Question

If the average ago of retirement for a random sample of 87 retired persons is 66 years with a standard deviation of 2 years.a. b. c. d.e.f.g. Find the probability of finding a random sample of 87 retired people in which the average age of retirement is 66.5 or more. Find all values to 3 decimal places.

Answer 1

Answer:

0.01 = 1% probability of finding a random sample of 87 retired people in which the average age of retirement is 66.5 or more.

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 normally distributed samples are 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 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.

In this question, we have that:

$\mu = 66, \sigma = 2, n = 87, s = \frac{2}{\sqrt{87}} = 0.214$

Find the probability of finding a random sample of 87 retired people in which the average age of retirement is 66.5 or more.

This probability is 1 subtracted by the pvalue of Z when X = 66.5. So

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

By the Central Limit Theorem

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

$Z = \frac{66.5 - 66}{0.214}$

$Z = 2.336$

$Z = 2.336$ has a pvalue of 0.99.

1 - 0.99 = 0.01

0.01 = 1% probability of finding a random sample of 87 retired people in which the average age of retirement is 66.5 or more.