Question

If you take a sample of 9 pacemakers from the company, what is the probability that the mean lifetime of components in this sample will be longer than 99 months

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.9878 = 98.78% probability that the mean lifetime of components in this sample will be longer than 99 months.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• The mean is of 9 years, hence, in months, $\mu = 9(12) = 108$

• The standard deviation is of 1 year, hence, in months, $\sigma = 12$.

• Sample of 9 pacemakers, hence $n = 9, s = \frac{12}{\sqrt{9}} = 4$.

The probability is 1 subtracted by the p-value of Z when X = 99, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{99 - 108}{4}$

$Z = -2.25$

$Z = -2.25$ has a p-value of 0.0122.

1 - 0.0122 = 0.9878.

0.9878 = 98.78% probability that the mean lifetime of components in this sample will be longer than 99 months.

A similar problem is given at https://brainly.com/question/24663213