Chapter 15, Problem 15.41E

(a)

To determine

To explain: The accuracy of the probability obtained in part (b) of Exercise 15.39 using normal approximation when compared to the exact probability in the given problem.

Answer to Problem 15.41E

The probability obtained in part (b) of Exercise 15.39 using normal approximation is under estimated compared with exact answer.

Explanation of Solution

Given info:

The probability that the Joe’s average winnings between 0.5 and 0.7 is 0.4961. The number of bets is 14,000.

Justification:

Here, the exact probability that the Joe’s average winning between 0.5 and 0.7 is 0.4961. From the part (b) of Exercise 15.39 using normal approximation the Joe’s average winnings is between 0.5 and 0.7 is 0.4648.

From the above two probabilities, it can be observed that the probability obtained by using normal approximation is less than the exact probability. Thus, the probability obtained in part (b) of Exercise 15.39 by using normal approximation is under estimated.

(b)

To determine

To explain: The accuracy of the probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation when compared to the exact probability in the given problem.

Answer to Problem 15.41E

The probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation is less than the exact probability.

Explanation of Solution

Given info:

The data shows that the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.4048. The number of bets is 3,500.

Calculation:

From the part (a) of the Exercise 15.39 the population mean is $\mu =0.60$ and standard deviation is $\sigma =18.96$.

Central limit theorem:

It states that as the sample size n becomes large the sampling distribution of $\overline{x}$ follows normal distribution with mean $\mu$ and standard deviation $\frac{\sigma }{\sqrt{n}}$.

Mean of the Sampling distribution of $\overline{x}$:

The sample mean $\overline{x}$ of sample size n has mean $\mu$. Thus, the mean of the sampling distribution is 0.60.

Standard deviation of the Sampling distribution of $\overline{x}$:

The standard deviation can be calculated by using the formula $\frac{\sigma }{\sqrt{n}}$.

Substitute 18.96 for $\sigma$ and 3,500 for n

$\begin{array}{c}\frac{\sigma }{\sqrt{n}}=\frac{18.96}{\sqrt{3,500}}\\ =\frac{18.96}{59.1608}\\ =0.3205\end{array}$

Thus, the standard deviation of $\overline{x}$ is 0.3205.

Probability using normal approximation:

$\begin{array}{c}P\left(0.5<\overline{x}<0.7\right)=P\left(\frac{0.5-0.6}{\frac{18.96}{\sqrt{3,500}}}<\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{0.7-0.6}{\frac{18.96}{\sqrt{3,500}}}\right)\\ =P\left(\frac{-0.1}{0.3205}<Z<\frac{0.1}{0.3205}\right)\\ =P\left(-0.31<Z<0.31\right)\\ =P\left(Z<0.31\right)-P\left(Z<-0.31\right)\end{array}$

                            $\begin{array}{c}=0.6217-0.3783\left[\begin{array}{l}\text{From}\text{Table}\text{A:}\text{Standard}\text{Normal}\\ \text{Cumulative}\text{Proportions}\end{array}\right]\\ =0.2434\end{array}$

Thus, the probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation is 0.2434.

From the given problem the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.4048. From the above results it can be observed that probability 0.4048 is greater than 0.2434.

Hence, the probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation is less than the exact probability.

(c)

To determine

To explain: The accuracy of the probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation when compared to the exact probability in the given problem.

Answer to Problem 15.41E

The probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation is less than the exact probability.

Explanation of Solution

Given info:

The data shows that the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.9629. The number of bets is 150,000.

Calculation:

Mean of the Sampling distribution of $\overline{x}$:

The sample mean $\overline{x}$ of sample size n has mean $\mu$. Thus, the mean of the sampling distribution is 0.60.

Standard deviation of the Sampling distribution of $\overline{x}$:

The standard deviation can be calculated by using the formula $\frac{\sigma }{\sqrt{n}}$.

Substitute 18.96 for $\sigma$ and 150,000 for n

$\begin{array}{c}\frac{\sigma }{\sqrt{n}}=\frac{18.96}{\sqrt{150,000}}\\ =\frac{18.96}{387.2983}\\ =0.0490\end{array}$

Thus, the standard deviation of $\overline{x}$ is 0.0490.

Probability using normal approximation:

$\begin{array}{c}P\left(0.5<\overline{x}<0.7\right)=P\left(\frac{0.5-0.6}{\frac{18.96}{\sqrt{150,000}}}<\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{0.7-0.6}{\frac{18.96}{\sqrt{150,000}}}\right)\\ =P\left(\frac{-0.1}{0.0490}<Z<\frac{0.1}{0.0490}\right)\\ =P\left(-2.04<Z<2.04\right)\\ =P\left(Z<2.04\right)-P\left(Z<-2.04\right)\end{array}$

                            $\begin{array}{c}=0.9793-0.0207\left[\begin{array}{l}\text{From}\text{Table}\text{A:}\text{Standard}\text{Normal}\\ \text{Cumulative}\text{Proportions}\end{array}\right]\\ =0.9586\end{array}$

Thus, the probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation is 0.9586.

From the given problem the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.9629. From the above results it can be observed that probability 0.9586 is less than 0.9629.

Hence, the probability of $\overline{x}$ is between 0.5 and 0.7 using normal approximation is less than the exact probability.