"A manufacturer of automobile batteries claims that the average length of life for its grade A battery is 55 months. Suppose the standard deviation of the life length is known to be 3 months, and the frequency distribution of the life length data is known to be mound shaped. Approximately, what percentage of the manufacturer`s grade A batteries will last more than 49 months?"

or P(x>49) is about 97.725% (or being less precise 97.5% using the empirical rule).

Step-by-step explanation:

We solve this question using the following information:

Having all this information, we can solve the question.

First Step: Use formula [1] to find the z-score of the raw score x = 49 months.

This means that the raw score is represented by a z-score of , which tells us that it is two standard deviations below the population mean.

Second Step: Consult this value in the cumulative standard normal table for z = 2 and apply the formula [2] to find the corresponding probability.

For a z = 2, the probability is 0.97725.  

Then

But we are not asked for P(z<-2) but for P(z>-2) = P(x>49). This probability is the complement of the previous result, that is

That is, the "percentage of the manufacturer's grade A batteries will last more than 49 months" is

or about 97.725%

A graph below shows this result.

Notice that if we had used the 68-95-99.7 rule (also known as the empirical rule), that is, in a normal distribution, the interval between one standard deviation below and above the mean contains, approximately, 68% of the observations; the interval between two standard deviations below and above the mean contains, approximately, 95% of the observations; and the interval between three standard deviations below and above the mean contains, approximately, 99.7% of the observations, we could have concluded that 2.5 % of the manufacturer's grade A batteries will last less than 49 months, and, as a result, 1 - 0.025 = 0.975 or 97.5% will last more than 49 months.

We can conclude that with a less precise answer (but faster) because of the symmetry of the normal distribution, that is, 1 - 0.95 = 0.05. At both extremes we have 0.05/2 = 0.025 or 2.5% and we were asked for P(x>49) = P(z>-2) (see the graph below).

or