Question

A statistics instructor designed an exam so that the grades would be roughly normally distributed with a mean of and a standard deviation of 1015. In the space provided, draw a sketch of this distribution and label the mean and, using the standard deviation, two additional points on the horizontal axis. (2 pts)

Answer 1

Answer:

a) Figure attached

b) $P(X\geq 90)=P(\frac{X-\mu}{\sigma}\geq \frac{90-\mu}{\sigma})=P(Z \geq\frac{90-75}{10})=P(z>1.5)$

And we can find this probability with the normal standard table and with the complement rule:

$P(z\geq 1.5)=1-P(z$

Step-by-step explanation:

A statistics instructor designed an exam so that the grades would be roughly normally distributed with a mean of μ=75 and a standard deviation of σ= 10.

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".  

Part a

Figure attached.

Part b

What proportion of students are expected to earn grades of ≥90?

We are interested on this probability

$P(X\ geq 90)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X\geq 90)=P(\frac{X-\mu}{\sigma}\geq \frac{90-\mu}{\sigma})=P(Z \geq\frac{90-75}{10})=P(z>1.5)$

And we can find this probability with the normal standard table and with the complement rule:

$P(z\geq 1.5)=1-P(z$