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Evgen [1.6K]
3 years ago
7

Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg

e 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

Mathematics
1 answer:
Fofino [41]3 years ago
7 0

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 <em>X</em> = number of college women on a diet.

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

The sample of women selected is, <em>n</em> = 267.

The random variable thus follows a Binomial distribution with parameters <em>n</em> = 267 and <em>p</em> = 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 <em>z</em>-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.

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Answer:

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The clerk's error is using 9.5 as sales tax amount instead of finding 9.5% of the total items

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