Consider a binomial experiment with n = 20 and p = .70. a. Compute f (12). b. Compute f (16). c. Compute P ( x ≥ 16). d. Compute P ( x ≤ 15). e. Compute E ( x ). f. Compute Var ( x ) and σ .

Question

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Answer 1

Textbook Question

Chapter 5.5, Problem 33E

Consider a binomial experiment with n = 20 and p = .70.

a. Compute f(12).
b. Compute f(16).
c. Compute P(x ≥ 16).
d. Compute P(x ≤ 15).
e. Compute E(x).
f. Compute Var (x) and σ.

a.

Expert Solution

To determine

Find the value of $f(12)$ .

Answer to Problem 33E

The value of $f(12)$ is 0.1144.

Explanation of Solution

Calculation:

The binomial experiment with $n=20 and p=0.70$ .

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f(x)=(nx)px(1−p)n−x, x=0,1,2,...,n$

Where, p is the probability of success.

Substitute values $n=20 and p=0.70 and x=12$ .

$f(12)=(2012)(0.70)12(1−0.70)20−12=20!12!(20−12)!(0.70)12(0.30)8=0.1144$

Thus, the probability of 12 successes is 0.1144.

b.

Expert Solution

To determine

Find the value of, $f(16)$ .

Answer to Problem 33E

The value of, $f(16)$ is 0.1304.

Explanation of Solution

Calculation:

Consider $n=20 and p=0.70 and x=16$ .

$f(16)=(2016)(0.70)16(1−0.70)20−16=20!16!(20−16)!(0.70)16(0.30)4=0.1304$

Thus, the probability of 16 successes is 0.1304.

c.

Expert Solution

To determine

Find the value of, $P(x≥16)$ .

Answer to Problem 33E

The value of $P(x≥16)$ is 0.2374.

Explanation of Solution

Calculation:

Here, $P(x≥16)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19 and x=20$ . That is,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)$

The probability $f(17)$ is obtained by substituting the values $n=20 and p=0.70 and x=17$ in the binomial probability formula. That is,

The probability value of $f(17)$ is,

$f(17)=(2017)(0.70)17(1−0.70)20−17=20!17!(20−17)!(0.70)17(0.30)3=0.0716$

The probability value of $f(18)$ is,

$f(18)=(2018)(0.70)18(1−0.70)20−18=20!18!(20−18)!(0.70)18(0.30)2=0.0278$

The probability value of $f(19)$ is,

$f(19)=(2019)(0.70)19(1−0.70)20−19=20!19!(20−19)!(0.70)19(0.30)1=0.0068$

The probability value of $f(20)$ is,

$f(20)=(2020)(0.70)20(1−0.70)20−20=20!20!(20−20)!(0.70)20(0.30)0=0.0008$

The required probability becomes,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Thus, the probability of at least 16 successes is 0.2374.

d.

Expert Solution

To determine

Find the value of, $P(x≤15)$ .

Answer to Problem 33E

The value of $P(x≤15)$ is 0.7626.

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P(x≤15)=1−P(x≥16)$

Substituting the value from part (c),

$P(x≤15)=1−P(x≥16)=1−0.2374=0.7626$

Thus, the probability of at most 15 successes is 0.7626.

e.

Expert Solution

To determine

Compute the expected value.

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$E(x)=μ=np$

Substituting the values $n=20 and p=0.70$

$E(x)=20×0.70=14$

The expected value of the binomial random variable is 14.

f.

Expert Solution

To determine

Compute the variance and standard deviation.

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

$σ2=np(1−p)$

Substituting the values $n=20 and p=0.70$

$σ2=np(1−p)=20×0.7(1−0.7)=4.2$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$σx=np(1−p)=0.42=2.0494$

The standard deviation of the binomial random variable is 2.0494.

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Answer 2

Textbook Question

Chapter 5.5, Problem 33E

Consider a binomial experiment with n = 20 and p = .70.

a. Compute f(12).
b. Compute f(16).
c. Compute P(x ≥ 16).
d. Compute P(x ≤ 15).
e. Compute E(x).
f. Compute Var (x) and σ.

a.

