Answer to Question #150338 in Statistics and Probability for Amy bautro

Question

John crewing agency has an average of 200 applicants per week for the position of deck cadet with a standard deviation of 12. A random sample of size 4 weeks will be taken.
A) determine the mean and standard error of the randomly selected weeks.
B) find the percentage of the randomly selected weeks to have a sample mean more than 210 applicants.
C) find the percentage of the randomly selected weeks to have a sample mean less than 185 applicants.
D) find the percentage of the randomly selected weeks to have a sample mean within 8 applicants of the population mean.

Answer 1

$\mu=200$

$\sigma=12$

$n =4$

$\mu_4=\mu\cdot n=800$

$\sigma_4=\sigma\cdot n=48$

$SE=\frac{\sigma_4}{\sqrt n}=\frac{48}{\sqrt4}=24$

More than 210:

$P(X>210)=P(Z>\frac{210-200}{12})=P(Z>0.83)=1-P(Z<0.83)=$

$=1-0.7967=0.2033$

Less than 185:

$P(X<185)=P(Z<\frac{185-200}{12})=P(Z<-1.25)=0.1056$

Within 8:

$P(192<X<208)=P(\frac{192-200}{12}<Z<\frac{208-200}{12})=P(-0.67<Z<0.67)=$

$=P(Z<0.67)-P(Z<-0.67)=0.7486-0.2514=0.4972$