Answer:
a) Figure attached
b) [tex]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)[/tex]
And we can find this probability with the normal standard table and with the complement rule:
[tex]P(z\geq 1.5)=1-P(z<1.5)=1-0.933=0.0668 [/tex]
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
[tex]P(X\ geq 90)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]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)[/tex]
And we can find this probability with the normal standard table and with the complement rule:
[tex]P(z\geq 1.5)=1-P(z<1.5)=1-0.933=0.0668 [/tex]
