Mathematics, 21.02.2020 01:20 jlugo1441

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

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Using the normal distribution and the central limit theorem, it is found that the probability that 75% or more of the women in the sample have been on a diet is of 0.0371.

Normal Probability Distribution

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

$Z = \frac{X - \mu}{\sigma}$

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 a proportion p in a sample of size n, the mean is $\mu = p$ and the standard error is $s = \sqrt{\frac{p(1 - p)}{n}}$ .

In this problem:

70% of college women have been on a diet within the past 12 months, hence p = 0.7

.A sample survey interviews an SRS of 267 college women, hence n = 267

The mean and the standard error are given by:

$\mu = p = 0.7$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{267}} = 0.028$

The probability that 75% or more of the women in the sample have been on a diet is 1 subtracted by the p-value of Z when X = 0.75, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.75 - 0.7}{0.028}$

Z = 1.78

Z = 1.78 has a p-value of 0.9629.

1 - 0.9629 = 0.0371.

To learn more about the normal distribution and the central limit theorem, you can take a look at link

Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey int

Answer from: dianaarcos2675

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.

Answer from: Quest

your answer is 228

step-by-step explanation:

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Answer from: Quest

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