Question

In order to get promoted to the next grade, a student is required to score above 55 in the final test. Assume the scores to be normally distributed with a mean of 60 and a standard deviation of 5.5. Calculate the percentage of students who will be promoted to the next grade.

Answer 1

Answer:

$P(X>55)=P(\frac{X-\mu}{\sigma}>\frac{55-\mu}{\sigma})=P(Z>\frac{55-60}{5.5})=P(z>-0.909)$

And we can find this probability with the complement rule:

$P(z>-0.909)=1-P(z$

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

$P(z>-0.909)=1-P(z$

So then we expect about 81.8% of students that will be promoted

Step-by-step explanation:

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

$X \sim N(60,5.5)$  

Where $\mu=60$ and $\sigma=5.5$

We want to find the following probability:

$P(X>55)$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

Using the z score formula we got:

$P(X>55)=P(\frac{X-\mu}{\sigma}>\frac{55-\mu}{\sigma})=P(Z>\frac{55-60}{5.5})=P(z>-0.909)$

And we can find this probability with the complement rule:

$P(z>-0.909)=1-P(z$

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

$P(z>-0.909)=1-P(z$

So then we expect about 81.8% of students that will be promoted