• Sep 8th 2009, 04:39 AM
underachiever
I have been given this question for a tut but am unsure of how to answer it as it doesnt seem to have enough information for me, any help is greatly appreciated:

A company that markets user assembled furniture sells a computer desk that is advertised with the claim "less than an hour to assemble". However, through postpurchase surveys the company has learned that only 25% of their customers succeeded in building the desk in under an hour; 5% said it took them over 2 hours.

(a)Using this information and assuming that the desk assembly time follows the normal probability model, show that the mean and the standard deviation for the assembly time are 77.45 and 25.87 minutes respectively.

Probability is not a strong point for me
thanks
• Sep 8th 2009, 04:49 AM
mr fantastic
Quote:

Originally Posted by underachiever
I have been given this question for a tut but am unsure of how to answer it as it doesnt seem to have enough information for me, any help is greatly appreciated:

A company that markets user assembled furniture sells a computer desk that is advertised with the claim "less than an hour to assemble". However, through postpurchase surveys the company has learned that only 25% of their customers succeeded in building the desk in under an hour; 5% said it took them over 2 hours.

(a)Using this information and assuming that the desk assembly time follows the normal probability model, show that the mean and the standard deviation for the assembly time are 77.45 and 25.87 minutes respectively.

Probability is not a strong point for me
thanks

Given: Pr(X < 1) = 0.25 and Pr(X > 2) = 0.05.

You have at least two options:

Option 1: Calculate Pr(X < 1) and Pr(X > 2) using the given mean and standard deviation and obtain 0.25 and 0.05.

Option 2: Calculate the mean and standard deviation directly from the given information. To do this, first calculate (using tables, technology or however you've been taught how to do it) the Z-values a and b such that Pr(Z < a) = 0.25 and Pr(Z > 2) = 0.05. Then:

$\displaystyle a = \frac{1 - \mu}{\sigma}$ ... (1)

$\displaystyle b = \frac{2 - \mu}{\sigma}$ ... (2)

Substitute the values of a and b and solve equations (1) and (2) simultaneously for the mean and the standard deviation.
• Sep 8th 2009, 05:32 AM
underachiever
Can anyone elaborate on option 1? I have checked both of my intro to stats textbooks and cant seem to find it anywhere
Thanks
• Sep 8th 2009, 05:36 AM
mr fantastic
Quote:

Originally Posted by underachiever
Can anyone elaborate on option 1? I have checked both of my intro to stats textbooks and cant seem to find it anywhere
Thanks

Have you been taught how to calculate probabilities using a normal distribution? That is what option 1 requires. It is certain to be in your textbook - probably in a Chapter called The Normal Distribution.
• Sep 8th 2009, 12:51 PM
underachiever
Thank you for your help I will look again, I am sorry my uni seem to be skimming things due to time constraints
• Sep 10th 2009, 06:04 PM
kristen11087
help!
i also need her help with this question, ive gone over my notes and textbook and listened to lectures again. I understand how to work this out with a probability calculator but I have no idea how to prove this question in hand written working out.
• Sep 11th 2009, 03:47 AM
mr fantastic
Quote:

Originally Posted by kristen11087
i also need her help with this question, ive gone over my notes and textbook and listened to lectures again. I understand how to work this out with a probability calculator but I have no idea how to prove this question in hand written working out.

You will find many threads in the Probablity and Statistics subforum that explain this in great detail.