Question

The heights of men in the USA are normally distributed with a mean of 68 inches and a standard deviation of 4 inches. A random sample of five men is selected. What is the probability that the sample mean is greater than 69.5 inches?

Answer 1

Answer:

$P(\bar X >69.5) =P(Z \frac{69.5-68}{\frac{4}{\sqrt{5}}}) = P(Z>0.839)$

And we can use the complement rule with the normal standard table or excel and we got:

$P(Z>0.839) =1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(68,4)$  

Where $\mu=68$ and $\sigma=4$

We select a sample of size n =5. Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We want this probability:

$P(\bar X >69.5)$

And we can use the z score formula given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got:

$P(\bar X >69.5) =P(Z \frac{69.5-68}{\frac{4}{\sqrt{5}}}) = P(Z>0.839)$

And we can use the complement rule with the normal standard table or excel and we got:

$P(Z>0.839) =1-P(z$

Answer 2

Answer:

20.05% probability that the sample mean is greater than 69.5 inches

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 problem, we have that:

$\mu = 68, \sigma = 4, n = 5, s = \frac{4}{\sqrt{5}} = 1.79$

What is the probability that the sample mean is greater than 69.5 inches?

This is 1 subtracted by the pvalue of Z when X = 69.5. So

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

By the Central Limit Theorem

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

$Z = \frac{69.5 - 68}{1.79}$

$Z = 0.84$

$Z = 0.84$ has a pvalue of 0.7995

1 - 0.7995 = 0.2005

20.05% probability that the sample mean is greater than 69.5 inches