Question

A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100. Using the eight-part symmetry of the area under a normal curve, what is the probability that a randomly chosen exam score is above 300?

The probability is___

Answer 1

The probability is 0.977

Explanation:

We know that the average
`\mu` is:

`\mu=500`

The standard deviation
`\sigma` is:

`\sigma=100`

The Z-score is:

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

We seek to find

`P(x>300)`

The Z-score is:

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

`Z=(300-500)/(100)`

`Z=-2`

The score of Z = -2 means that 300 is -2 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 1 deviations from the mean has percentage of 2.35% for Z<-2

So

`P(z>-2)=1-P(Z<-2)`

`P(z>-2)=1-0.0235`

`P(z>-2)=0.9765`