As a part of the application process, most graduate programs require applicants

Question

As a part of the application process, most graduate programs require applicants to submit their Graduate Record Examinations (GRE) score (S). The GRE test is a standardized multiple choice test designed to measure the intellectual aptitude of a person. The Educational Testing Service (ETS) who administers the test reported that the average (scaled) score on the quantitative section of the test is 150 with a the appropriate notation. probabilities MUST be reported in 4 decimal digits. standard deviation of 6 points. They also claim that S follows closely a normal distribution. Use standardization approach!!

(a) What percentage of those who take the GRE test score at least 159 on the quantitative section?

(b) What is the probability that a randomly selected student will have a (scaled) GRE score on the quantitative section that exceeds 145 but is less than 163?

(c) Suppose that a program considers for admission only students whose score on the quantitative section places them among the top 4%. What is the minimum score one has to score in order to be considered for admission by the particular program?

Explanation / Answer

MEAN = 150

STANDARD DEVIATION = 6

AS THE DISTRIBUTION IS NORMAL

THE FORMULA TO BE USED

Z = (X-MEAN)/STANDARD DEVIATION

A) P(X>159) =

For x = 159, z = (159 - 150) / 6 = 1.5

Hence P(x > 159) = P(z > 1.5) = [total area] - [area to the left of 1.5]

1 - [area to the left of 1.5]

now from the z table we will take the value of z score = 1.5

    = 1 - 0.9332 = 0.0668 = 6.68%

B)P(145<X<163) =

) For x = 145 , z = (145 - 150) / 6 = -0.83 and for x = 163, z = (163 - 150) / 6 = 2.16

Hence P(145 < x < 163) = P(-0.83 < z < 2.16) = [area to the left of z = 2.16] - [area to the left of -0.83]

= 0.9846 - 0.2033 = 0.7813

C) HERE THE FORMULA TO BE USED

X = MEAN+ZSCORE* STANDARD DEVIATION

4% AMONG THE TOP MEAN 96%

THE ZSCORE FOR 96% = 2.05

THEREFORE

X = 150+2.05*6 = 162.3

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