 12.1 oz and a standard deviation of 0.2 oz. a) What is the probability of one can less than 12 oz
Answer:
30.85% probability of one can less than 12 oz
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
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]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 12.1, \sigma = 0.2[/tex]
a) What is the probability of one can less than 12 oz
This is the pvalue of Z when X = 12. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12 - 12.1}{0.2}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085
30.85% probability of one can less than 12 oz
We have a biased coin (probability of heads is equal to 1/4). Consider the following 2 step process: In the first step we flip the coin until we get a heads. Let X denote the trial on which the first heads occurs. In the second step we flip the coin X more times. Let Y be the number of heads in the second step. (a) For each non-negative integer, k, what is the probability that X = k? (b) Conditioned on the event {X = k}. What is the probability Y = 0? (c) Use the Law of Total Probability to compute the unconditional probability that Y = 0.
Answer:
See the attached pictures for answer.
Step-by-step explanation:
See the attached picture for detailed explanation.
g On any day, the probability of rain is 0.3. The occurrence of rain on any day is independent of the occurrence of rain on any other day. Calculate the probability that, starting with tomorrow, the second day ofrain will occur within 5 days
Answer:
0.249579
Step-by-step explanation:
P(rain) = 0.3
P(no rain) = 1 - 0.3 = 0.7
The event of rain falling a second time within the next 5 days is possible in these ways
1. Rain on days 1 and 2
2. Rain on days 1 and 3; none on day 2
3. Rain on days 1 and 4; none on days 2 and 3
4. Rain on days 1 and 5; none on days 2, 3 and 4
5. Rain on days 1 and 6; none on day 2, 3, 4 and 5
[tex]P(\text{second rain within 5 days}) = 0.3^2+0.3^2\times0.7+0.3^2\times0.7^2+0.3^2\times0.7^3+0.3^2\times0.7^4 = 0.3^2(1+0.7+0.7^2+0.7^3+0.7^4)= 0.09\times2.7731=0.249579[/tex]
Solve 0 = (x – 4)2 – 1 by graphing the related function.
What are the solutions to the equation?
3 and 5 is the answer
AND THAT'S JUST ON PERIOD POOH!
Answer:
Therefore, the solutions of the quadratic equations are:
[tex]x=5,\:x=3[/tex]
The graph is also attached.
Step-by-step explanation:
The solution of the graph could be obtained by finding the x-intercept.
[tex]y=\left(x-\:4\right)^2-1[/tex]
Finding the x-intercept by substituting the value y = 0
so
[tex]y=\left(x-\:4\right)^2-1[/tex]
[tex]\:0\:=\:\left(x\:-\:4\right)^2\:-\:1[/tex] ∵ y = 0
[tex]\left(x-4\right)^2-1=0[/tex]
[tex]\left(x-4\right)^2-1+1=0+1[/tex]
[tex]\left(x-4\right)^2=1[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]\mathrm{Solve\:}\:x-4=\sqrt{1}[/tex]
[tex]\mathrm{Apply\:rule}\:\sqrt{1}=1[/tex]
[tex]x-4=1[/tex]
[tex]x=5[/tex]
[tex]\mathrm{Solve\:}\:x-4=-\sqrt{1}[/tex]
[tex]\mathrm{Apply\:rule}\:\sqrt{1}=1[/tex]
[tex]x-4=-1[/tex]
[tex]x=3[/tex]
So, when y = 0, then x values are 3, and 5.
Therefore, the solutions of the quadratic equations are:
[tex]x=5,\:x=3[/tex]
The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.
Answer:
3 and 5 is the answer
Step-by-step explanation:
A supervisor must split 60 hours of overtime between five people. One employee must be assigned twice the number of hours as each of the other four employees. How many hours of overtime will be assigned to each employee?
Solution:
Given that,
A supervisor must split 60 hours of overtime between five people
One employee must be assigned twice the number of hours as each of the other four employees
Let "x" be the number of hours overtime per person.
One person does 2x hours of overtime
Which means,
(number of hours overtime per person)(6 person) = 60 hours
6x = 60
x = 10
Thus, 4 people do 10 hours and one person does 20 hours
(A) In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have:
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient.
(B) Now that you have one of the two variables in Part (A) isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers..
