Question

Replacement times for Timex watches are normally distributed with a mean of 10.2 years and a standard deviation of 3.1 years (Based on data from “Getting Things Fixed”, Consumer Reports).

a. What percentage of watches should last between 8 and 9 years?

b. At least how many years should the longest 25% of all watches last?

c. If Timex wants to provide a warranty so that only 1% of the watches will be replaced before the warranty expires, what is the time length of the warranty?

d. Suppose a certain jeweler has received a carton containing 48 Timex watches. Let y represent the mean lifetime of these watches. What is the distribution of this sample mean, y ? Specify the model, mean, standard deviation, and justification for your answers.

e. What is the probability that the mean lifetime of this carton of watches will be 10 years?

Answer 1

Let X years be the replacement times for any given Timex watch. X is normally distributed with mean and standard deviation

a. What percentage of watches should last between 8 and 9 years?

the probability that any given watch would last between 8 and 9 years

ans: The percentage of watches that would last between 8 and 9 years is 10.94%

b. At least how many years should the longest 25% of all watches last?

Let 25% of the watches last more than q years. This is same as the probability that any given watch lasts more than q years is 0.25

P(X>q)=0.25

In terms of z scores, we need

Using the standard normal tables, we get for z=0.67, P(Z<0.67) = 0.75

That is

P(Z>0.67)=0.25

We need

P(X>q)=0.25

We can equate the z score of q to 0.67 and get

ans: The longest 25% of all watches should last at least 12.28 years

c. If Timex wants to provide a warranty so that only 1% of the watches will be replaced before the warranty expires, what is the time length of the warranty?

Let 1% of the watches need replacement before q years. This is same as the probability that any given watch needs replacement before q years is 0.01.

That is

P(X<q)=0.01

In terms of the z scores, we need

We know that the area to the left of mean (which is 0) of a standard normal curve is 0.5, that is P(Z<0)=0.5. Here, we need a probability which is less than 0.5, hence the z value must be negative.

That is, we need

Using the standard normal tables, for z=2.33, we get P(Z<2.33)=0.99

That is

P(Z<-2.33)=0.01

We need

P(X<q)=0.01

We can equate the z score of q to -2.33 and get

ans: If Timex wants to provide a warranty so that only 1% of the watches will be replaced before the warranty expires, the time length of the warranty needs to be 2.98 years

d. Suppose a certain jeweler has received a carton containing 48 Timex watches. Let y represent the mean lifetime of these watches. What is the distribution of this sample mean, y ? Specify the model, mean, standard deviation, and justification for your answers.

Let Y represent the mean lifetime of n=48 watches. Since the sample size is greater than 30, using the central limit theorem, we know that the distribution of sample mean is normal with mean and standard deviation (also called standard error of mean)

ans: Since the sample size n=48 is greater than 30, using the central limit theorem, we can say that the distribution of this sample mean, Y is normal with mean 10.2 years and standard deviation 0.4474 years.

e. What is the probability that the mean lifetime of this carton of watches will be 10 years?

Note: The point probability of a continuous random variable is always 0. Hence the probability that the mean lifetime of this carton of watches will be equal to 10 years is 0.

the probability that the mean lifetime of this carton of watches will be at least 10 years is

ans: the probability that the mean lifetime of this carton of watches will be at least 10 years is 0.6736 and the the probability that the mean lifetime of this carton of watches will be at most 10 years is 0.3264.  The probability that the mean lifetime of this carton of watches will be equal to 10 years is 0.