Question

In: Statistics and Probability

Suppose the following data represent the ratings​ (on a scale from 1 to​ 5) for a...

Suppose the following data represent the ratings​ (on a scale from 1 to​ 5) for a certain smart phone​ game, with 1 representing a poor rating. Complete parts​ (a) through​ (d) below. Stars Frequency 1 2963 2 2372 3 4696 4 4393 5 10 comma 585 ​(a) Construct a discrete probability distribution for the random variable x. Stars​ (x) ​P(x) 1 nothing 2 nothing 3 nothing 4 nothing 5 nothing ​(Round to three decimal places as​ needed.) ​(b) Graph the discrete probability distribution. Choose the correct graph below. A. 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 A histogram has a horizontal axis labeled from 0 to 4 in intervals of 1 and a vertical axis labeled from 0 to 0.5 in intervals of 0.1 has five vertical lines positioned on the horizontal axis tick marks. The approximate heights of the vertical lines are as follows, with the horizontal coordinate listed first and the line height listed second: 1, 0.42; 2, 0.18; 3, 0.19; 4, 0.1; 5, 0.12. B. 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 A histogram has a horizontal axis labeled from 0 to 4 in intervals of 1 and a vertical axis labeled from 0 to 0.5 in intervals of 0.1 five vertical lines positioned on the horizontal axis tick marks. The approximate heights of the vertical lines are as follows, with the horizontal coordinate listed first and the line height listed second: 1, 0.12; 2, 0.1; 3, 0.19; 4, 0.18; 5, 0.42. C. 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 A histogram has a horizontal axis labeled from 0 to 4 in intervals of 1 and a vertical axis labeled from 0 to 0.5 in intervals of 0.1 five vertical lines positioned on the horizontal axis tick marks. The approximate heights of the vertical lines are as follows, with the horizontal coordinate listed first and the line height listed second: 1, 0.1; 2, 0.18; 3, 0.12; 4, 0.19; 5, 0.42. D. 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 A histogram has a horizontal axis labeled from 0 to 4 in intervals of 1 and a vertical axis labeled from 0 to 0.5 in intervals of 0.1 five vertical lines positioned on the horizontal axis tick marks. The approximate heights of the vertical lines are as follows, with the horizontal coordinate listed first and the line height listed second: 1, 0.19; 2, 0.18; 3, 0.42; 4, 0.12; 5, 0.1. ​(c) Compute and interpret the mean of the random variable x. The mean is nothing stars. ​(Round to one decimal place as​ needed.) Which of the following interpretations of the mean is​ correct? A. As the number of experiments​ decreases, the mean of the observations will approach the mean of the random variable. B. The observed value of an experiment will be less than the mean of the random variable in most experiments. C. The observed value of an experiment will be equal to the mean of the random variable in most experiments. D. As the number of experiments​ increases, the mean of the observations will approach the mean of the random variable. ​(d) Compute the standard deviation of the random variable x. The standard deviation is nothing stars. ​(Round to one decimal place as​ needed.)

Solutions

Expert Solution

(a)

Total frequency = 2963 + 2372 + 4696 + 4393 + 10585 = 25009

x P(x)
1 2963 / 25009 = 0.118
2 2372 / 25009 = 0.095
3 4696 / 25009 = 0.188
4 4393 / 25009 = 0.176
5 10585 / 25009 = 0.423

(b)

Based on above the discrete probability, the correct graph is,

B. 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 A histogram has a horizontal axis labeled from 0 to 4 in intervals of 1 and a vertical axis labeled from 0 to 0.5 in intervals of 0.1 five vertical lines positioned on the horizontal axis tick marks. The approximate heights of the vertical lines are as follows, with the horizontal coordinate listed first and the line height listed second: 1, 0.12; 2, 0.1; 3, 0.19; 4, 0.18; 5, 0.42.

(c)

Mean, E(x) = = 1 * 0.118 + 2 * 0.095 + 3 * 0.188 + 4 * 0.176 + 5 * 0.423 = 3.7

D. As the number of experiments​ increases, the mean of the observations will approach the mean of the random variable.

(d)

E(x2) = = 12 * 0.118 + 22 * 0.095 + 32 * 0.188 + 42 * 0.176 + 52 * 0.423 = 15.581

Variance = E(x2) - E(x)2 = 15.581 - 3.72 = 1.891

Standard deviation = = 1.4


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