Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed s 30 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 30 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 34 unemployed individuals for a follow-up study Find the probability that a single randomly selected value is less than 31 Pix<31)- Find the probability that a sample of size n-34 is randomly selected with a mean less than 3I PIM<31)- Enter your answers as numbers accurate to 4 decimal places. Points possible: 1 Submilt Next 1/23/2019 BANG &OLUFSE 5 6
Right answer needed please.

Answer 1

Let's consider X be the length of unemployment in weeks.

Given mean of X $\mu=30$

standard deviation is $\sigma=4$

n=34

a) The probability that the single randomly selected value is less than 31 is given by

$P(X<31)=P(\frac{{X}-\mu}{{\sigma }}<\frac{{31}-\mu}{{\sigma }})$

$P(X<31)=P(Z<\frac{{31}-30}{{4 }})$

$P(X<31)=P(Z<\frac{1}{{4 }})$

$P(X<31)=P(Z<0.25)$

$P(X<31)=0.5987$ ### By using z table

b) The probability that the sample of size 34 is randomly selected with mean less than 31 is given by

$P(M<31)=P(\frac{M-\mu}{\frac{\sigma }{\sqrt{n}}}<\frac{31-\mu}{\frac{\sigma }{\sqrt{n}}})$

$P(M<31)=P(Z<\frac{31-30}{\frac{4 }{\sqrt{34}}})$

$P(M<31)=P(Z<\frac{1}{\frac{4 }{5.83}})$

$P(M<31)=P(Z<\frac{1}{0.6861})$

$P(M<31)=P(Z<1.46)$

$P(M<31)=0.9279$ ### By using the z table.

The probability that the sample of size 34 is randomly selected with mean less than 31 =0.9279