Due Tue 07/17/2018 11:59 pm The graph illustrates the distribution of test score

Question

Due Tue 07/17/2018 11:59 pm The graph illustrates the distribution of test scores taken by College Algebra students. The maximum possible score on the test was 130, while the mean score was 75 and the standard deviation was 12 39 51 63 75 87 11 Distribution of Test Scores What is the approximate percentage of students who scored between 39 and 75 on the test? What is the approximate percentage of students who scored between 51 and 99 on the test? What is the approximate percentage students who scored between 63 and $7 on the test? What is the approximate percentage of studests who scored between 75 and 87 on the test? Get help Video Video Paints This is attempt 1 of 1 o a D LL 5 6 8

Explanation / Answer

Question 1

We have to find P(39<X<75)

P(39<X<75) = P(X<75) – P(X<39)

For X=75

Z = (75 – 75)/12 = 0

P(Z<0) = P(X<75) = 0.50

For X=39

Z = (39 – 75) / 12 = -3

P(Z<-3) = P(X<39) = 0.00135

P(39<X<75) = P(X<75) – P(X<39)

P(39<X<75) = 0.500000 - 0.00135

P(39<X<75) = 0.49865

Required percentage = 49.87%

Question 2

We have to find P(51<X<99)

P(51<X<99) = P(X<99) – P(X<51)

For X=99

Z = (99 – 75)/12 = 2

P(Z<2) = P(X<99) = 0.97725

For X=51

Z = (51 – 75) / 12 = -2

P(Z<-2) = P(X<51) = 0.02275

P(51<X<99) = 0.97725 - 0.02275 = 0.9545

Required percentage = 95.45%

Question 3

We have to find P(63<X<87)

P(63<X<87) = P(X<87) – P(X<63)

Z for X=87

Z = (87 – 75) / 12 = 1

P(Z<1) = P(X<87) = 0.841345

Z for X = 63

Z = (63 – 75) / 12 = -1

P(Z<-1) = P(X<63) = 0.158655

P(63<X<87) = 0.841345 - 0.158655 = 0.682689

Required percentage = 68.27%

Question 4

We have to find P(75<X<87)

P(75<X<87) = P(X<87) – P(X<75)

Z for X=87

Z for X=87

Z = (87 – 75) / 12 = 1

P(Z<1) = P(X<87) = 0.841345

Z for X=75

Z = (75 – 75)/12 = 0

P(Z<0) = P(X<75) = 0.50

P(75<X<87) = 0.841345 – 0.50000 = 0.341345

Required percentage = 34.13%

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