# finding number of standard deviations between sample and population mean

• August 8th 2010, 10:13 PM
Jskid
finding number of standard deviations between sample and population mean
Suppose that the mean mileage of all pickup trucks of a particular make is unknown and will be estimated from a random sample of 36 trucks. The sample mean mileage was found to be 6 litres per 100 km. Assume that the standard deviation of the mileages of all trucks is .5 litres per 100 km. There is a 0.95 probability that the mean of the sample will be within z SDs of the unknown mileage. What is z?

I believe I find the z-score associated with 0.95 but I'm not sure if I need to do something with the sample mean or SD.
• August 9th 2010, 02:52 AM
Quote:

Originally Posted by Jskid
Suppose that the mean mileage of all pickup trucks of a particular make is unknown and will be estimated from a random sample of 36 trucks. The sample mean mileage was found to be 6 litres per 100 km. Assume that the standard deviation of the mileages of all trucks is .5 litres per 100 km. There is a 0.95 probability that the mean of the sample will be within z SDs of the unknown mileage. What is z?

I believe I find the z-score associated with 0.95 but I'm not sure if I need to do something with the sample mean or SD.

A 0.95 probability that the sample mean is within z standard deviations
means that you need to locate the two 2.5% regions at the tails.

Hence, you need the z-score corresponding to 0.975 instead.
Also, as you have the "population" SD

$Z=\displaystyle\huge\frac{sample\ mean-\mu}{\frac{\sigma}{\sqrt{36}}}$

where $\mu$ is the population mean.

The above calculation is taken for sample mean greater than population mean.
The graph is symmetrical however.
• August 9th 2010, 07:17 PM
Jskid
The area under the normal distribution curve for 0.25% is z=-1.96 and for 97.25% is z=1.96. The answer to my question is 1.96. So why do I need $Z=\displaystyle\huge\frac{sample\ mean-\mu}{\frac{\sigma}{\sqrt{36}}}$?
• August 10th 2010, 03:38 AM
The area under the normal distribution curve for 0.25% is z=-1.96 and for 97.25% is z=1.96. The answer to my question is 1.96. So why do I need $Z=\displaystyle\huge\frac{sample\ mean-\mu}{\frac{\sigma}{\sqrt{36}}}$?