Hi I need help w/ sampling distribution.

#1
HI,
I'm taking statistics and i was trying to solve a problem, but I couldn't solve it. I was wondering if you could help me out.
Here's the question:
An industrial process produces batches of a chemical whose impuirty levels follow a normal distribution with standard deviation 1.5 grams per hundred grams of the chemical. A random sample of 64 batches is selected in order to estimate the population mean impurity level.
1. The probability is 0.05 that the sample mean impurity level exceeds the population mean by how much? [ Find b such that P( mean> µX + b) = 0.05.



2. The probability is 0.15 that the sample mean
impurity level differs from the population mean by how much ? [ Findb such that P(µX - b < < µX + b) = 1 - 0.15. That is, P( > µX + b) = 0.15/2.]



PS. We need to find the mean first right? I don't know how to find the mean.

Edit
The answer for
#1 is 0.31
#2 is 0.27
 
Last edited:

JohnM

TS Contributor
#2
CORRECTIONS BELOW IN RED - sorry!

Michelle,

Welcome to the forum.

You don't need to find the mean in this problem, since you are asked to find the number of standard errors away from the mean that correspond to probability levels .05 and .15.

Using the standard normal distribution, find the z-score that corresponds to the upper-tail probability of .05, and then multiply that z-score by the standard error of the mean (1.5 / sqrt(64)). That will tell you how far away from the mean it is.

Follow the same logic for a probability of .15.

JohnM
 
#3
JohnM said:
CORRECTIONS BELOW IN RED - sorry!

Michelle,

Welcome to the forum.

You don't need to find the mean in this problem, since you are asked to find the number of standard errors away from the mean that correspond to probability levels .05 and .15.

Using the standard normal distribution, find the z-score that corresponds to the upper-tail probability of .05, and then multiply that z-score by the standard error of the mean (1.5 / sqrt(64)). That will tell you how far away from the mean it is.

Follow the same logic for a probability of .15.

JohnM
Hello,
I was able to find the z score (standard normal) with the graphing calculator, but how can i find it manually? [This is how i found it invnorm(.05.0,1)=1.644 1.644*(1.5 / sqrt(64))]
z= sample mean- pop mean/ (sd/sqrtn))
is this the right formula?