Chapter 11.3, Problem 15PPS

(a)

To determine

To find: The expected number of upsets

Answer to Problem 15PPS

The expected number of upsets is 4

Explanation of Solution

Given;

The given data is as follows:

  

Calculation:

The expected value is calculated as,

  $\begin{array}{c}E\left(X\right)=\sum \left[X.P\left(X\right)\right]\\ =\left(X_{1}P\left(X_1\right)+\mathrm{........}+X_{n}P\left(X_n\right)\right)\\ =\left[\begin{array}{l}0\left(\frac{1}{32}\right)+1\left(\frac{1}{32}\right)+2\left(\frac{1}{32}\right)+3\left(\frac{1}{32}\right)+\\ 4\left(\frac{1}{32}\right)+5\left(\frac{1}{32}\right)+6\left(\frac{1}{32}\right)+7\left(\frac{1}{32}\right)\end{array}\right]\\ =4.34375\end{array}$

Round off the value to the nearest integer.

  $E\left(X\right)=4$

Thus, the expected number of upsets is 4

(b)

To determine

To find: The standard deviation.

Answer to Problem 15PPS

The standard deviation is 1.90

Explanation of Solution

The formula for the variance is,

  ${\sigma }^2=\sum \left[{\left[X-E\left(X\right)\right]}^{2}P\left(X\right)\right]$

The formula for the standard deviation is,

  $\begin{array}{c}\sigma =\sqrt{{\sigma }^2}\\ =\sqrt{\sum \left[{\left[X-E\left(X\right)\right]}^{2}P\left(X\right)\right]}\end{array}$

  

Table 1

Thus, the standard deviation is 1.90

(c)

To determine

To find: The relative frequency table for 50 trials.

Answer to Problem 15PPS

The relative frequency table for 50 trials is shown in Table 2.

Explanation of Solution

Draw the possible table for 50 trials.

Prize, X	Frequency	P(X)
0	1	0.02
1	2	0.04
2	1	0.02
3	8	0.16
4	9	0.18
5	20	0.40
6	7	0.14
7	2	0.04
8	0	0.00

Table 2

Thus, the relative frequency table for 50 trials is shown in Table 2.

(d)

To determine

To find: The graph for experimental probability distribution.

Answer to Problem 15PPS

The graph for experimental probability distribution is shown in figure 1.

Explanation of Solution

From Table (2), the graph is plotted as,

  

Figure 1

Thus, the graph for experimental probability distribution is shown in figure 1.