Question

Consider a normal distribution of values with a mean of 32 and a standard deviation of 1.5. Find the probability that a value is less than 36.8.99.93%99.87%98.43%93.73%

Answer 1

Finding Z:

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

Replacing the values given:

$Z=\frac{36.8-32}{1.5}=3.2$

Then...

$P(X<36.8)=P(Z<3.2)$

Using the Standard Normal Table, we can see that:

$P(Z<3.2)=0.9993$

Answer: 99.93%