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

Answer:

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

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

Step-by-step 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,

$P(Z\leq \frac{x-\mu}{\sigma})\times 100$

$=P(Z\leq \frac{300-510}{100})\times 100$

$=P(Z\leq \frac{-210}{100})\times 100$

$=P(Z\leq-2.1)\times 100$

$=0.0179\times 100$

$=1.79\%$

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, $P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90$

$P(Z< \frac{t-\mu}{\sigma})=0.90$

$\frac{t-\mu}{\sigma}=1.28$

$\frac{t-510}{100}=1.28$

$t-510=128$

$t=128+510$

$t=638$

Score puts someone in the 90th percentile is 638.