
Concept explainers
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.

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.

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.

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.

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.

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.

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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Chapter 5 Solutions
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