Question

Suppose that the scores on a test have a normal distribution with mean 24 and standard deviation 4. What is the proportion of scores less than 28?

Answer 1

Answer: 0.8413

Step-by-step explanation:

Given : The scores on a test have a normal distribution with mean 24 and standard deviation 4.

i.e. $\mu= 24$  and $\sigma= 4$

Let x denotes the scores on the test.

Then, the probability that a student score less than 28 will be :-

$P(x<28)=P(\dfrac{x-\mu}{\sigma}<\dfrac{28-24}{4})\\\\=P(z<1)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=0.8413 \ \ [\text{By z-table}]$

Hence, the the proportion of scores less than 28 is 0.8413 .