Mathematics, 15.04.2020 00:44, xxkeyxx51

"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?"

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answered: gabriellestaleyga16

$\\ P(z-2) = 0.97725$ 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:

We are dealing here with normally distributed data, that is "the frequency distribution of the life length data is known to be mound-shaped".The normal distribution is defined by two parameters: the population mean ( $\\ \mu$ ) and the population standard deviation ( $\\ \sigma$ ). In this case, we have that $\\ \mu = 55$ months, and $\\ \sigma = 3$ months.To find the probabilities, we have to use the standard normal distribution, which has $\\ \mu = 0$ and $\\ \sigma = 1$ . The probabilities for this distribution are collected in the standard normal table, available in Statistics books or on the Internet. We can also use statistics programs to find these probabilities.For most cases, we need to use the cumulative standard normal table, and for this we have to previously "transform" a raw score (x) into a z-score using the next formula: $\\ z = \frac{x - \mu}{\sigma}$ [1]. A z-score tells us the distance from the mean that a raw score is from it in standard deviations units. If this value is negative, the raw score is below the mean. Conversely, a positive value indicates that it is above the mean.The cumulative standard normal table is made for positive values of z. Since the normal distribution is symmetrical around the mean, we can find the negative values of z using this formula: $\\ P(z$ [2].

Having all this information, we can solve the question.

The percentage of the manufacturer's grade A batteries that will last more than 49 months

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

$\\ z = \frac{49 - 55}{3}$

$\\ z = \frac{-6}{3}$

$\\ z = -2$

This means that the raw score is represented by a z-score of $\\ z = -2$ , 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

$\\ P(z$

$\\ P(z2)$

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

$\\ P(z-2) = 1 - P(z$

$\\ P(z-2) = 1 - 0.02275$

$\\ P(z-2) = 0.97725$

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

$\\ P(z-2) = 0.97725$ 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).