Expert Solution

To determine

Find the value of $f(12)$ .

Answer to Problem 33E

The value of $f(12)$ is 0.1144.

Explanation of Solution

Calculation:

The binomial experiment with $n=20 and p=0.70$ .

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f(x)=(nx)px(1−p)n−x, x=0,1,2,...,n$

Where, p is the probability of success.

Substitute values $n=20 and p=0.70 and x=12$ .

$f(12)=(2012)(0.70)12(1−0.70)20−12=20!12!(20−12)!(0.70)12(0.30)8=0.1144$

Thus, the probability of 12 successes is 0.1144.

b.

Expert Solution

To determine

Find the value of, $f(16)$ .

Answer to Problem 33E

The value of, $f(16)$ is 0.1304.

Explanation of Solution

Calculation:

Consider $n=20 and p=0.70 and x=16$ .

$f(16)=(2016)(0.70)16(1−0.70)20−16=20!16!(20−16)!(0.70)16(0.30)4=0.1304$

Thus, the probability of 16 successes is 0.1304.

c.

Expert Solution

To determine

Find the value of, $P(x≥16)$ .

Answer to Problem 33E

The value of $P(x≥16)$ is 0.2374.

Explanation of Solution

Calculation:

Here, $P(x≥16)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19 and x=20$ . That is,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)$

The probability $f(17)$ is obtained by substituting the values $n=20 and p=0.70 and x=17$ in the binomial probability formula. That is,

The probability value of $f(17)$ is,

$f(17)=(2017)(0.70)17(1−0.70)20−17=20!17!(20−17)!(0.70)17(0.30)3=0.0716$

The probability value of $f(18)$ is,

$f(18)=(2018)(0.70)18(1−0.70)20−18=20!18!(20−18)!(0.70)18(0.30)2=0.0278$

The probability value of $f(19)$ is,

$f(19)=(2019)(0.70)19(1−0.70)20−19=20!19!(20−19)!(0.70)19(0.30)1=0.0068$

The probability value of $f(20)$ is,

$f(20)=(2020)(0.70)20(1−0.70)20−20=20!20!(20−20)!(0.70)20(0.30)0=0.0008$

The required probability becomes,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Thus, the probability of at least 16 successes is 0.2374.

d.

Expert Solution

To determine

Find the value of, $P(x≤15)$ .

Answer to Problem 33E

The value of $P(x≤15)$ is 0.7626.

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P(x≤15)=1−P(x≥16)$

Substituting the value from part (c),

$P(x≤15)=1−P(x≥16)=1−0.2374=0.7626$

Thus, the probability of at most 15 successes is 0.7626.

e.

Expert Solution

To determine

Compute the expected value.

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$E(x)=μ=np$

Substituting the values $n=20 and p=0.70$

$E(x)=20×0.70=14$

The expected value of the binomial random variable is 14.

f.

Expert Solution

To determine

Compute the variance and standard deviation.

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

$σ2=np(1−p)$

Substituting the values $n=20 and p=0.70$

$σ2=np(1−p)=20×0.7(1−0.7)=4.2$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$σx=np(1−p)=0.42=2.0494$

The standard deviation of the binomial random variable is 2.0494.

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Answer 3

Textbook Question

Chapter 5.5, Problem 33E

Consider a binomial experiment with n = 20 and p = .70.

a. Compute f(12).
b. Compute f(16).
c. Compute P(x ≥ 16).
d. Compute P(x ≤ 15).
e. Compute E(x).
f. Compute Var (x) and σ.

a.

Expert Solution

To determine

Find the value of $f(12)$ .

Answer to Problem 33E

The value of $f(12)$ is 0.1144.

Explanation of Solution

Calculation:

The binomial experiment with $n=20 and p=0.70$ .

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f(x)=(nx)px(1−p)n−x, x=0,1,2,...,n$

Where, p is the probability of success.

Substitute values $n=20 and p=0.70 and x=12$ .

