Chapter 19, Problem 19.63SE

(a)

To determine

To find: The mean and standard deviation of people who drop out the Rewards Program within four weeks of this sample size.

Answer to Problem 19.63SE

The value of mean is 54 people.

The value of standard deviation is 6.65 people.

Explanation of Solution

Given info:

Americans spend more than $30 billion annually on variety of weight loss products and services. In a study, 18% of those who began dropped out in first four weeks from the Rewards Program at Jenny Craig in 2005. Also, simple random sample contained 300 people at the beginning the program.

Calculation:

Define the random variable “X” as the number of people who dropped out in first four weeks from the Rewards Program. Here, there were three hundred (n) people beginning of the program. Also, with the probability of “people who dropped from the program” (p) is 0.18.

There are two possible outcomes (people who dropped in first four week and who did not drop out in first four weeks). Also, each person is independent of the other. Hence, X follows binomial distribution with $n=300,p=0.18$.

Mean:

The mean is calculated by using the following formula:

$\mu =np$

Substitute n as 300 and p as 0.18.

$\begin{array}{c}\mu =\left(300\right)\left(0.18\right)\\ =54\end{array}$

Thus, the value of mean is 54 people.

Standard deviation:

The standard deviation is calculated by using the following formula:

$\sigma =\sqrt{np\left(1-p\right)}$

Substitute n as 300, p as 0.18 and q as $0.82\left(=1-0.18\right)$

$\begin{array}{c}\sigma =\sqrt{np\left(1-p\right)}\\ =\sqrt{\left(300\right)\left(0.18\right)\left(0.82\right)}\\ =\sqrt{44.28}\\ =6.6543\end{array}$

Thus, the value of standard deviation is 6.65 people.

(b)

To determine

To find: The probability that at least 235 people in the sample will still be in the Rewards Program after the first four weeks using normal approximation.

To obtain: The exact binomial probability and compare the two results.

Answer to Problem 19.63SE

The normal approximation of the probability that at least 235 people in the sample will still be in the Rewards Program after the first four weeks is 0.9505.

The exact probability obtaining at least 235 obtaining at least 235 people in the sample will be in Rewards Program after first four weeks is 0.9554.

The exact probability and approximate probability differ by 0.0049.

Explanation of Solution

Calculation:

Let X be the people who dropped out of the program,

Rule of thumb:

• If the sample size n is large, the binomial distribution can be approximated to normal $N\left(np,\sqrt{np\left(1-p\right)}\right)$.
• Normal approximation can be used when the sample size n is large, in such a way that $np\ge 10\text{and}n\left(1-p\right)\ge 10$

Here, $n=300,p=0.18$.

$\begin{array}{c}np=\left(300\right)\left(0.18\right)\\ =54\\ \ge 10\end{array}$

Also,

$\begin{array}{c}n\left(1-p\right)=\left(300\right)\left(1-0.18\right)\\ =300\times 0.82\\ =246\\ \ge 10\end{array}$

Thus, the two conditions are satisfied when the sample size n is large, so that the binomial distribution can be safely approximated to normal distribution.

The probability that at least 235 people in the sample will be in Rewards Program after first four weeks, then 65 $\left(=300-235\right)$ people would drop out.

The approximate probability obtaining not more than 65 people in the sample will drop out from Rewards Program after first four weeks is mentioned below:

$\begin{array}{c}P\left(X\le 65\right)=P\left(\frac{X-\mu }{\sigma }\le \frac{65-\mu }{\sigma }\right)\\ =P\left(\frac{X-54}{6.6543}\le \frac{65-54}{6.6543}\right)\\ =P\left(z\le \frac{11}{6.6543}\right)\\ =P\left(z\le 1.6531\right)\end{array}$

From the TABLE A: Standard Normal Cumulative Proportions, and the area to the left of $z\le 1.6531$ are 0.9505.

$\begin{array}{c}P\left(X\le 65\right)=P\left(z\le 1.6531\right)\\ =0.9505\end{array}$

Thus, the probability that at least 235 people in the sample will still be in the Rewards Program after the first four weeks is 0.9505.

Exact probability:

The exact probability obtaining at least 235 people in the sample will be in Rewards Program after first four weeks is as follws:

Software procedure:

Step-by-step procedure to obtain the ‘Binomial probability’ using the MINITAB software:

• Choose Calc > Probability Distributions > Binomial Distribution.
• Choose Cumulative Probability.
• Enter Number of trials as 300 and Event probability as 0.18.
• In Input constant, enter 65.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(X\le 65\right)=0.9554$

Thus, the exact probability to obtain at least 235 people in the sample will be in Rewards Program after first four weeks is 0.9554.

Justification:

Here, exact probability obtaining at least 235 people in the sample will be in Rewards Program after first four weeks is 0.9554 and approximate probability of obtaining at least 235 people in the sample will be in Rewards Program after first four weeks is 0.9505.

Therefore, the exact probability is greater than the approximate probability by $0.0049\left(=0.9554-0.9505\right)$.