Chapter 11.3, Problem 16PPS

(a)

To determine

To find: The expected number of single ticket.

Answer to Problem 16PPS

The expected number of single ticket is $0.29

Explanation of Solution

Given;

The French Club sold 500 raffle tickets for $1 each.

The win for first, 2 second and 5 third is $100, $10 and $5

Calculation:

The expected value is calculated as,

  $\begin{array}{c}E\left(X\right)=\sum \left[X.P\left(X\right)\right]\\ =\left(X_{1}P\left(X_1\right)+\mathrm{........}+X_{n}P\left(X_n\right)\right)\\ =100\left(\frac{1}{500}\right)+10\left(\frac{1}{250}\right)+5\left(\frac{1}{100}\right)\\ =\$0.29\end{array}$

Thus, the expected number of single ticket is $0.29

(b)

To determine

To find: The standard deviation.

Answer to Problem 16PPS

The standard deviation is 4.53

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,

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

X	P(X)	${\left[X-E\left(X\right)\right]}^2$	${\left[X-E\left(X\right)\right]}^{2}P\left(X\right)$
$100	0.00200	9942.0841	19.8841682
$10	0.00400	94.2841	0.3771364
$5	0.01000	22.1841	0.221841
$0	0.98400	0.0841	0.0827544
		$\sum \left[{\left[X-E\left(X\right)\right]}^{2}P\left(X\right)\right]$	20.5659
		$\sqrt{\sum \left[{\left[X-E\left(X\right)\right]}^{2}P\left(X\right)\right]}$	4.53

Table 1

Thus, the standard deviation is 4.53

(c)

To determine

To find: The raffle in which participation is done.

Answer to Problem 16PPS

The raffle in which participation is done is Glee Club.

Explanation of Solution

The Glee Club is less risky than the French Club as the Glee Club has almost half the standard deviation than the French Club.

Thus, the raffle in which participation is done is Glee Club.