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Suppose 40% of the employees in a very large corporation are women. If a random sample of 30 employees is chosen from the corporation, let X be the number of women in the sample. a) For a specific x,...

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Suppose 40% of the employees in a very large corporation are women.

If a random sample of 30 employees is chosen from the corporation, let X be

the number of women in the sample.

a) For a specific x, the R function pbinom(x, 25, 0.3) computes P{X ≤ x}.

Use it to evaluate P{X ≤ 17}, P{X ≤ 6}, and hence P{7 ≤ X ≤ 17}.

b) Find µ = E(X) and σ = SD(X). Use the normal approximation to evaluate P{7 ≤ X ≤ 17}. That is, take Z = (X − µ)/σ to be approximately

standard normal. It is best to start with P{6.5 <><>

c) Now suppose the proportion π of women in the corporation is unknown.

A random sample of 30 employees has 20 women. Do you believe π is as

small as 0.4? Explain.

d) In the circumstances of part (c), use formula (1.2) to find an approximate

95% confidence interval for π.

Hints and comments: For (a) and (b), about 0.96; you should give 4-place accuracy.

The margin of error in (d) is about 0.17. Example 1.5 shows that the actual coverage

probability of the confidence interval in (d) may differ substantially from 95%; a

better confidence interval in this case is based on the Agresti-Coull adjustment of

Problem 1.16: (0.486, 0.808).

Answered 273 days After Dec 07, 2021

Solution

Monica answered on Sep 07 2022
71 Votes
Solution
Given:
40% of the employees in a very large corporation are women.
If the random sample of 30 employees is chosen from the corporation.
Let X is the number of women in the sample.
a)
It is asked to find the probability of
R code:
pbinom(17,30,0.4)
[1] 0.9788
Thus, the required probability is equal to 0.9788

R code
pbinom(6,30,0.4)
[1] 0.0172
Thus, the required probability is equal to 0.0172
It is asked to find the probability of
R code:
pbinom(17,30,0.4)-pbinom(7,30,0.4)
[1] 0.9352
Thus, the probability is equal to 0.9352
) Continuity co
ection factor
Binomial distribution is approximated to the normal distribution...
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