The following are the means and standard deviations of some well-known standardized tests, referred to as Test A, Test B, and Test C. All three yield normal distributions.TestMeanStandard DeviationTest A500100Test B1508Test C7015A score of 450 on Test A corresponds to what score on Test B? A score of 625 on Test A corresponds to what score on Test C?

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Description: The following are the means and standard deviations of some well-known standardized tests, referred to as Test A, Test B, and Test C. All three yield normal distributions.TestMeanStandard DeviationTest A500100Test B1508Test C7015A score of 450 on Tes

1. Given that Test Mean Standard Deviation A 500 100 B 150 8 C 70 15 A score of 450…

Answered: The following are the means and…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

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ISBN:9780079039897

Author:Carter

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Chapter10: Statistics

Section10.4: Distributions Of Data

Problem 19PFA

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The following are the means and standard deviations of some well-known standardized tests, referred to as Test A, Test B, and Test C. All three yield normal distributions.

Test	Mean	Standard Deviation
Test A	500	100
Test B	150	8
Test C	70	15

A score of 450 on Test A corresponds to what score on Test B?
A score of 625 on Test A corresponds to what score on Test C?

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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Step 1

Given that

Test	Mean	Standard Deviation
A	500	100
B	150	8
C	70	15

A score of 450 on Test A:

$z s c o r e o f A = z s c o r e o f B \frac{s c o r e - m e a n}{s \tan d a r d d e v i a t i o n} = \frac{s c o r e - m e a n}{s \tan d a r d d e v i a t i o n} \frac{450 - 500}{100} = \frac{s c o r e - 150}{8} - 0.5 = \frac{s c o r e - 150}{8} (- 0.5 * 8) + 150 = S c o r e o f B S c o r e o f B = 146$

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