$f(12)=(2012)(0.70)12(1−0.70)20−12=20!12!(20−12)!(0.70)12(0.30)8=0.1144$

Thus, the probability of 12 successes is 0.1144.

b.

Expert Solution

To determine

Find the value of, $f(16)$ .

Answer to Problem 33E

The value of, $f(16)$ is 0.1304.

Explanation of Solution

Calculation:

Consider $n=20 and p=0.70 and x=16$ .

$f(16)=(2016)(0.70)16(1−0.70)20−16=20!16!(20−16)!(0.70)16(0.30)4=0.1304$

Thus, the probability of 16 successes is 0.1304.

c.

Expert Solution

To determine

Find the value of, $P(x≥16)$ .

Answer to Problem 33E

The value of $P(x≥16)$ is 0.2374.

Explanation of Solution

Calculation:

Here, $P(x≥16)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19 and x=20$ . That is,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)$

The probability $f(17)$ is obtained by substituting the values $n=20 and p=0.70 and x=17$ in the binomial probability formula. That is,

The probability value of $f(17)$ is,

$f(17)=(2017)(0.70)17(1−0.70)20−17=20!17!(20−17)!(0.70)17(0.30)3=0.0716$

The probability value of $f(18)$ is,

$f(18)=(2018)(0.70)18(1−0.70)20−18=20!18!(20−18)!(0.70)18(0.30)2=0.0278$

The probability value of $f(19)$ is,

$f(19)=(2019)(0.70)19(1−0.70)20−19=20!19!(20−19)!(0.70)19(0.30)1=0.0068$

The probability value of $f(20)$ is,

$f(20)=(2020)(0.70)20(1−0.70)20−20=20!20!(20−20)!(0.70)20(0.30)0=0.0008$

The required probability becomes,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Thus, the probability of at least 16 successes is 0.2374.

d.

Expert Solution

To determine

Find the value of, $P(x≤15)$ .

Answer to Problem 33E

The value of $P(x≤15)$ is 0.7626.

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P(x≤15)=1−P(x≥16)$

Substituting the value from part (c),

$P(x≤15)=1−P(x≥16)=1−0.2374=0.7626$

Thus, the probability of at most 15 successes is 0.7626.

e.

Expert Solution

To determine

Compute the expected value.

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$E(x)=μ=np$

Substituting the values $n=20 and p=0.70$

$E(x)=20×0.70=14$

The expected value of the binomial random variable is 14.

f.

Expert Solution

To determine

Compute the variance and standard deviation.

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

$σ2=np(1−p)$

Substituting the values $n=20 and p=0.70$

$σ2=np(1−p)=20×0.7(1−0.7)=4.2$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$σx=np(1−p)=0.42=2.0494$

The standard deviation of the binomial random variable is 2.0494.

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Answer 4

Textbook Question

Chapter 5.5, Problem 33E

Consider a binomial experiment with n = 20 and p = .70.

a. Compute f(12).
b. Compute f(16).
c. Compute P(x ≥ 16).
d. Compute P(x ≤ 15).
e. Compute E(x).
f. Compute Var (x) and σ.

a.

Expert Solution

To determine

Find the value of $f(12)$ .

Answer to Problem 33E

The value of $f(12)$ is 0.1144.

Explanation of Solution

Calculation:

The binomial experiment with $n=20 and p=0.70$ .

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f(x)=(nx)px(1−p)n−x, x=0,1,2,...,n$

Where, p is the probability of success.

Substitute values $n=20 and p=0.70 and x=12$ .

$f(12)=(2012)(0.70)12(1−0.70)20−12=20!12!(20−12)!(0.70)12(0.30)8=0.1144$

Thus, the probability of 12 successes is 0.1144.

b.

Expert Solution

To determine

Find the value of, $f(16)$ .

Answer to Problem 33E

The value of, $f(16)$ is 0.1304.

Explanation of Solution

Calculation:

Consider $n=20 and p=0.70 and x=16$ .

$f(16)=(2016)(0.70)16(1−0.70)20−16=20!16!(20−16)!(0.70)16(0.30)4=0.1304$

Thus, the probability of 16 successes is 0.1304.

c.

