What is the probability that the sample mean will be within 1 hour of the population

May 2016
3
0
minnesota
The mean television viewing time for Americans is 15 hours per week (Money, November 2003). Suppose a sample of 60 Americans is taken to further investigate viewing habits. Assume the population standard deviation for weekly viewing time is σ = 4 hours.

a. What is the probability that the sample mean will be within 1 hour of the population mean?


b. What is the probability that the sample mean will be within 45 minutes of the population mean?
 

romsek

MHF Helper
Nov 2013
6,746
3,037
California
Re: What is the probability that the sample mean will be within 1 hour of the populat

if you have an underlying population distribution with mean $\mu$ and standard deviation $\sigma$

and you take $N$ samples of that distribution and compute the mean of the sample, $\bar{X}$

then, provided $N$ is large enough $\bar{X}$ has an approximately normal distribution with mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{N}}$

In the given problem $N$ is large enough for this to be true.

given this you should be able to answer the questions.
 
May 2016
3
0
minnesota
Re: What is the probability that the sample mean will be within 1 hour of the populat

Thanks.
 
May 2016
3
0
minnesota
Re: What is the probability that the sample mean will be within 1 hour of the populat


This is what I have so far

4/60=






0.0667
P[(14<x<16)=P(14-15)/0.0667<(x-15)/0.0667<(16-15)/0.0667]
P(14-15)/0.0667=-14.9925
P(16-15)/0.0667=14.9925P[-14.9925 <Z < 14.9925]
 
May 2016
30
0
Wisconsin
Re: What is the probability that the sample mean will be within 1 hour of the populat

Hey, can someone tell me how to start a new thread Please? I'm a newbie!
 

romsek

MHF Helper
Nov 2013
6,746
3,037
California
Re: What is the probability that the sample mean will be within 1 hour of the populat


This is what I have so far

4/60=






0.0667
P[(14<x<16)=P(14-15)/0.0667<(x-15)/0.0667<(16-15)/0.0667]
P(14-15)/0.0667=-14.9925
P(16-15)/0.0667=14.9925P[-14.9925 <Z < 14.9925]
you scale $\sigma$ by $\dfrac {1}{\sqrt{N}}$

not $\dfrac{1}{N}$ as you have done.

otherwise it looks correct. See if you can use a table or some software to get the actual probability.