Question

Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 college women. What is the probability that 75% or more of the women in the sample have been on a diet? P ( ^ p ≥ 0.75 ) = 0.9629 P ( ^ p ≥ 0.75 ) = 0.9728 P ( ^ p ≥ 0.75 ) = 0.1214 P ( ^ p ≥ 0.75 ) = 0.0272 P ( ^ p ≥ 0.75 ) = 0.0371

Answer 1

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let X = number of college women on a diet.

The probability of a woman being on diet is, P (X) = p = 0.70.

The sample of women selected is, n = 267.

The random variable thus follows a Binomial distribution with parameters n = 267 and p = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions ($\hat p$) follows a Normal distribution.  

The mean of this distribution is:

$\mu_{\hat p} = p = 0.70$

The standard deviation of this distribution is: $\sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028$

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

$P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z$

**Use the z-table for the probability.

$P(\hat p \geq 0.75)=1-P(Z$

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.