A sample survey interviews an SRS of 209 college women. Suppose that 69% of all college women have been on a diet within the last 12 months. What is the probability that 72% or more of the women in the sample have been on a diet

Using the Normal Distribution and the Central Limit Theorem, it is found that there is a 0.1736 = 17.36% probability that 72% or more of the women in the sample have been on a diet.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It 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, which is the percentile of X.
By the Central Limit Theorem, for sampling distribution of the sample proportion of size n, with an estimate of p, the mean is [tex]\mu = p[/tex] and the standard error is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex].

In this problem, [tex]p = 0.69, n = 209[/tex], hence, the mean and the standard error are given by:

[tex]\mu = p = 0.69[/tex]

[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.69(0.31)}{209}} = 0.032[/tex]

The probability that 72% or more of the women in the sample have been on a diet is 1 subtracted by the p-value of Z when X = 0.72, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.72 - 0.69}{0.032}[/tex]

[tex]Z = 0.94[/tex]

[tex]Z = 0.94[/tex] has a p-value of 0.8264.

1 - 0.8264 = 0.1736.

0.1736 = 17.36% probability that 72% or more of the women in the sample have been on a diet.

A similar problem is given at https://brainly.com/question/15413688