Answer:
C = 17/7 or 2 3/7
D = -4/7
Step-by-step explanation:
3C+4D=5
2C+5D=2
2C = 2 - 5D
C = 1 - 2.5D
3(1 - 2.5D) + 4D = 5
3 - 7.5D + 4D = 5
-3.5D = 2
D = -4/7
C = 1 - 2.5(-4/7)
C = 1 + 10/7 = 17/7
A small social network contains seven people who are network friends with six other people in the network, one person who is network friends with five other people in the network, and five people who are network friends with four other people in the network. The rest are network friends with three other people in the network. The network contains 50 pairs of network friends.
(a) How many people are network friends with three other people in the network?
(b) How many people are in the network?
Answer:
(a) 11 people
(b) 24 people
Step-by-step explanation:
Part (a)
Let X be the people who are network friends with three other people in the network.
The total pairings = 50
Since [tex]v[/tex]∈[tex]V[/tex],
[tex](7)(6) + (1)(5) + (5)(4) + (X)(3) = (2)(50)[/tex]
Solve for X to find the people who are network friends with three other people,
[tex]67 + 3X = 100[/tex]
[tex]X = 11[/tex]
Answer: There are 11 people who are network friends with three other people in the network.
Part (b)
Total people in the network,
[tex]= 7 + 1 + 5 + X\\ = 13 + 11\\ = 24[/tex]
Answer: There are a total of 24 people in the network.
There are 6 people who are network friends with three other people in the network. The total number of people in the network is 19.
Explanation:This problem involves understanding the concept of pairs in network friends. Each friendship forms a pair. So, if one person has 6 friends, this would contribute 6 pairs. Here, there are 7 people with 6 friends, 1 person with 5 friends, and 5 people with 4 friends. Therefore, these people contribute (7x6) + (1x5) + (5x4) = 42 + 5 + 20 = 67 pairs. But, we're told there are only 50 pairs in total. The reason for this discrepancy is that each pair is being counted twice. When we divide the total pairs (67) by 2, we get 33.5 but since the number of pairs must be an integer, we subtract this half pair to get 33 pairs for the listed people. That means, there must be 50 - 33 = 17 pairs remaining for the people who are network friends with three other people. Given that each of these people contributes 3 pairs, there must be 17 / 3 = 5.67 people, but since we're talking about whole people, we need to round this up to 6. Therefore, there are 6 people who are network friends with three other people in the network. Now, adding up all the people in the network, we have 7 (people with 6 friends) + 1 (person with 5 friends) + 5 (people with 4 friends) + 6 (people with 3 friends) = 19 people in total.
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Maya is planning a bridal shower for her best friend. At the party, she wants to serve 3 beverages, 3 appetizers, and 3 desserts, but she does not have time to cook. She can choose from 12 bottled drinks, 12 frozen appetizers, and 10 prepared desserts at the supermarket. How many different ways can Maya pick the food and drinks to serve at the bridal shower
Using the combinations formula C(n, r), the number of ways Maya can choose the food and drinks for the bridal shower is computed as: C(12,3) for drinks * C(12,3) for appetizers * C(10,3) for desserts.
Explanation:The subject of this question is combinations in Mathematics. Maya has choices of 12 bottled drinks, 12 frozen appetizers, and 10 prepared desserts. She wants to pick 3 of each, and the order of selection does not matter. We can use the combination formula to calculate the number of ways she can do this:
For the bottled drinks, the number of combinations can be calculated as C(12,3).For the frozen appetizers, the number of combinations can be calculated as C(12,3).For the prepared desserts, the number of combinations can be calculated as C(10,3).So the total number of different ways Maya can pick the food and drinks to serve at the bridal shower is C(12,3) * C(12,3) * C(10,3).
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A drawer contains 4 different pairs of gloves. Suppose we choose 3 gloves randomly, what is the probability that there is no matching pair?
Answer:
The probability that there is no matching pair is 4/7 = 0.5714286.
Step-by-step explanation:
For the first glove we have no restrictions. For the second glove, we have 6 gloves that works for us and 1 that doesnt work (the one that matches the first glove), hence we have 6 possibilities out of 7. Once we pick a good second glove, for the last glove, we only have 4 cases that doesnt match the other two pairs, out of 6 total. This means that the probability that there is no matching pair is 6/7*4/6 = 4/7.
Oslo Company's target quality characteristic, T, for one of its key components is set at 82. Using the Taguchi Quality Loss Function (QLF) the company has determined the cost coefficient, k, to be $6,000. What is the estimated loss, L(x), if the value of the quality characteristic, x, is 85
Answer:
The estimated loss is $5,400 if the value of the quality characteristic,x is 85.
