Chapter 7.6, Problem 14P

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.

(a) What is the probability that 50 or more will buy vanilla?

(b) What is the probability that 12 or more will buy chocolate?

(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.)

(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).

To determine

To find: The probability that 50 or more will buy vanilla.

Answer to Problem 14P

Solution: The probability that 50 or more will buy vanilla is $P(r\ge 50)=\mathrm{0.1587.}$

Explanation of Solution

Calculation: Using $n=175,p=0.25,q=1-p=\mathrm{0.75.}$

Thus,

$\begin{array}{l}np=(175)(0.25)\\ =43.75\\ nq=(175)(0.75)\\ =131.25\end{array}$

The condition $np>5\text{and}nq>5$ is satisfied. Thus, it is appropriate to use the normal approximation for binomial distribution.

Now, we calculate $\mathrm{P}(r>180)$ using continuity correction factor for normal distribution.

$\begin{array}{l}\mathrm{P}(r\ge 50)=P(x\ge r-0.5)\\ =P(x\ge 50-0.5)\\ =P(x\ge 49.5)\end{array}$

Now, mean and standard deviation are as follows:

$\begin{array}{l}\mu =np\\ =(175)(0.25)\\ =43.75\\ \sigma =\sqrt{npq}\\ =\sqrt{(175)(0.25)(0.75)}\\ =\sqrt{32.81}\\ =5.73\end{array}$

Further, the z-value corresponding to x is

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{49.5-43.75}{5.73}\\ =1.003\end{array}$

$\begin{array}{l}\mathrm{P}(x\ge 49.5)=P(z\ge 1.003)\\ =1-P(z\le 1.003)\\ =1-0.8413\\ =0.1587\end{array}$

Thus, the desired probability is 0.1587.

To determine

To find: The probability that 12 or more will by chocolate.

Answer to Problem 14P

Solution: The probability that 12 or more will by chocolate is $\mathrm{P}(r\ge 12)=0.8686$

Explanation of Solution

Calculation: The probability of sale one chocolate ice cream is $p=0.09$.

Using, $n=175,p=0.09,q=1-p=\mathrm{0.91.}$

$\begin{array}{l}np=(175)(0.09)\\ =15.75\\ nq=(175)(0.91)\\ =159.25\end{array}$

The condition $np>5\text{and}nq>5$ is satisfied. Thus, it is appropriate to use the normal approximation for binomial distribution.

Now, we calculate $\mathrm{P}(r\ge 12)$ using continuity correction factor for normal distribution.

$\begin{array}{l}\mathrm{P}(r\ge 12)=P(x\ge r-0.5)\\ =P(x\ge 12+0.5)\\ =P(x\ge 11.5)\end{array}$

Now, mean and standard deviation are as follows:

$\begin{array}{l}\mu =np\\ =(175)(0.09)\\ =15.75\\ \sigma =\sqrt{npq}\\ =\sqrt{(175)(0.09)(0.91)}\\ =\sqrt{14.33}\\ =3.79\end{array}$

Further, the z-value corresponding to x is

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{11.5-15.75}{3.79}\\ =-1.121\end{array}$

$\begin{array}{l}\mathrm{P}(x\ge 11.5)=P(z\ge -1.121)\\ =1-P(z\le -1.121)\\ =1-0.1314\\ =0.8686\end{array}$

Thus, the desired probability is 0.8686.

To determine

To find: The probability that someone is buying any one flavor either vanilla or chocolate.

Answer to Problem 14P

Solution: probability that someone is buying any one flavor either vanilla or chocolate is 0.3175.

Explanation of Solution

Calculation: The probability that someone is buying any one flavor either vanilla or chocolate is

Let A represents vanilla flavor and B represents chocolate flavor.

$\mathrm{P}(A\cup B)=P(A)+P(B)-P(A\cap B)$

Here, A and B are not mutually exclusive events and the choice to buy one flavor is independent of the choice to buy another flavor.

$\begin{array}{l}P(A\cap B)=P(A)P(B)\\ P(A\cup B)=P(A)+P(B)-P(A)P(B)\\ =0.25+0.09-(0.25)(0.59)\\ =0.3175\end{array}$

To determine

To find: The probability that between 50 and 60 customers will buy chocolate or vanilla ice-cream.

Answer to Problem 14P

Solution: The probability that between 50 and 60 customers will buy chocolate or vanilla ice-cream is $P(50\le r\le 60)=0.6246$

Explanation of Solution

Calculation: From part (c) the probability of buying chocolate or vanilla ice cream is $p=0.3175$ using $n=175,p=0.3175,q=1-p=\mathrm{0.6825.}$

Now, we calculate $\mathrm{P}(50\le r\le 60)$ using continuity correction factor for normal distribution.

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

Now, mean and standard deviation are as follows:

$\begin{array}{l}\mu =np\\ =(175)(0.3175)\\ =55.56\\ \sigma =\sqrt{npq}\\ =\sqrt{(175)(0.3175)(0.6825)}\\ =\sqrt{37.92}\\ =6.16\end{array}$

Further,

$\begin{array}{l}P(49.5\le r\le 60.5)=P\left(\frac{49.5-55.56}{6.16}\le \frac{x-\mu }{\sigma }\le \frac{60.5-55.56}{6.16}\right)\\ =P(-0.98\le z\le 0.80)\\ =P(z\le 0.80)-P(z\le -0.98)\\ =0.7881-0.1635\\ \text{}=0.6246\end{array}$

Thus, the desired probability is 0.6246.