Chapter 6, Problem 78CE

Although television HDTV converters are tested before they are placed in the installer’s truck, the installer knows that 20 percent of them still won’t work properly. The driver must install eight converters today in an apartment building. (a) Ten converters are placed in the truck. What is the probability that the driver will have enough working converters? *(b) How many boxes should the driver load to ensure a 95 percent probability of having enough working converters?

To determine

Find the probability that the driver have enough working converters.

Answer to Problem 78CE

The probability that the driver have enough working converters is 0.6778.

Explanation of Solution

Calculation:

The given information is that the 20% of the converters would not work properly. The driver must install the 8 converters in an apartment in one day. There are 10 converters in the truck.

In total of 10 converters, 2 would not work properly. The driver has to install 8 converters, it means any 2 or fewer are not working properly then the driver can get enough working converters.

Binomial Distribution:

The random variable X is the number of success in n trails. The random variable X follows binomial distribution with parameters n and $\pi$.

The PDF of the distribution is,

$P\left(X=x\right)=\frac{n!}{x!\left(n-x\right)!}{\pi }^x{\left(1-\pi \right)}^{n-x}$ for X = 0, 1, 2, 3, 4, …, n.

Here, the random variable X is the number of not working converters, n is 10 and the probability of no shows is 0.20.

The probability of the driver have enough working converters is $P\left(X\le 2\right)$.

Probability = BINOM.DIST (numbers, trails, probability, cumulative).

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (2,10,0.20,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 2\right)$ is 0.6778.

Thus, the probability that the driver have enough working converters is 0.6778.

To determine

Find the number of boxes should be loaded to get 95% have enough working converters.

Answer to Problem 78CE

The number of boxes should be loaded to get 95% have enough working converters is 13.

Explanation of Solution

Calculation:

Assume the random variable X represents the number of working converters and the probability for the event is 0.80.

Calculate the number of boxes need to get the 95% have enough working inverters,$P\left(X\ge 8\right)\ge 0.95$ for various values of n.

The value of $P\left(X\ge 8\right)=1-P\left(X\le 7\right)$.

Probability = BINOM.DIST (7, n, 0.80, 1).

For $n=10$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (7,10,0.80,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 7\right)$ for n as 10 is 0.3222.

$\begin{array}{c}P\left(X\ge 8\right)=1-P\left(X\le 7\right)\\ =1-0.3222\\ =0.6778\end{array}$

For $n=11$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (7,11,0.80,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 7\right)$ for n as 11 is 0.1611.

$\begin{array}{c}P\left(X\ge 8\right)=1-P\left(X\le 7\right)\\ =1-0.1611\\ =0.8389\end{array}$

For $n=12$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (7,12,0.80,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 7\right)$ for n as 12 is 0.0726.

$\begin{array}{c}P\left(X\ge 8\right)=1-P\left(X\le 7\right)\\ =1-0.0726\\ =0.9274\end{array}$

For $n=13$.

Software procedure:

Step by step procedure to obtain the probability by using EXCEL software:

• Open new EXCEL file.
•  In cell A1, enter the formula as “=BINOM.DIST (7,13,0.80,1)”.

Output using EXCEL software is given as follows:

From the output, the value of $P\left(X\le 7\right)$ for n as 13 is 0.0300.

$\begin{array}{c}P\left(X\ge 8\right)=1-P\left(X\le 7\right)\\ =1-0.0300\\ =0.9700\end{array}$

Thus, the number of boxes should be loaded to get 95% have enough working converters is 13.