Question

Purchasing decision. A building contractor has decided to purchase a load of the factory-reject aluminum siding as long as the average number of flaws per piece of siding in a sample of size 35 from the factory's reject pile is 2.1 or less. If it is known that the number of flaws per piece of siding in the factory's reject pile has a Poisson probability distribution with a mean of 2.5, find the approximate probability that the contractor will not purchase a load of siding

Short answer

The probability that the contractor will not purchase a load of siding is 0.9332.

Step-by-step solution

Given information

The company's policy is to purchase a load when the factory rejects a pile is 2.1 or less.

The random variable x is defined as the number of flaws per piece.

Provided that random variable x has a Poisson distribution with a mean of 2.5.The probability that the contractor will not purchase a load of siding is calculated using the normal distribution.

Calculating the probability

The Poisson distribution has the same mean and variance.

Here$\overline{x}$is the average number of flaws having meanlocalid="1659736038381" $=2.5$and variance,localid="1659736032501" $=2.5$

Hence,

localid="1659736046989" ${\mu }_{\overline{x}}=2.5$,

localid="1659736043170" $\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{\sqrt{2.5}}{\sqrt{35}}\\ & =& 0.2678\\ & & \end{array}$

The contractor will not purchase if the average number of flaws is 2.1 or more.The associated normal curve is drawn as follows:

Therefore, the required probability is,

$\begin{array}{rcl}P\left(\overline{x}⩾2.1\right)& =& P\left(\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overrightarrow{x}}}⩾\frac{2.1-2.5}{0.2673}\right)\\ & =& P\left(z⩾-1.4967\right)\\ & =& 0.5+A\\ & & \end{array}$

$$\begin{gathered} =0.5+0.4332 \\ =0.9332 \end{gathered}$$

Hence, the contractor's probability of not purchasing a load of siding is 0.9332.