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=(x-\mu)/(\sigma)`

Replacing the values given:

`Z=(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%