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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 18 / Problem 32E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: The ideas of average and root-mean-square value

Chapter 18, Problem 32E

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QUESTION:

Problem 32E

The ideas of average and root-mean-square value can be applied to any distribution. A class of 150 students had the following scores on a 100-point quiz:

Score	Number of Students
10	11
20	12
30	24
40	15
50	19
60	10
70	12
80	20
90	17
100	10

(a) Find the average score for the class. (b) Find the root-means square score for the class.

Questions & Answers

QUESTION:

Problem 32E

The ideas of average and root-mean-square value can be applied to any distribution. A class of 150 students had the following scores on a 100-point quiz:

Score	Number of Students
10	11
20	12
30	24
40	15
50	19
60	10
70	12
80	20
90	17
100	10

(a) Find the average score for the class. (b) Find the root-means square score for the class.

ANSWER:

Solution 32E

Step 1:

The average score will be,

Score( ${X}_{i}$ )	Number of Students( ${F}_{i}$ )	${F}_{i}{X}_{i}$
10	11	110
20	12	240
30	24	720
40	15	600
50	19	950
60	10	600
70	12	840
80	20	1600
90	17	1530
100	10	1000
$\sum\limits_{i}^{\ }{X}_{i}=$	$\sum\limits_{i}^{\ }{F}_{i}=150$	$\sum\limits_{i}^{\ }{F}_{i}{X}_{i}=8190$

We know that, the average is,

$average=\frac{\sum\limits_{i}^{\ }{F}_{i}{X}_{i}}{\sum\limits_{i}^{\ }{F}_{i}}=\frac{8190}{150}=54.6$

So, the average score of the class is 54.6.

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