Question

A population has a mean of 53 and a standard deviation of 21. A sample of 49 observations will be taken. The probability that the sample mean will be greater than 57.95 is Select one: a. .4505 b. .0495

Answer 1

Answer:

b. .0495

Step-by-step explanation:

To solve this question, we need to know the concepts of the normal probability distribution and of 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 $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

In this problem, we have that:

$\mu = 53, \sigma = 21, n = 49, s = \frac{21}{\sqrt{49}} = 3$

The probability that the sample mean will be greater than 57.95

This is 1 subtracted by the pvalue of Z when X = 57.95. So

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

By the Central Limit Theorem

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

$Z = \frac{57.95 - 53}{3}$

$Z = 1.65$

$Z = 1.65$ has a pvalue of 0.9505.

So the probability that the sample mean will be greater than 57.95 is 1-0.9505 = 0.0495

So the correct answer is:

b. .0495