Explanation Given: Scores 90 − 99 80 − 89 70 − 79 60 − 69 50 − 59 Students 4 8 12 4 2 Approach: P ( X = x ) = N ( x ) N where N ( x ) = Number of favourable outcomes to x and N = Total number of outcomes . Let X denote the random variable that measures the scores of students in mathematics examination. Taking X to be the midpoint of the group interval, the probability distribution is as follows: x 94.5 84.5 74.5 64.5 54.5 P ( X = x ) 4 30 8 30 12 30 4 30 2 30 Table ( 1 ) Mean, μ = E [ X ] = ∑ x i P i …(1) where x i is score and P i is the respective probability. Standard Deviation, σ = ∑ P i ( x i − μ ) 2 …(2) Calculation: From the table (1) x 1 = 94

Finite Mathematics for the Managerial, Life, and Social Sciences-Custom Edition

11th Edition

ISBN: 9781305283831

Author: Tan

Publisher: Cengage Learning

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Chapter 8.3, Problem 38E

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The mean and the standard deviation of the given data.

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Chapter 8.3, Problem 37E

Chapter 8.3, Problem 39E

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Chapter 8 Solutions

Finite Mathematics for the Managerial, Life, and Social Sciences-Custom Edition

Ch. 8.1 - What is a random variable? Give an example.Ch. 8.1 - Prob. 2CQ Ch. 8.1 - Prob. 3CQ Ch. 8.1 - Three balls are selected at random without...Ch. 8.1 - 2.A coin is tossed four times. Let the random...Ch. 8.1 - A die is rolled repeatedly until a 6 falls...Ch. 8.1 - Cards are selected one at a time without...Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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The mean and the standard deviation of the given data.

Solution Summary: The author explains that the mean and standard deviation of the given data are 77.167 and 10.625.

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To find: The mean and the standard deviation of the given data.

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Chapter 8 Solutions

To find:

The mean and the standard deviation of the given data.