Step-by-step explanation:
We are given the following in the question:
Target value,T = 82
Cost coefficient, k = $6,000
x = 85
The Taguchi Quality Loss Function (QLF) is given by:
[tex]L(x) = k(x -T)^2[/tex]
where L(x) is the loss, k is the cost coefficient, T is the target value or the mean value.
We have to estimate the loss if the value of the quality characteristic is 85.
We put x = 85
[tex]L(85) = 6000(85 -82)^2\\L(85) =54000[/tex]
Thus, the estimated loss is $5,400 if the value of the quality characteristic,x is 85.
Which of the following is the cheapest route to visit each city using the "Brute Force Method"
starting from A and ending at A.
Answer:
ACDBA, $900Step-by-step explanation:
Since there are 4 cities, there are (4-1)! = 6 possible routes. Half of those are the reverse of the other half, so there are 6/2 = 3 different possible routes. All of those are listed among the answer choices, along with their cost. All you need to do is choose the answer with the lowest cost:
ACDBA, $900
__
At $960, the other two routes are higher cost.
From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Calculate the mean and variance for the defective transistors. Mean = Variance = (Please round answers to 4 decimal places.)
Answer:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
We can calculate the expected value with the following formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
Given that,
A company has found that in carton of transistors: 92% contain no defective transistors,
3% contain one defective transistor, 3% contain two defective transistors,
and 2% contain three defective transistors.
We have to find,
Calculate the mean and variance for the defective transistors. Mean = Variance.
According to the question,
Let, X the random variable who represent the number of defective transistors.
The following probability distribution for X,
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
To calculate the expected value with the following formula:
[tex]E(X) = \sum^{n}_{i=1} X_i. P.(X_i)[/tex]
On substitute all the values in the formula,
[tex]E(X) = 0\times0.92 + 1\times0.03+2\times0.03 + 3\times0.02\\\\E(X) = 0.15[/tex]
To find the variance first the second moment given by:
[tex]E(X^2) = \sum^{n}_{i=1} X_i^2. P.(X_i)[/tex]
On substitute all the values in the formula;
[tex]E(X) = 0^2\times0.92 + 1^2\times0.03+2^2\times0.03 + 3^2\times0.02\\\\E(X) = 0.330[/tex]
Therefore, The variance is calculated with this formula:
[tex]Var(X) = E(X)^2 - (E(X))^2 = 0.33 - (0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance,
[tex]Standard \ deviation = \sqrt{0.0375} = 0.5545[/tex]
Hence, The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
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The number of hours needed to paint a house, h, varies inversely with the number of painters, n. Four painters need 6 hours to paint a house. How many hours would it take 3 painters to paint the house?
Answer:
It would take 3 painters, a total number of 8 hours, to paint the house
Step-by-step explanation:
Let the number of hours be denoted by N, and the number of painters be denoted by P. The inverse relationship can be expressed as:
N α [tex]\frac{1}{P}[/tex]
Removing the proportionality symbol (α), introduces a constant of proportionality, which can be assumed to be C, in this example.
So, N = [tex]\frac{C}{P}[/tex]
When N = 6, P = 4
6 = [tex]\frac{C}{4}[/tex]
C = 24
So that, N = [tex]\frac{24}{P}[/tex]. This is the formula to use for determination of the other variable when one of them is known.
Now, given 3 painters, the number of hours needed can be obtained as follows:
N = [tex]\frac{24}{3}[/tex] = 8 hours
A 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia). At a summer camp there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other. What is the probability that at least one of them suffers from arachnophobia
Answer:
Probability that at least one of them suffers from arachnophobia is 0.5160.
Step-by-step explanation:
We are given that a 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia).
Also, At a summer camp there are 10 teenagers sleeping in each tent.
Firstly, the binomial probability is given by;
[tex]P(X=r) =\binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....[/tex]
where, n = number of trials(teenagers) taken = 10
r = number of successes = at least one
p = probability of success and success in our question is % of
the teenagers suffering from arachnophobia, i.e. 7%.
