Chapter 6.4, Problem 21E

The car is in the shop: Among automobiles of a certain make, 23% require service during a one-year warranty period. A dealer sells 87 of these vehicles.

1. a. Approximate the probability that 25 or fewer of these vehicles require repairs.
2. b. Approximate the probability that more than 17 vehicles require repairs.
3. c. Approximate the probability that the number of vehicles that require repairs is between 15 and 20, exclusive.

To determine

Find the probability 25 or fewer of one-year warranty period vehicles require repairs by using Approximation.

Answer to Problem 21E

The probability 25 or fewer of one-year warranty period vehicles require repairs by using Approximation is 0.9190.

Explanation of Solution

Calculation:

The given information is that there is 23% of vehicles that require a service during a one-year warranty period and the dealer sells 87 of these vehicles.

Normal approximation to the Binomial:

If $np\ge 10$ and $n\left(1-p\right)\ge 10$ where n is the number of trials and p is the success probability then the binomial random variable X follows approximately normal distribution with mean ${\mu }_X=np$ and standard deviation ${\sigma }_X=\sqrt{np\left(1-p\right)}$.

Let X represents the number of vehicles require service during a one-year warranty period. Therefore, the probability 25 or fewer of one-year warranty period vehicles require repairs represents $P\left(X\le 25\right)$.

From the given information, it can be observed that the sample size n is 87 and the success probability is 0.23 $\left(=\frac{23}{100}\right)$.

Requirement check:

Condition 1: $np\ge 10$

Condition 2: $n\left(1-p\right)\ge 10$

Condition 1: $np\ge 10$

Substitute 87 for n and 0.23 for p in np,

$\begin{array}{c}np=\left(87\right)\left(0.23\right)\\ =20.01\end{array}$

Thus, the requirement $np\left(=20.01\right)\ge 10$ is satisfied.

Condition 2: $n\left(1-p\right)\ge 10$

Substitute 87 for n and 0.23 for p in $n\left(1-p\right)$,

$\begin{array}{c}n\left(1-p\right)=\left(87\right)\left(1-0.23\right)\\ =87\left(0.77\right)\\ =66.99\end{array}$

Thus, the requirement of $n\left(1-p\right)\left(=66.99\right)\ge 10$ is satisfied.

Here, both the requirements are satisfied. Thus, it is appropriate to use the normal approximation.

For mean ${\mu }_X$:

Substitute 87 for n and 0.23 for p in the formula of ${\mu }_X$,

$\begin{array}{c}{\mu }_X=\left(87\right)\left(0.23\right)\\ =20.01\end{array}$

Thus, the value of ${\mu }_X$ is 20.01.

For standard deviation ${\sigma }_X$:

Substitute 87 for n and 0.23 in the formula of ${\sigma }_X$,

$\begin{array}{c}{\sigma }_X=\sqrt{87\left(0.23\right)\left(1-0.23\right)}\\ =\sqrt{87\left(0.23\right)\left(0.77\right)}\\ =\sqrt{15.4077}\\ \approx 3.92526\end{array}$

Thus, the value of ${\sigma }_X$ is 3.92526.

Continuity correction:

The continuity correction is the adjustment when approximating a discrete distribution with a continuous distribution. The areas under the normal curve to use when the continuity correction applied to the probabilities are as follows:

• If $P\left(a\le X\le b\right)$ then find area between $a-0.5$ and $b+0.5$.
• $P\left(X\le b\right)$ then find area to the left of $b+0.5$.
• $P\left(X\ge a\right)$ then find area to the right of $a-0.5$.
• $P\left(X=a\right)$ then find area between $a-0.5$ and $a+0.5$.

Here, the probability $P\left(X\le 25\right)$ represents $P\left(X\le b\right)$. Therefore, probability $P\left(X\le 25\right)$ is the area to the left of 25.5 $\left(=25+0.5\right)$.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 20.01 and Standard deviation as 3.92526.
• Click the Shaded Area tab.
• Choose X value and Left Tail for the region of the curve to shade.
• Enter the X value as 25.5.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability 25 or fewer of one-year warranty period vehicles require repairs by using Approximation is 0.9190.

To determine

Find the probability that more than 17 vehicles require repairs by using Approximation.

Answer to Problem 21E

The probability that more than 17 vehicles require repairs by using Approximation is 0.7387.

Explanation of Solution

Calculation:

The probability that more than 17 vehicles require repairs represents $P\left(X>17\right)$.

Here, probability $P\left(X>17\right)$ is equal to $P\left(X\ge 18\right)$.

Also, it can be observed that the probability $P\left(X\ge 18\right)$ represents $P\left(X\ge a\right)$ by using continuity correction. Therefore, probability $P\left(X\ge 18\right)$ is the area to the right of 17.5 $\left(=18-0.5\right)$.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 20.01 and Standard deviation as 3.92526.
• Click the Shaded Area tab.
• Choose X value and Right Tail for the region of the curve to shade.
• Enter the X value as 17.5.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability that more than 17 vehicles require repairs by using Approximation is 0.7387.

To determine

Find the probability that the number of vehicles that require repairs is between 15 and 20, exclusive by using approximation.

Answer to Problem 21E

The probability that the number of vehicles that require repairs is between 15 and 20, exclusive by using approximation is 0.3230.

Explanation of Solution

Calculation:

The probability that the number of vehicles that require repairs is between 15 and 20, exclusive is $P\left(15<X<20\right)$.

Here, the probability $P\left(15<X<20\right)$ is equal to $P\left(16\le X\le 19\right)$.

Also, it can be observed that the probability $P\left(16\le X\le 19\right)$ represents $P\left(a\le X\le b\right)$ by using continuity correction. Therefore, probability $P\left(16\le X\le 19\right)$ is the area between 15.5 $\left(=16-0.5\right)$ and 19.5 $\left(=19+0.5\right)$.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 20.01 and Standard deviation as 3.92526.
• Click the Shaded Area tab.
• Choose X value and Middle for the region of the curve to shade.
• Enter the X value 1 as 15.5 and X value 2 as 19.5.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability that the number of vehicles that require repairs is between 15 and 20, exclusive by using approximation is 0.3230.