Question

The mean quantitative score on a standardized test for female​ college-bound high school seniors was 550...

The mean quantitative score on a standardized test for female​ college-bound high school seniors was

550

The scores are approximately Normally distributed with a population standard deviation of

50

A scholarship committee wants to give awards to​ college-bound women who score at the

96TH

percentile or above on the test. What score does an applicant​ need? Complete parts​ (a) through​ (g) below.The mean quantitative score on a standardized test for female​ college-bound high school seniors was

550

The scores are approximately Normally distributed with a population standard deviation of

50

A scholarship committee wants to give awards to​ college-bound women who score at the

96

percentile or above on the test. What score does an applicant​ need? Complete parts​ (a) through​ (g) below.

A.

The 96th percentile has 96​% of the area to the left because it is higher than 96​% of the scores. The table above gives the areas to the left of​ z-scores. Therefore, we look for 0.9600 in the interior part of the table. Use the Normal table given above to locate the area closest to 0.9600. Then report the​ z-score for that area.

Homework Answers

Answer #1

Solution:-

Given that,

mean = = 550

standard deviation = = 50

Using standard normal table,

P(Z > z) = 96%

= 1 - P(Z < z) = 0.96  

= P(Z < z) = 1 - 0.96

= P(Z < z ) = 0.04

= P(Z < -1.75 ) = 0.04  

z = -1.75

Using z-score formula,

x = z * +

x = -1.75 * 50 + 550

x = 462.5

test score = 463

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