Let X = Number of teenagers suffering from arachnophobia
So, X ~ [tex]Binom(n= 10,p=0.07)[/tex]
So, probability that at least one of them suffers from arachnophobia
= P(X >= 1) = 1 - probability that none of them suffers from arachnophobia
= 1 - P(X = 0) = 1 - [tex]\binom{10}{0}0.07^{0}(1-0.07)^{10-0}[/tex]
= 1 - (1 * 1 * [tex]0.93^{10}[/tex] ) = 1 - 0.484 = 0.5160 .
Therefore, Probability that at least one of them suffers from arachnophobia is 0.5160 .
Bob can row 13mph in still water. The total time to travel downstream and return upstream to the starting point is 2.6 hours. If the total distance downstream and back is 32 miles. Determine the speed of the river (current speed)
Answer:
"s" is the river speed 16 miles is the one-way distance
Time = distance / velocity
2.6 = [16 / (13 + s) ] + [16 / (13 -s) ]
That solves to s = 3 miles per hour
Step-by-step explanation:
Answer: the speed of the current is 3 mph
Step-by-step explanation:
Let x represent the speed of the river current.
Bob can row 13mph in still water. Assuming that while rowing upstream, he rowed against the current, this means that his total speed upstream is (13 - x) mph
Assuming that while rowing downstream, he rowed with the current, this means that his total speed downstream is (13 + x) mph.x
If the total distance downstream and back is 32 miles. Assuming the distance upstream and downstream is the same, then the distance upstream is 32/2 = 16 miles. Distance downstream is also 16 miles.
Time = Distance/speed
Time taken to row upstream is
16/(13 - x)
Time taken to row downstream is
16/(13 + x)
If total time spent is 2.6 hours, it means that
16/(13 - x) + 16/(13 + x) = 2.6
Cross multiplying, it becomes
16(13 + x) + 16(13 - x) = 2.6(13 + x)(13 - x)
= 208 + 16x + 208 - 16x = 2.6(169 - 13x + 13x - x²)
416 = 2.6(169 - x²)
416/2.6 = 169 - x²
160 = 169 - x²
x² = 169 - 160 = 9
x =√9
x = 3
In the last homework, you analyzed the equation ˙x = cx − x 3 graphically, for c positive, zero and negative. Now perform a formal linear stability analysis, and compare your results with a graphical analysis
Answer:
Please find attached file for complete answer solution and explanation of same question.
Step-by-step explanation:
A stone is thrown horizontally with a speed of 15 m/s from the top of a vertical cliff at the edge of a lake. If the stone hits the water 2.0 s later, the height of the cliff is closest to
Answer:
20 m
Step-by-step explanation:
We are given that
Initial horizontal speed ,[tex]u_x=15 m/s[/tex]
Time, t=2 s
Initial vertical velocity, [tex]u_y=0[/tex]
We know that
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
Where [tex]g=-9.8m/s^2[/tex]
Using the formula
[tex]h=0(2)+\frac{1}{2}(-9.8)(2)^2[/tex]
[tex]h=-19.6 m\approx -20 m[/tex]
The negative sign indicates that the displacement of stone is in downward direction.
Hence, the height of the cliff is closest to 20 m
The University of Arkansas recently approved out of state tuition discounts for high school students from any state. The students must qualify by meeting certain standards in terms of GPA and standardized test scores. The goal of this new policy is to increase the geographic diversity of students from states beyond Arkansas and its border states. Historically, 90% of all new students came from Arkansas or a bordering state. Ginger, a student at the U of A, sampled 180 new students the following year and found that 157 of the new students came from Arkansas or a bordering state. Does Ginger’s study provide enough evidence to indicate that this new policy is effective with a level of significance 10%? What would be the correct decision?
a. Reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansasb. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
c. Reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
d. Fail to reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansas
Answer:
[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]
[tex]p_v =P(z<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [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 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
Step-by-step explanation:
Data given and notation n
n=180 represent the random sample taken
X=157 represent the students who came from Arkansas or a bordering state
[tex]\hat p=\frac{157}{180}=0.872[/tex] estimated proportion of students who came from Arkansas or a bordering state
[tex]p_o=0.9[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
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 the true proportion is higher or not than 0.9.:
Null hypothesis:[tex]p\geq 0.9[/tex]
Alternative hypothesis:[tex]p < 0.9[/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{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/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=0.1[/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<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [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 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack _____________.
Answer:
Reliability
Step-by-step explanation:
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack reliability.
Reliability can be defined as the extent to which an experiment or test provides the same result on repeated trials.
