Chapter 3, Problem 39CE

To determine

Check whether each row of given distribution is valid or invalid.

Answer to Problem 39CE

The expected value and standard deviation of total winnings when a gambler bets $3 on a single round are –0.081 and 2.9989.

Explanation of Solution

From the given grade distribution, row (a) values are 0.3, 0.3, 0.3, 0.2, and 0.1.

The valid probability distribution should have a total probability equal to 1 and probability values should be greater than or equal to zero.

Here, the sum of probability is calculated as follows:

$\begin{array}{c}\text{Total probability}=0.3+0.3+0.3+0.2+0.1\\ =1.2\\ >1\end{array}$

Thus, row (a) is an invalid probability distribution since the total probability is greater than 1.

From the given grade distribution, row (b) values are 0, 0, 1, 0, and 0.

The valid probability distribution should have total probability equal to 1 and probability values should be greater than or equal to zero.

Here, the sum of probability is calculated as follows:

$\begin{array}{c}\text{Total probability}=0+0+1+0+0\\ =1\end{array}$

Thus, row (b) is a valid probability distribution since the total probability is equal than 1.

From the given grade distribution, row (c) values are 0.3, 0.3, 0.3, 0, and 0.

The valid probability distribution should have a total probability equal to 1.

Here, the sum of probability is calculated as follows:

$\begin{array}{c}\text{Total probability}=0.3+0.3+0.3+0+0\\ =0.9\\ <1\end{array}$

Thus, row (c) is an invalid probability distribution since the total probability is less than 1.

From the given grade distribution, row (d) values are 0.3, 0.5, 0.2, 0.1, and –0.1.

The valid probability distribution should have a total probability equal to 1 and probability values should be greater than or equal to zero.

Here, the negative probability value is given.

Thus, row (d) is an invalid probability distribution since the negative probability value is found.

From the given grade distribution, row (e) values are 0.2, 0.4, 0.2, 0.1, and 0.1.

The valid probability distribution should have a total probability equal to 1 and probability values should be greater than or equal to zero.

Here, the sum of probability is calculated as follows:

$\begin{array}{c}\text{Total probability}=0.2+0.4+0.2+0.1+0.1\\ =1\end{array}$

Thus, row (e) is a valid probability distribution since the total probability is equal than 1.

From the given grade distribution, row (f) values are 0, –0.1, 1.1, 0, and 0.

The valid probability distribution should have a total probability equal to 1 and probability values should be greater than or equal to zero.

Here, the negative probability value is given.

Thus, row (f) is an invalid probability distribution since negative probability value is found.