Question

Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve. (a) If the average test score is 510 with a standard deviation of 100 points, what percentage of students scored below 300? Enter as a percentage to the nearest tenth of a percent. % (b) What score puts someone in the 90th percentile? Start by finding z such that P(Z < z) = 0.90, then find what test score has that z value. Round to the nearest whole score:

Answer 1

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean
`\mu=510`,

Standard deviation
`\sigma=100`

Sample mean
`x=300`

Percentage of students scored below 300 is given by,

`P(Z\leq (x-\mu)/(\sigma))* 100`

`=P(Z\leq (300-510)/(100))* 100`

`=P(Z\leq (-210)/(100))* 100`

`=P(Z\leq-2.1)* 100`

`=0.0179* 100`

`=1.79\%`

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

`P(x\leq t)=0.90`

Now,
`P((x-\mu)/(\sigma) < (t-\mu)/(\sigma))=0.90`

`P(Z< (t-\mu)/(\sigma))=0.90`

`(t-\mu)/(\sigma)=1.28`

`(t-510)/(100)=1.28`

`t-510=128`

`t=128+510`

`t=638`

Score puts someone in the 90th percentile is 638.