Question

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x-bar is used to estimate μ. What is the probability that the sample mean will be within ±15 of the population mean? State your answer as a decimal with 4 decimal places.

Answer 1

If a population has mean : and standard deviation ; for large samples (n>30), By Central limit theorem

Sampling distribution of sample mean : follows normal distribution with mean : = and standard deviation : =

For the given problem,

=200 ; =50 ; n=100

Therefore

follows normal distribution with =200 and =

probability that the sample mean will be within ±15 of the population mean = P(200-15 < < 200+15)

Z-score for (200-15) = (200-15-200)/5 =-3

Z-score for (200+15) = (200+15-200)/5 =3

P(200-15 < < 200+15) = P(-3<Z<3) = P(Z<3) -P(Z<-3)

From standard normal tables P(Z<3)= 0.9987; P(Z<-3) = 0.0013

P(Z<3) -P(Z<-3) =0.9987-0.0013=0.9974

P(200-15 < < 200+15) = P(-3<Z<3) = P(Z<3) -P(Z<-3) =0.9974

Probability that the sample mean will be within ±15 of the population mean = 0.9974