The Graduate Record Examination (GRE) is a test required for admission to many U

Question

The Graduate Record Examination (GRE) is a test required for admission to many US graduate schools. Students’ scores on the quantitative portion of the GRE follow a normal distribution with a mean of 150 and a standard deviation of 8.8. In addition to other qualifications, a score of at least 152 is required for admission to a particular graduate school.

GRE scores continued:

Using either by hand calculations or technology, answer the following. There may be some small variation between by hand and technology calculations. (Hint: by hand calculations need to convert statement of x to statement of z).

What proportion of combined GRE scores can be expected to be over 160? (hint: in Excel use NORM.DIST function)

What proportion of combined GRE scores can be expected to be under 160? (hint: in Excel use NORM.DIST function)

What proportion of combined GRE scores can be expected to be between 155 and 160? (hint: in Excel use NORM.DIST function)

What is the probability that a randomly selected student will score over 145 points? (hint: in Excel use NORM.DIST function)

What is the probability that a randomly selected student will score less than 150 points? (hint: in Excel use NORM.DIST function)

Explanation / Answer

mena = 150 , s = 8.8

a)
P(X > 160)

z = ( x -mean) /s
= (160 - 150) / 8.8
= 1.136

P(X > 160) = P(z > 1.136) = 0.1279
By using standard table

b)
P(X < 160)

z = ( x -mean) /s
= (160 - 150) / 8.8
= 1.136

P(X < 160) = P(z < 1.136) = 0.8721
By using standard table

c) P(155 < X < 160)

= P[(x - mean) / s < z < ( x -mean) /s]
= P[(155 - 150) / 8.8 < z < (160 - 150) / 8.8]
= P(0.568 < z < 1.136)
P(155 < X < 160) = P(0.568 < z < 1.136) = 0.1571

d)
P(X > 145)

z = ( x -mean) /s
= (145 - 150) / 8.8
= -0.568

P(X > 145) = P(z >-0.568) = 0.715
By using standard table

e)
P(X < 150)

z = ( x -mean) /s
= (150 - 150) / 8.8
= 0

P(X < 150) = P(z < 0) = 0.5
By using standard table

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