Expert Solution

To determine

Find the value of, $P(x≥16)$ .

Answer to Problem 33E

The value of $P(x≥16)$ is 0.2374.

Explanation of Solution

Calculation:

Here, $P(x≥16)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19 and x=20$ . That is,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)$

The probability $f(17)$ is obtained by substituting the values $n=20 and p=0.70 and x=17$ in the binomial probability formula. That is,

The probability value of $f(17)$ is,

$f(17)=(2017)(0.70)17(1−0.70)20−17=20!17!(20−17)!(0.70)17(0.30)3=0.0716$

The probability value of $f(18)$ is,

$f(18)=(2018)(0.70)18(1−0.70)20−18=20!18!(20−18)!(0.70)18(0.30)2=0.0278$

The probability value of $f(19)$ is,

$f(19)=(2019)(0.70)19(1−0.70)20−19=20!19!(20−19)!(0.70)19(0.30)1=0.0068$

The probability value of $f(20)$ is,

$f(20)=(2020)(0.70)20(1−0.70)20−20=20!20!(20−20)!(0.70)20(0.30)0=0.0008$

The required probability becomes,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Thus, the probability of at least 16 successes is 0.2374.

d.

Expert Solution

To determine

Find the value of, $P(x≤15)$ .

Answer to Problem 33E

The value of $P(x≤15)$ is 0.7626.

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P(x≤15)=1−P(x≥16)$

Substituting the value from part (c),

$P(x≤15)=1−P(x≥16)=1−0.2374=0.7626$

Thus, the probability of at most 15 successes is 0.7626.

e.

Expert Solution

To determine

Compute the expected value.

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$E(x)=μ=np$

Substituting the values $n=20 and p=0.70$

$E(x)=20×0.70=14$

The expected value of the binomial random variable is 14.

f.

Expert Solution

To determine

Compute the variance and standard deviation.

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

$σ2=np(1−p)$

Substituting the values $n=20 and p=0.70$

$σ2=np(1−p)=20×0.7(1−0.7)=4.2$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$σx=np(1−p)=0.42=2.0494$

The standard deviation of the binomial random variable is 2.0494.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Find the value of $f(12)$ .

Answer to Problem 33E

The value of $f(12)$ is 0.1144.

Explanation of Solution

Calculation:

The binomial experiment with $n=20 and p=0.70$ .

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f(x)=(nx)px(1−p)n−x, x=0,1,2,...,n$

Where, p is the probability of success.

Substitute values $n=20 and p=0.70 and x=12$ .

$f(12)=(2012)(0.70)12(1−0.70)20−12=20!12!(20−12)!(0.70)12(0.30)8=0.1144$

Thus, the probability of 12 successes is 0.1144.

b.

Expert Solution

To determine

Find the value of, $f(16)$ .

Answer to Problem 33E

The value of, $f(16)$ is 0.1304.

Explanation of Solution

Calculation:

Consider $n=20 and p=0.70 and x=16$ .

$f(16)=(2016)(0.70)16(1−0.70)20−16=20!16!(20−16)!(0.70)16(0.30)4=0.1304$

Thus, the probability of 16 successes is 0.1304.

c.

Expert Solution

To determine

Find the value of, $P(x≥16)$ .

Answer to Problem 33E

The value of $P(x≥16)$ is 0.2374.

Explanation of Solution

Calculation:

Here, $P(x≥16)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19 and x=20$ . That is,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)$

The probability $f(17)$ is obtained by substituting the values $n=20 and p=0.70 and x=17$ in the binomial probability formula. That is,

The probability value of $f(17)$ is,

$f(17)=(2017)(0.70)17(1−0.70)20−17=20!17!(20−17)!(0.70)17(0.30)3=0.0716$

The probability value of $f(18)$ is,

$f(18)=(2018)(0.70)18(1−0.70)20−18=20!18!(20−18)!(0.70)18(0.30)2=0.0278$

The probability value of $f(19)$ is,

$f(19)=(2019)(0.70)19(1−0.70)20−19=20!19!(20−19)!(0.70)19(0.30)1=0.0068$

The probability value of $f(20)$ is,

$f(20)=(2020)(0.70)20(1−0.70)20−20=20!20!(20−20)!(0.70)20(0.30)0=0.0008$

The required probability becomes,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Thus, the probability of at least 16 successes is 0.2374.

d.