The huge difference in Dylan's IQ test scores in a short period of time suggests that the Whaddayanno IQ test is not reliable.
g Suppose that when a certain lake is stocked with fish, the birth and death rates ˇ and ı are both inversely propor- (a) Show that ????1 p????2 P.t/D 2ktC P0 ; where k is a constant. (b) If P0 D 100 and after 6 months there are 169 fish in the lake, how many will there be after 1 year?
Answer:
k=1
P(12)=256
Step-by-step explanation:
The model for population involving birth and death is given as:
[tex]\frac{dP}{dt}=(b-d)P[/tex]
where b=birth rate and d=death rate.
If the birth and death rate are inversely proportional to [tex]\sqrt{P}[/tex]
[tex]b=\frac{A}{\sqrt{P} } \\d=\frac{B}{\sqrt{P} }[/tex] where A and B are constants of variation.
Substituting b and d into our model
[tex]\frac{dP}{dt}=(\frac{A}{\sqrt{P} }-\frac{B}{\sqrt{P} })P\\\frac{dP}{dt}=\frac{k}{\sqrt{P} }P\\[/tex] where A-B=k, another constant
Simplifying using indices
[tex]\frac{dP}{dt}={k}P^{1-\frac{1}{2} }\\\frac{dP}{dt}={k}P^\frac{1}{2} \\[/tex]
Next, we Separate Variables and Integrate both sides
[tex]\frac{dP}{\sqrt{P} }={k}dt\[/tex]
[tex]\int\frac{dP}{\sqrt{P} }=\int{k}dt\[/tex]
[tex]2P^{1/2} =kt+C[/tex] where C is the constant of integration
[tex]P(t) =(\frac{kt}{2} +C)^2[/tex]
When t=0, P(t)=[tex]P_0[/tex], C=[tex]\sqrt{P_0}[/tex]
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex] as required.
(b)If [tex]P_0[/tex]=100, t=6 months, P(t)=169
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]169 =(\frac{6k}{2} +\sqrt{100})^2\\\sqrt{169} =3k+100\\13=3k+10\\3k=13-10=3\\k=1[/tex]
Since we have found the constant k, we can then calculate the population after 1 year. Note that we use 12 months since we used month earlier to get k.
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]P(12) =(\frac{1X12}{2} +\sqrt{100})^2\\=(6+10)^2=256[/tex]
Therefore the population after a year is 256.
The population of fish after 1 year would be 285.
Let P be the population of fish at any time t months.
We are given:
Initial population: [tex]P_0[/tex] = 100 fishPopulation after 6 months: P(6) = 169 fishWe need to determine the population after 12 months.
The logistic growth model can be expressed as:
dP/dt = rP(1 - P/K)
where r is the intrinsic growth rate and K is the carrying capacity of the environment. Here, since the birth and death rates are inversely proportional to √P, we can infer that the growth rate might be proportional to √P, leading us to use a simplified logistic growth equation.
Given P(6) = 169, let’s find the intrinsic growth rate , r. From population changes between 0 and 6 months:
Let’s assume the logistic growth function in the form:
P(t) = [tex]P_0 e^{rt}[/tex]
We solve for r:
169 = [tex]100 e^{6r}[/tex]
1.69 = [tex]e^{6r}[/tex]
Taking natural logarithms:
ln(1.69) = 6r
0.524 = 6r
r = 0.0873/month
Now, to find P(12):
P(12) = [tex]100 e^{0.0873*12}[/tex]
= [tex]100 e^{1.048}[/tex]
= 100 * 2.853
= 285.3 fish
So, the population after 1 year is approximately 285 fish.
A sample of 100 items has a population standard deviation LaTeX: \:\sigma\:\:σof 5.1 and a sample mean of 21.6. What is the value of the point estimate for the population mean?
Answer:
The point estimate of population mean is 21.6.
Step-by-step explanation:
A point estimate is a numerical value that is used to estimate the value of the parameter under study.
The point estimate is calculated using the sample values.
For example, sample ([tex]\bar x[/tex]) is the point estimate of population mean (μ), sample standard deviation (s) is the point estimate of population standard deviation (σ), sample proportion ([tex]\hat p[/tex]) is the point estimate of population proportion (p).
It is provided that the sample mean of 100 items is,
[tex]\bar x=21.6[/tex]
Then the point estimate of population mean is:
[tex]\mu=\bar x=21.6[/tex]
Thus, the point estimate of population mean is 21.6.
