Q R2.6. Acing the GRE The Graduate Recor... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 2: Q R2.6. (page 148)

Acing the GRE The Graduate Record Examinations (GREs) are widely used to help predict the performance of applicants to graduate schools. The scores on the GRE Chemistry test are approximately Normal with a mean = of $694$ and a standard deviation $= 112$
a. Approximately what percent of test-takers earn a score less than 500 or greater than 900 on the GRE Chemistry test?
b. Estimate the $99 t h$ percentile score on the GRE Chemistry Test.

Short Answer

Expert verified

Part (a) The percentage is $7.47 %$

Part (b) The percentile is $954.96$

Step by step solution

Part (a) Step 1: Given information

Population mean $(渭) = 694$

Standard deviation $(蟽) = 112$

Part (a) Step 2: Calculation

The random variable $X$ displays the earning score according to the normal distribution.

The percentage of tests with scores of less than $500$ or more than $900$ can be computed as follows:

$P (X < 500 o r X > 900) = P (\frac{x 鈭� 渭}{蟽} < \frac{500 鈭� 渭}{蟽}) + P (\frac{x 鈭� 渭}{蟽} > \frac{900 鈭� 渭}{蟽}) = P (Z < \frac{500 鈭� 694}{112}) + P (Z > \frac{900 鈭� 694}{112}) = P (Z < 鈭� 1.73) + P (Z > 1.84) = 0.0747$

Thus, the required percentage is $7.47 %$

Part (b) Step 1: Calculation

The $z -$ score for the probability of $0.99$ is $2.33$ using the usual normal table.

The test's $99 t h$ percentile can be determined as follows:

$x = 渭 + Z 脳蟽 = 694 + (2.33) 脳 112 = 954.96$

Therefore, the required percentile is $954.96$

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n	Mean	SD	Min	Q1	Med	Q3	Max
50	51742.44	8210.64	36641	46071	50009	57020	71836

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Short Answer

Step by step solution

Part (a) Step 1: Given information

Part (a) Step 2: Calculation

Part (b) Step 1: Calculation

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