Expert Solution

To determine

Find the value of, $P(x≤15)$ .

Answer to Problem 33E

The value of $P(x≤15)$ is 0.7626.

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P(x≤15)=1−P(x≥16)$

Substituting the value from part (c),

$P(x≤15)=1−P(x≥16)=1−0.2374=0.7626$

Thus, the probability of at most 15 successes is 0.7626.

e.

Expert Solution

To determine

Compute the expected value.

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$E(x)=μ=np$

Substituting the values $n=20 and p=0.70$

$E(x)=20×0.70=14$

The expected value of the binomial random variable is 14.

f.

Expert Solution

To determine

Compute the variance and standard deviation.

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

$σ2=np(1−p)$

Substituting the values $n=20 and p=0.70$

$σ2=np(1−p)=20×0.7(1−0.7)=4.2$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$σx=np(1−p)=0.42=2.0494$

The standard deviation of the binomial random variable is 2.0494.

Answer 8

a.

Expert Solution

To determine

Find the value of $f(12)$ .

Answer to Problem 33E

The value of $f(12)$ is 0.1144.

Explanation of Solution

Calculation:

The binomial experiment with $n=20 and p=0.70$ .

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f(x)=(nx)px(1−p)n−x, x=0,1,2,...,n$

Where, p is the probability of success.

Substitute values $n=20 and p=0.70 and x=12$ .

$f(12)=(2012)(0.70)12(1−0.70)20−12=20!12!(20−12)!(0.70)12(0.30)8=0.1144$

Thus, the probability of 12 successes is 0.1144.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find the value of $f(12)$ .

Answer 13

Answer to Problem 33E

The value of $f(12)$ is 0.1144.

Answer 14

Explanation of Solution

Calculation:

The binomial experiment with $n=20 and p=0.70$ .

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f(x)=(nx)px(1−p)n−x, x=0,1,2,...,n$

Where, p is the probability of success.

Substitute values $n=20 and p=0.70 and x=12$ .

$f(12)=(2012)(0.70)12(1−0.70)20−12=20!12!(20−12)!(0.70)12(0.30)8=0.1144$

Thus, the probability of 12 successes is 0.1144.

Answer 15

b.

Expert Solution

To determine

Find the value of, $f(16)$ .

Answer to Problem 33E

The value of, $f(16)$ is 0.1304.

Explanation of Solution

Calculation:

Consider $n=20 and p=0.70 and x=16$ .

$f(16)=(2016)(0.70)16(1−0.70)20−16=20!16!(20−16)!(0.70)16(0.30)4=0.1304$

Thus, the probability of 16 successes is 0.1304.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Find the value of, $f(16)$ .

Answer 20

Answer to Problem 33E

The value of, $f(16)$ is 0.1304.

Answer 21

Explanation of Solution

Calculation:

Consider $n=20 and p=0.70 and x=16$ .

$f(16)=(2016)(0.70)16(1−0.70)20−16=20!16!(20−16)!(0.70)16(0.30)4=0.1304$

Thus, the probability of 16 successes is 0.1304.

Answer 22

c.

Expert Solution

To determine

Find the value of, $P(x≥16)$ .

Answer to Problem 33E

The value of $P(x≥16)$ is 0.2374.

Explanation of Solution

Calculation:

Here, $P(x≥16)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19 and x=20$ . That is,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)$

The probability $f(17)$ is obtained by substituting the values $n=20 and p=0.70 and x=17$ in the binomial probability formula. That is,

The probability value of $f(17)$ is,

$f(17)=(2017)(0.70)17(1−0.70)20−17=20!17!(20−17)!(0.70)17(0.30)3=0.0716$

The probability value of $f(18)$ is,

$f(18)=(2018)(0.70)18(1−0.70)20−18=20!18!(20−18)!(0.70)18(0.30)2=0.0278$

The probability value of $f(19)$ is,

$f(19)=(2019)(0.70)19(1−0.70)20−19=20!19!(20−19)!(0.70)19(0.30)1=0.0068$

The probability value of $f(20)$ is,

$f(20)=(2020)(0.70)20(1−0.70)20−20=20!20!(20−20)!(0.70)20(0.30)0=0.0008$

The required probability becomes,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Thus, the probability of at least 16 successes is 0.2374.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Find the value of, $P(x≥16)$ .

