Question

The scores for a particular examination are normally distributed with a mean of 68.5% and a standard deviation of 8.2%. What is the probability that a student who wrote the examination had a mark between 80% and 100%? Give your answer to the nearest hundredth.

Answer 1

`P(80/100<x<100/100)=0.08`

Explanation:

We are given that

Mean,
`\mu=68.5`%=68.5/100

Standard deviation,
`\sigma=8.2`%=8.2/100

We have to find the probability that a student who wrote the examination had a mark between 80% and 100%.

`P(80/100<x<100/100)=P((80/100-68.5/100)/(8.2/100)<(x-\mu)/(\sigma)<(100/100-68.5/100)/(8.2/100))`

`P(80/100<x<100/100)=P(1.40<Z<3.84)`

We know that

`P(a<Z<b)=P(Z<b)-P(Z<a)`

Using the formula

`P(80/100<x<100/100)=P(Z<3.84)-P(Z<1.40)`

`P(80/100<x<100/100)=0.99994-0.91924`

`P(80/100<x<100/100)=0.0807\approx 0.08`