# If the life expectancy is 75 with a standard deviation of 7 years, with a random sample of 49...

## Question:

If the life expectancy is 75 with a standard deviation of 7 years, with a random sample of 49 individuals what is the probability that the sample mean will be larger than 77 years?

## Sample Mean:

In statistics, while we may be given an overall average of a value for a population, we may want to determine the probability of that value in smaller samples. This value would be known as the sample mean.

There is a 2.28% chance that the sample mean will be larger than 77 years.

In written form, we are looking for the probability:

{eq}P(\bar{X} > 77) {/eq}

Let's determine the z value for this sample probability. Considering the data to follow a normal distribution, the equation for this z value is:

{eq}z = \dfrac{\bar{x} - \mu_x}{\sigma / \sqrt{N}} {/eq}, where:

• {eq}\bar{x} {/eq} is the value we're comparing the sample mean to
• {eq}\mu_x {/eq} is the expected mean given
• {eq}\sigma {/eq} is the standard deviation
• N is the sample size

Let's plug in and solve for z:

{eq}z = \dfrac{\bar{x} - \mu_x}{\sigma / \sqrt{N}}\\ z = \dfrac{77 - 75}{7 / \sqrt{49}}\\ z = 2 {/eq}

So now we're looking for {eq}P(Z > 2) {/eq}. If we consult a standard Z chart, we'll see that the probability of P(Z < 2) is 0.9772. However, because we want P(Z > 2), we'll subtract this value by 1:

{eq}P(\bar{x} > 77) = P(Z > 2) = 1- P(Z < 2) = 1- 0.9772\\ \boxed{P(\bar{x} > 77) = 0.0228} {/eq}