Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 10.9 years and a standard deviation of 1.5 years.

Question

Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 10.9 years and a standard deviation of 1.5 years.

Find the probability that a randomly selected quartz time piece will have a replacement time less than 6.3 years?

P(X < 6.3 years) = 

Enter your answer accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

If the company wants to provide a warranty so that only 3.9% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty?

warranty =  years

Answer 1

Hi Bella,

This problem uses the classic equation in introductory statistics: z=x-mu/sigma where:

z=z-score (explained below)

x=Data point given, 6.3 years in this case

mu=average or mean

sigma=standard deviation

You will also see this equation given as z=x-x-bar/s. This is essentially the same, but it is for sample, not population, data. Anyway, assuming you do not have statistical software, step 1 is to find z:

z=(6.3-10.9)/1.5

z=-3.067

After you find z, navigate to a z-score table or use one provided by your instructor. A good one can be found at ztable.net, which is what I use for the rest of this problem. Now, navigate on the table column to -3, then on the table row to either 0.06 or 0.07. You will arrive at a value of either 0.00111 or 0.00107. Your problem specified four decimal places, so both are recorded as 0.0011.

P(x<6.3 yrs.)=0.0011

This is extremely low, and if you think about it, this makes sense because the company would not want to replace the quartz that quickly. It would not be cost-effective.

The second part of the question is essentially the same type of problem in reverse. You are given 3.9 percent, which, expressed as a decimal probability, is 0.039. Search the interior of the z-table mentioned above for values close to this. You will get a vertical value of about -1.7 and horizontal value between 0.06 and 0.07. Add these together and you get a z-score between -1.76 and -1.77, which I’ll call -1.765. So:

z= -1.765

mu=10.9

sigma=1.5

Then, apply that classic equation z=x-mu/sigma:

-1.765=(x-10.9)/1.5

Solve for x; multiply both sides by 1.5

-2.6475=x-10.9

x=8.2525 years, about 8 years, 3 months

Answers may vary based on differences in z-tables and statistical software. I hope this helps you.