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:

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 ,

Standard deviation

Sample mean

Percentage of students scored below 300 is given by,

Percentage of students scored below 300 is 1.79%.

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

90th percentile is such that,

Now,

Score puts someone in the 90th percentile is 638.

the first one

step-by-step explanation:

just id the test.

5+2x=a

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