Question

If the mean number of years between asbestos exposure and death is found to be 25 years with a standard deviation of 7 years in a particular county, what is the probability that the mean number of years between asbestos exposure and death will be between 22 and 29 years in a sample of 50 people taken from that community

Answer 1

Answer:

0.9988 = 99.88% probability that the mean number of years between asbestos exposure and death will be between 22 and 29 years.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean number of years between asbestos exposure and death is found to be 25 years with a standard deviation of 7 years in a particular county

This means that $\mu = 25, \sigma = 7$

Sample of 50:

This means that $n = 50, s = \frac{7}{\sqrt{50}}$

What is the probability that the mean number of years between asbestos exposure and death will be between 22 and 29 years?

This is the p-value of Z when X = 29 subtracted by the p-value of Z when X = 22.

X = 29

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

By the Central Limit Theorem

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

$Z = \frac{29 - 25}{\frac{7}{\sqrt{50}}}$

$Z = 4.04$

$Z = 4.04$ has a p-value of 1.

X = 22

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

$Z = \frac{22 - 25}{\frac{7}{\sqrt{50}}}$

$Z = -3.03$

$Z = -3.03$ has a p-value of 0.0012.

1 - 0.0012 = 0.9988

0.9988 = 99.88% probability that the mean number of years between asbestos exposure and death will be between 22 and 29 years.