Chapter 6.6, Problem 13P

(a)

To determine

Find the probability that more than 180 take your free sample.

Answer to Problem 13P

The probability that more than 180 take your free sample 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 take free samples.

The day you are offering free samples, 317 customers pass by your counter. That is, $n=317$, 60% of all customers will take free samples, That is, $p=0.60$, and the about 37% would buy what they have sampled given that those who take the free samples. That is, $P\left(\text{Buy}|\text{Sampled}\right)=0.37$.

Checking conditions:

$\begin{array}{c}np=317\left(0.60\right)\\ =190.2\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =317\left(1-0.60\right)\\ =317\left(0.40\right)\\ =126.8\\ >5\end{array}$

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,

$\begin{array}{c}\mu =np\\ =317\left(0.60\right)\\ =190.2\end{array}$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{317\left(0.60\right)\left(1-0.60\right)}\\ =\sqrt{76.08}\\ =8.7224\end{array}$

The probability that more than 180 take your free sample is,

$\begin{array}{c}P\left(r>180\right)=P\left(r\ge 181\right)\\ \approx P\left(x\ge 181-0.5\right)\\ =P\left(x\ge 180.5\right)\end{array}$

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 190.2, and Standard deviation as 8.7224.
• Click the Shaded Area tab.
• Choose X Value and Right tail, for the region of the curve to shade.
• Enter the X value as 180.5.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the probability is 0.8669.

Hence, the probability that more than 180 take your free sample is 0.8669.

(b)

To determine

Find the probability that fewer than 200 take your free sample.

Answer to Problem 13P

The probability that fewer than 200 take your free sample is 0.8568.

Explanation of Solution

Calculation:

The probability that fewer than 200 take your free sample is,

$\begin{array}{c}P\left(r<200\right)=P\left(r\le 199\right)\\ \approx P\left(x\le 199+0.5\right)\\ =P\left(x\le 199.5\right)\end{array}$

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 190.2, and Standard deviation as 8.7224.
• Click the Shaded Area tab.
• Choose X Value and Left tail, for the region of the curve to shade.
• Enter the X value as 199.5.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the probability is 0.8568.

Hence, the probability that fewer than 200 take your free sample is 0.8568.

(c)

To determine

Find the probability that a customer takes a free sample and buys the product.

Answer to Problem 13P

The probability that a customer takes a free sample and buys the product is 0.222.

Explanation of Solution

Calculation:

Conditional probability:

The conditional probability formula is,

$P\left(A|B\right)=\frac{P\left(A\text{ and }B\right)}{P\left(B\right)}$

The probability that a customer takes a free sample and buys the product is,

$\begin{array}{c}P\left(\text{Buy }and\text{ Sampled}\right)=P\left(\text{Sampled}\right)\times P\left(\text{Buy}|\text{Sampled}\right)\\ =0.60\times 0.37\\ =0.222\end{array}$

Hence, the probability that a customer takes a free sample and buys the product is 0.222.

(d)

To determine

Find the probability that between 60 and 80 customers would take the free sample and buy the product.

Answer to Problem 13P

The probability that between 60 and 80 customers would take the free sample and buy the product is 0.8436.

Explanation of Solution

Calculation:

Let r denotes the number of customer takes a free sample and buys the product.

The day you are offering free samples, 317 customers pass by your counter. That is, $n=317$, and the probability of success for each trail is $p=0.222$.

Checking conditions:

$\begin{array}{c}np=317\left(0.222\right)\\ =70.374\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =317\left(1-0.222\right)\\ =317\left(0.778\right)\\ =246.626\\ >5\end{array}$

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,

$\begin{array}{c}\mu =np\\ =317\left(0.222\right)\\ =70.374\end{array}$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{317\left(0.222\right)\left(1-0.222\right)}\\ =\sqrt{54.750972}\\ =7.3994\end{array}$

The probability that between 60 and 80 customers would take the free sample and buy the product is,

$\begin{array}{c}P\left(60\le r\le 80\right)\approx P\left(60-0.5\le r\le 80+0.5\right)\\ =P\left(59.5\le x\le 80.5\right)\end{array}$

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 70.374, and Standard deviation as 7.3994.
• Click the Shaded Area tab.
• Choose X Value and Middle, for the region of the curve to shade.
• Enter the X value 1 as 59.5, the X value 2 as 80.5.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the probability is 0.8436.

Hence, the probability that between 60 and 80 customers would take the free sample and buy the product is 0.8436.