Respuesta :
Answer:
[tex]\mu = 17.5[/tex]
[tex]\sigma = 4[/tex]
n = 50
A) show the sampling distribution of x, the sample mean average for a sample of 50 unemployment individuals.
We will use central limit theorem
So, mean of sampling distribution = [tex]\mu_{\bar{x}}=\mu = 17.5[/tex]
Standard deviation of sampling distribution = [tex]\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}= 0.5656[/tex]
B) What is the probability that a simple random sample of 50 unemployment individuals will provide a sample mean within one week of the population mean?
A sample mean within one week of the population mean means [tex]x-\mu = 1[/tex]
So, [tex]P(|x-\mu|<1)=P(-1<x-\mu<1)[/tex]
=[tex]P(\frac{-1}{\frac{4}{\sqrt{50}}}<\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} <\frac{1}{\frac{4}{\sqrt{50}}})[/tex]
=[tex]P(-1.77<Z<1.77[/tex]
=[tex]P(z<1.77)-P(z<-1.77)[/tex]
=0.9616-0.0384
=0.9232
The probability that a simple random sample of 50 unemployment individuals will provide a sample mean within one week of the population mean is 0.9232.
C) What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within a half week of the population mean?
A sample mean within one week of the population mean means [tex]x-\mu = 1[/tex]
So, [tex]P(|x-\mu|<0.5)=P(-0.5<x-\mu<0.5)[/tex]
=[tex]P(\frac{-1-0.5}{\frac{4}{\sqrt{50}}}<\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} <\frac{0.5}{\frac{4}{\sqrt{50}}})[/tex]
=[tex]P(-0.88<Z<0.88[/tex]
=[tex]P(z<0.88)-P(z<0.88)[/tex]
=0.8106-0.1894
=0.6212
The probability that a simple random sample of 50 unemployed individuals will provide a sample mean within a half week of the population mean is 0.6212
