# Working with sampling distribution and probability.

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A southern state has an unemployment rate of 6%. The state conducts monthly surveys in order to track the unemployment rate. In a recent month, a random sample of 700 people showed that 35 were unemployed.

a) If the true unemployment rate is 6%, describe the sampling distribution of p^.

b) Find P(p^ >= 0.05)

c) Assume the population proposal, p, is unknown. Describe the sampling distribution of p^ based on the most recent sample.

d) Find the probability that the sample proportion will lie within 0.05 of the true proportion of people who are unemployed.

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##### Solution Summary

A southern state has an unemployment rate of 6%. The state conducts monthly surveys in order to track the unemployment rate. In a recent month, a random sample of 700 people showed that 35 were unemployed.

a) If the true unemployment rate is 6%, describe the sampling distribution of p^.

b) Find P(p^ >= 0.05)

c) Assume the population proposal, p, is unknown. Describe the sampling distribution of p^ based on the most recent sample.

d) Find the probability that the sample proportion will lie within 0.05 of the true proportion of people who are unemployed.

##### Solution Preview

a) If the true unemployment rate is 6%, then let X be the number of people who were unemployed. Obviously,

X~b(700,0.06), that is a binomial distribution with n=700, and p=0.06. So p^=(X1+X2+...+X700)/700 approximately follows normal distribution, where

<br> E(p^)={E(X1)+E(X2)+...+E(X700)}/700=0.06,

<br>and

<br> ...

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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