Chapter 6, Problem 13RE

To determine

Find the probability that a package of 1,060 rods contains 1,000 or more that are good.

Answer to Problem 13RE

The probability that a package of 1,060 rods contains 1,000 or more that are good is 0.8547.

Explanation of Solution

Calculation:

The given information is that there is 5% of the rods are defective and the customer orders 1,000 rods and a package of 1,060 rods are shipped with the guarantee of at least 1,000 rods are good.

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 denotes the number of rods that are good.

The probability of rods are good is 0.95$\left(=1-0.05\right)$.

Requirement check:

Condition 1: $np\ge 10$

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

Condition 1: $np\ge 10$

Substitute 1,060 for n and 0.95 for p in np,

$\begin{array}{c}np=\left(1,060\right)\left(0.95\right)\\ =1,007\end{array}$

Thus, the requirement $np\left(=1,007\right)\ge 10$ is satisfied.

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

Substitute 1,060 for n and 0.75 for p in $n\left(1-p\right)$,

$\begin{array}{c}n\left(1-p\right)=\left(1,060\right)\left(1-0.95\right)\\ =1,060\left(0.05\right)\\ =53\end{array}$

Thus, the requirement of $n\left(1-p\right)\left(=53\right)\ge 10$.

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

For mean ${\mu }_X$:

Substitute 1,060 for n and 0.95 for p in the formula of ${\mu }_X$,

$\begin{array}{c}{\mu }_X=\left(1,060\right)\left(0.95\right)\\ =1,007\end{array}$

Thus, the value of ${\mu }_X$ is 1,007.

For standard deviation ${\sigma }_X$:

Substitute 1,060 for n and 0.95 in the formula of ${\sigma }_X$,

$\begin{array}{c}{\sigma }_X=\sqrt{1,060\left(0.95\right)\left(1-0.95\right)}\\ =\sqrt{1,060\left(0.95\right)\left(0.05\right)}\\ =\sqrt{50.35}\\ \approx 7.09577\end{array}$

Thus, the value of ${\sigma }_X$ is 7.09577.

Here, the probability that a package of 1,060 rods contains 1,000 or more that are good represent $P\left(X\ge 1,000\right)$.

Continuity correction:

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

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

Here, the probability $P\left(X\ge 1,000\right)$ represents $P\left(X\ge a\right)$. Therefore, probability $P\left(X\ge 1,000\right)$ is the area to the right of 999.5$\left(=1,000-0.5\right)$.

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 1,007 and Standard deviation as 7.09577.
• Click the Shaded Area tab.
• Choose X value and Right Tail for the region of the curve to shade.
• Enter the X value as 999.5.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability that a package of 1,060 rods contains 1,000 or more that are good is 0.8547.