Question

Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give two exams, one to a class of size 25 and the other to a class of size 64. Approximate the probability that the average test score in the class of size 25 exceeds 80.

Answer 1

Answer:

$P(\bar X >80)=P(Z>2.143)=1-P(z$

Step-by-step explanation:

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".

Let X the random variable that represent the Student scores on exams given by a certain instructor, we know that X have the following distribution:

$X \sim N(\mu=74, \sigma=14)$

The sampling distribution for the sample mean is given by:

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

The deduction is explained below we have this:

$E(\bar X)= E(\sum_{i=1}^{n}\frac{x_i}{n})= \sum_{i=1}^n \frac{E(x_i)}{n}= \frac{n\mu}{n}=\mu$

$Var(\bar X)=Var(\sum_{i=1}^{n}\frac{x_i}{n})= \frac{1}{n^2}\sum_{i=1}^n Var(x_i)$

Since the variance for each individual observation is $Var(x_i)=\sigma^2$ then:

$Var(\bar X)=\frac{n \sigma^2}{n^2}=\frac{\sigma}{n}$

And then for this special case:

$\bar X \sim N(74,\frac{14}{\sqrt{25}}=2.8)$

We are interested on this probability:

$P(\bar X >80)$

And we have already found the probability distribution for the sample mean on part a. So on this case we can use the z score formula given by:

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

Applying this we have the following result:

$P(\bar X >80)=P(Z>\frac{80-74}{\frac{14}{\sqrt{25}}})=P(Z>2.143)$

And using the normal standard distribution, Excel or a calculator we find this:

$P(Z>2.143)=1-P(z$