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

Answers: 3

Show answers

Answers

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

hello from mrbilldoesmath!

answer: x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb

Answer from: Quest

e. hinduismt religion first emerged on the indian subcontent

Another question on Mathematics

Mathematics, 21.06.2019 13:50

Find the slope of the line y =8/9x + 1.

Answers: 2

Answer

Mathematics, 21.06.2019 15:00

Listed in the table is the percentage of students who chose each kind of juice at lunchtime. use the table to determine the measure of the central angle you would draw to represent orange juice in a circle graph

Answers: 1

Answer

Mathematics, 21.06.2019 17:30

The ksp expression for the system pbcl2(s) pb+2(aq) + 2 cl-(aq) is: [pbc12][pb+2][2cl-] [pb+2][cl-]2 [pb+2][2cl-]2 [pbcl2][pb+2][cl-]2 [pb+2][2cl-]2[pbcl2]

Answers: 1

Answer

Mathematics, 21.06.2019 23:30

Plzzz answer both questions and i will give you 100 points

Answers: 1

Answer

You know the right answer?

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

Questions