Chapter 6.6, Problem 14P

In the following problems, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Ice Cream: Flavors What’s your favorite ice cream flavor? For people who buy ice cream, the all-time favorite is still vanilla. About 25% of ice cream sales are vanilla. Chocolate accounts for only 9% of ice cream sales. (See reference in Problem 8.) Suppose that 175 customers go to a grocery store in Cheyenne, Wyoming today to buy ice cream.

1. (a) What is the probability that 50 or more will buy vanilla?
2. (b) What is the probability that 12 or more will buy chocolate?
3. (c) A customer who buys ice cream is not limited to one container or one flavor. What is the probability that someone who is buying ice cream will buy chocolate or vanilla? Hint: Chocolate flavor and vanilla flavor are not mutually exclusive events. Assume that the choice to buy one flavor is independent of the choice to buy another flavor. Then use the multiplication rule for independent events, together with the addition rule for events that are not mutually exclusive, to compute the requested probability. (See Section 4.2.)
4. (d) What is the probability that between 50 and 60 customers will buy chocolate or vanilla ice cream? Hint: Use the probability of success computed in part (c).

8. Insurance: Claims Do you try to pad an insurance claim to cover your deductible? About 40% of all U.S. adults will try to pad their insurance claims! (Source: Are You Normal?, by Bernice Kanner, St. Martin’s Press.) Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed in the next few days. What is the probability that

1. (a) half or more of the claims have been padded?
2. (b) fewer than 45 of the claims have been padded?
3. (c) from 40 to 64 of the claims have been padded?
4. (d) more than 80 of the claims have not been padded?

To determine

Find the probability that 50 or more would buy vanilla.

Answer to Problem 14P

The probability that 50 or more would buy vanilla is 0.8669.

Explanation of Solution

Calculation:

Conditions for normal approximation to the binomial:

For a binomial experiment with n number of trails, r number of success, probability of success for each trail p and probability of failure $q=1-p$, the r has binomial distribution which is approximated to a normal distribution if,

• $np>5$
• $nq>5$

Mean:

The mean formula for the binomial distribution using normal approximation is,

$\mu =np$

In the formula n denotes number of trails, and p denotes probability of success.

Standard deviation:

The standard deviation formula for the binomial distribution using normal approximation is,

$\sigma =\sqrt{npq}$

In the formula $q=1-p$, n denotes number of trails, and p denotes probability of success.

Conditions for Continuity correction:

The continuity correction is used for converting the discrete random variable r denoting the number of success to continuous normal random variable x,

• The value x is obtained by subtracting 0.5 from r when r is the left point for an interval. That is, $x=r-0.5$.
• The value x is obtained by adding 0.5 from r when r is the right point for an interval. That is, $x=r+0.5$.

Z score:

The number of standard deviations the original measurement x is from the value of mean $\mu$ is measured using the z-score or z value. The formula for z score is,

$z=\frac{x-\mu }{\sigma }$

In the formula, x is the raw score, $\mu$ is the mean and $\sigma$ is the standard deviation.

Let r denotes the number of customers would buy vanilla.

There are 175 customers go to a grocery store in Cheyenne, Wyoming today to buy ice cream. That is, $n=175$, and about 25% of ice cream sales are vanilla. That is, $p=0.25$.

Checking conditions:

$\begin{array}{c}np=175\left(0.25\right)\\ =43.75\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =175\left(1-0.25\right)\\ =175\left(0.75\right)\\ =131.25\\ >5\end{array}$

It can be observed that two of the conditions $np>5$, $nq>5$ are satisfied by the binomial experiment. It is appropriate to use normal approximation to the binomial.

The mean is,

$\begin{array}{c}\mu =np\\ =175\left(0.25\right)\\ =43.75\end{array}$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{175\left(0.25\right)\left(1-0.25\right)}\\ =\sqrt{32.8125}\\ =5.7282\end{array}$

The probability that 50 or more would buy vanilla is,

$\begin{array}{c}P\left(r\ge 50\right)\approx P\left(x\ge 50-0.5\right)\\ =P\left(x\ge 49.5\right)\end{array}$

Step by step procedure to obtain standard normal curve using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 43.75, and Standard deviation as 5.7282.
• Click the Shaded Area tab.
• Choose X Value and Right tail, for the region of the curve to shade.
• Enter the X value as 49.5.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the probability is 0.1577.

Hence, the probability that more than 180 take your free sample is 0.1557.

