Question

Assuming that the laptop replacement times have a mean of 3.7 years and a standard deviation of 0.4 years, find the probability that 38 randomly selected laptops will have a mean replacement time of 3.5 years or less

Answer 1

Answer:

0.10% probability that 38 randomly selected laptops will have a mean replacement time of 3.5 years or less

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 3.7, \sigma = 0.4, n = 38, s = \frac{0.4}{\sqrt{38}} = 0.0649$

Find the probability that 38 randomly selected laptops will have a mean replacement time of 3.5 years or less

This is the pvalue of Z when X = 3.5. So

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

By the Central Limit Theorem

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

$Z = \fra{3.5 - 3.7}{0.0649}$

$Z = -3.08$

$Z = -3.08$ has a pvalue of 0.0010

0.10% probability that 38 randomly selected laptops will have a mean replacement time of 3.5 years or less