Question

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

Answer 1

Answer:

0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 800, \sigma = 40, n = 16, s = \frac{40}{\sqrt{16}} = 10$

Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

This probability is the pvalue of Z when $X = 775$. So

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

$Z = \frac{775 - 800}{10}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062. So there is a 0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.