To determine

Find the probability that 12 or more would buy chocolate.

Answer to Problem 14P

The probability that 12 or more would buy chocolate is 0.8692.

Explanation of Solution

Calculation:

Let r denotes the number of customers would buy chocolate.

There are 175 customers go to a grocery store in Cheyenne, Wyoming today to buy ice cream. That is, $n=175$, and chocolate accounts for only 9% of ice cream sales. That is, $p=0.09$.

Checking conditions:

$\begin{array}{c}np=175\left(0.09\right)\\ =15.75\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =175\left(1-0.09\right)\\ =175\left(0.91\right)\\ =159.25\\ >5\end{array}$

It can be observed that two of the conditions $np>5$, $nq>5$ are satisfied by the binomial experiment. It is appropriate to use normal approximation to the binomial.

The mean is,

$\begin{array}{c}\mu =np\\ =175\left(0.09\right)\\ =15.75\end{array}$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{175\left(0.09\right)\left(1-0.09\right)}\\ =\sqrt{14.3325}\\ =3.7858\end{array}$

The probability that 12 or more would buy chocolate is,

$\begin{array}{c}P\left(r\ge 12\right)\approx P\left(x\ge 12-0.5\right)\\ =P\left(x\ge 11.5\right)\end{array}$

Step by step procedure to obtain standard normal curve using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 15.75, and Standard deviation as 3.7858.
• Click the Shaded Area tab.
• Choose X Value and Right tail, for the region of the curve to shade.
• Enter the X value as 11.5.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the probability is 0.8692.

Hence, the probability that 12 or more would buy chocolate is 0.8692.

To determine

Find the probability that someone who is buying ice cream would buy chocolate or vanilla.

Answer to Problem 14P

The probability that someone who is buying ice cream would buy chocolate or vanilla is 0.3175.

Explanation of Solution

Calculation:

Union of two events:

For two events A and B, the probability of the union of two events is,

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

Here, buying vanilla and buying chocolate ice creams are not mutually exclusive because when a person buying vanilla also can buy chocolate ice cream.

The probability that someone who is buying ice cream would buy chocolate or vanilla is,

$\begin{array}{c}P\left(\text{chocolate }or\text{ vanilla}\right)=P\left(\text{chocolate}\right)+P\left(\text{vanilla}\right)-P\left(\text{chocolate }and\text{ vanilla}\right)\\ =0.09+0.25-\left(0.09\times 0.25\right)\\ =0.34-0.0225\\ =0.3175\end{array}$

Hence, the probability that someone who is buying ice cream would buy chocolate or vanilla is 0.3175.

To determine

Find the probability that between 50 and 60 customers would buy chocolate or vanilla ice cream.

Answer to Problem 14P

The probability that between 50 and 60 customers would buy chocolate or vanilla ice cream is 0.6262.

Explanation of Solution

Calculation:

Let r denotes the number of customer would buy chocolate or vanilla ice cream.

There are 175 customers go to a grocery store in Cheyenne, Wyoming today to buy ice cream. That is, $n=175$, and the probability of success for each trail is $p=0.3175$.

Checking conditions:

$\begin{array}{c}np=175\left(0.3175\right)\\ =55.5625\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =175\left(1-0.3175\right)\\ =175\left(0.6825\right)\\ =119.4375\\ >5\end{array}$

It can be observed that two of the conditions $np>5$, $nq>5$ are satisfied by the binomial experiment. It is appropriate to use normal approximation to the binomial.

The mean is,

$\begin{array}{c}\mu =np\\ =175\left(0.3175\right)\\ =55.5625\end{array}$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{175\left(0.3175\right)\left(1-0.3175\right)}\\ =\sqrt{37.9214}\\ =6.1580\end{array}$

The probability that between 50 and 60 customers would buy chocolate or vanilla ice cream is,

$\begin{array}{c}P\left(50\le r\le 60\right)\approx P\left(50-0.5\le r\le 60+0.5\right)\\ =P\left(49.5\le x\le 60.5\right)\end{array}$

Step by step procedure to obtain standard normal curve using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 55.5625, and Standard deviation as 6.1580.
• Click the Shaded Area tab.
• Choose X Value and Middle, for the region of the curve to shade.
• Enter the X value 1 as 49.5, the X value 2 as 60.5.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the probability is 0.6262.

Hence, the probability that between 50 and 60 customers would buy chocolate or vanilla ice cream is 0.6262.