Respuesta :
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 ([tex]\hat p[/tex]) follows a Normal distribution.
The mean of this distribution is:
[tex]\mu_{\hat p} = p = 0.70[/tex]
The standard deviation of this distribution is: [tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028[/tex]
Compute the probability that 75% or more of the women in the sample have been on a diet as follows:
[tex]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<1.79)[/tex]
**Use the z-table for the probability.
[tex]P(\hat p \geq 0.75)=1-P(Z<1.79)=1-0.96327=0.03673\approx0.037[/tex]
Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.
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 [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 a proportion p in a sample of size n, the mean is [tex]\mu = p[/tex] and the standard error is [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex].
In this problem:
- 70% of college women have been on a diet within the past 12 months, hence [tex]p = 0.7[/tex].
- A sample survey interviews an SRS of 267 college women, hence [tex]n = 267[/tex].
The mean and the standard error are given by:
[tex]\mu = p = 0.7[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{267}} = 0.028[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.75 - 0.7}{0.028}[/tex]
[tex]Z = 1.78[/tex]
[tex]Z = 1.78[/tex] 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 https://brainly.com/question/24663213