Which of the following represents a null hypothesis? Group of answer choices Class A high school basketball teams who employ a sports psychologist will have a higher proportion ofwins over the course of the season than comparable teams who do not employ a sports psychologist. There will be no difference in rate of skill improvement between college gymnasts who practice meditation and those who do not. Does incorporating relaxation exercises into the daily practice routine of college vocal majors enhance their performance confidence? None of the above
In factual tests, an invalid speculation recommends that there is no importance between factors in a bunch of noticed information. In the given decisions, the assertion shows no distinction in the pace of ability improvement between school gymnasts who practice contemplation and the people who don't address the invalid speculation.
Explanation:In factual theory testing, an invalid speculation is an assertion recommending that no measurable relationship and importance exists in a bunch of noticed information between factors. It expects that any noticed contrasts are because of possibility. So in the given choices, the invalid speculation is: 'There will be no distinction in the pace of expertise improvement between school gymnasts who practice contemplation and the people who don't.' This assertion proposes no effect of the variable (reflection) on the result (ability improvement), which is what an invalid theory addresses in a trial of importance.
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A movie theater offers 6 showings of a movie each day. A total of 1000 people come to see the movie on a particular day. The theater is interested in the number of people who attended each of the six showings. How many possibilities are there for the tallies for each showing for that day?
Answer:
2.03 × 10¹⁴ different possibilities.
Step-by-step explanation:
We want to know the different ways 1000 people can come out to watch a movie at 6 different times of the day.
This is a permutation and combination problem.
Dividing 1000 people amongst 6 movie showings = 1005C5 = 1005!/(1005-5)!5!
= (1005×1004×1003×1002×1001×1000!)/(1000!5!) = (1005×1004×1003×1002×1001)/5! = 2.03 × 10¹⁴ different possibilities.
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25∘ north of east. The road then turns due north for another 3.0 mi to the finish line. In miles, what is the straight-line distance from the starting point to the end of the race? Express your answer in miles.
Answer:
6.79 miles
Step-by-step explanation:
Consider the triangle ABC attached where A is the starting point and C is the finish point.
|AB|=5.2 miles
|BC|=3.0 Miles
≺ABC = 90+75 =165°
We are given the length of two sides and an angle not opposite any of the given sides.
The rule for solving this scenario is referred to as the Cosine Rule.
Note that the side opposite each angle is labelled using the corresponding small letter.
The Cosine Rule States that:
b²=a²+c²-2acCosB
|AC|²=3²+5.2²-(2X5.2XCos165°)
|AC|²=9+27.04-(-10.05)
|AC|²=9+27.04+10.05=46.09
|AC|=√46.09=6.79 miles
Therefore the distance on a straight line from A to C is 6.79 miles
3. It is known that 80% of all college professors have doctoral degrees. If 10 professors are randomly selected, find the probability that a. five have a doctoral degree b. fewer than 4 have doctoral degrees. c. At least 6 have doctoral degrees. d. Between 5 and 7 (inclusive) have doctoral degrees. e. What is the expected number of college professors with doctoral degrees
Answer:
a) [tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
b) [tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
c) [tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
d) [tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
e) [tex] E(X) = n*p = 10*0.8 =8[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=10, p=0.8)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
We want to find this probability:
[tex] P(X=5)[/tex]
And using the pmf we got:
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
Part b
[tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
Part c
For this case we can use the complement rule like this:
[tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
Part d
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
Part e
The expected number is given by:
[tex] E(X) = n*p = 10*0.8 =8[/tex]
This answer provides step-by-step explanations for finding probabilities of professors with doctoral degrees and the expected number of professors with doctoral degrees.
Expected number of college professors with a doctoral degree:
Find the probability that five have a doctoral degree: Using the binomial probability formula, [tex]P(X = 5) = 10C5 * 0.8^5 * 0.2^5[/tex]
Find the probability that fewer than 4 have a doctoral degree: Calculate P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
Find the probability that at least 6 have a doctoral degree: Calculate P(X ≥ 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)).
Find the probability that between 5 and 7 inclusive have a doctoral degree: Calculate P(5 ≤ X ≤ 7) = P(X = 5) + P(X = 6) + P(X = 7).
Expected number of professors with a doctoral degree: Multiply the total number of professors by the probability (0.8).