Question

An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than $2,400.$

Short answer

The probability that the sum of the 50 values is more than 2400 is$P\left(\sum X>2400\right)=0.0040.$

Step-by-step solution

Given information

Given,

The value of Mean is $45$.

The value of Standard deviation is$8$

$n=50$

Formula used:

$\sum X~N\left(\left(n\right)\left({\mu }_x\right),\left(\sqrt{n}\right)\left({\sigma }_x\right)\right)$

Find the chance that the total of the 50 values is greater than 2400.

According to the information provided, a sample size of 50 people was chosen at random from a population with a mean of 45 and a standard deviation of 8. Use a Ti-83 calculator to compute the chance that the total of 50 values is larger than 2400. To do so, go to $2^{\text{nd}}$, then DISTR, and then scroll down to the normalcdf option and fill in the required information. After that, press the calculator's ENTER key to get the desired result. The screenshot is as follows:

As a result, the likelihood of the sum of 50 values being more than 2400:

$$\begin{gathered} =1-0.99599 \\ =0.0040 \end{gathered}$$