Answer 27

Answer to Problem 33E

The value of $P(x≥16)$ is 0.2374.

Answer 28

Explanation of Solution

Calculation:

Here, $P(x≥16)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19 and x=20$ . That is,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)$

The probability $f(17)$ is obtained by substituting the values $n=20 and p=0.70 and x=17$ in the binomial probability formula. That is,

The probability value of $f(17)$ is,

$f(17)=(2017)(0.70)17(1−0.70)20−17=20!17!(20−17)!(0.70)17(0.30)3=0.0716$

The probability value of $f(18)$ is,

$f(18)=(2018)(0.70)18(1−0.70)20−18=20!18!(20−18)!(0.70)18(0.30)2=0.0278$

The probability value of $f(19)$ is,

$f(19)=(2019)(0.70)19(1−0.70)20−19=20!19!(20−19)!(0.70)19(0.30)1=0.0068$

The probability value of $f(20)$ is,

$f(20)=(2020)(0.70)20(1−0.70)20−20=20!20!(20−20)!(0.70)20(0.30)0=0.0008$

The required probability becomes,

$P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Thus, the probability of at least 16 successes is 0.2374.

Answer 29

d.

Expert Solution

To determine

Find the value of, $P(x≤15)$ .

Answer to Problem 33E

The value of $P(x≤15)$ is 0.7626.

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P(x≤15)=1−P(x≥16)$

Substituting the value from part (c),

$P(x≤15)=1−P(x≥16)=1−0.2374=0.7626$

Thus, the probability of at most 15 successes is 0.7626.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

Find the value of, $P(x≤15)$ .

Answer 34

Answer to Problem 33E

The value of $P(x≤15)$ is 0.7626.

Answer 35

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P(x≤15)=1−P(x≥16)$

Substituting the value from part (c),

$P(x≤15)=1−P(x≥16)=1−0.2374=0.7626$

Thus, the probability of at most 15 successes is 0.7626.

Answer 36

e.

Expert Solution

To determine

Compute the expected value.

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$E(x)=μ=np$

Substituting the values $n=20 and p=0.70$

$E(x)=20×0.70=14$

The expected value of the binomial random variable is 14.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

Compute the expected value.

Answer 41

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Answer 42

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$E(x)=μ=np$

Substituting the values $n=20 and p=0.70$

$E(x)=20×0.70=14$

The expected value of the binomial random variable is 14.

Answer 43

f.

Expert Solution

To determine

Compute the variance and standard deviation.

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

$σ2=np(1−p)$

Substituting the values $n=20 and p=0.70$

$σ2=np(1−p)=20×0.7(1−0.7)=4.2$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$σx=np(1−p)=0.42=2.0494$

The standard deviation of the binomial random variable is 2.0494.

Answer 44

f.

Expert Solution

Answer 45

f.

Expert Solution

Answer 46

Expert Solution

Answer 47

To determine

Compute the variance and standard deviation.

Answer 48

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Answer 49

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

$σ2=np(1−p)$

Substituting the values $n=20 and p=0.70$

$σ2=np(1−p)=20×0.7(1−0.7)=4.2$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$σx=np(1−p)=0.42=2.0494$

The standard deviation of the binomial random variable is 2.0494.

Answer 50

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Consider a binomial experiment with n = 20 and p = .70. a. Compute f (12). b. Compute f (16). c. Compute P ( x ≥ 16). d. Compute P ( x ≤ 15). e. Compute E ( x ). f. Compute Var ( x ) and σ .

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Answer to Problem 33E

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Answer to Problem 33E

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