Question

Establishing tolerance limits. The tolerance limits for a product's quality characteristic (e.g., length, weight, or strength) are the minimum or maximum values at which the product will operate properly. Tolerance limits are set by the engineering design function of the manufacturing operation (Total Quality Management, Vol. 11, 2000). The tensile strength of a particular metal part can be characterized as being normally distributed with a mean of 25 pounds and a standard deviation of 2 pounds. The part's upper and lower tolerance limits are 30 pounds and 21 pounds, respectively. A part that falls within the tolerance limits results in a profit of \(10. A part that falls below the lower tolerance limit costs the company \)2; a part that falls above the upper tolerance limit costs the company $1. Find the company’s expected profit per metal part produced.

Short answer

The company’s expected profit per part is $9.66.

Step-by-step solution

Given information

The tensile strength of a particular metal part is normally distributed with a mean of 25 pounds and a standard deviation of 2 pounds.

The part's upper and lower tolerance limits are 30 pounds and 21 pounds, respectively.

A part that falls within the tolerance limits results in a profit of $10. A part that falls below the lower tolerance limit costs the company $2; a part that falls above the upper tolerance limit costs the company $1

Calculating the company’s expected profit

Let,

$\begin{array}{rcl}P\left(21⩽X⩽30\right)& =& P\left(\frac{21-25}{2}⩽\frac{X-\mu }{\sigma }⩽\frac{30-25}{2}\right)\\ & =& P\left(-2⩽Z⩽2.5\right)\\ & =& P\left(Z⩽2.5\right)-P\left(Z⩽2\right)\\ & & \end{array}$

$$\begin{gathered} =0.9938-0.0228 \\ =0.9710 \end{gathered}$$

Now,

$\begin{array}{rcl}P\left(X⩽21\right)& =& P\left(Z⩽\frac{21-25}{2}\right)\\ & =& P\left(Z⩽-2\right)\\ & =& 0.0228\\ & & \end{array}$

$\begin{array}{rcl}P\left(X⩾30\right)& =& P\left(Z⩽\frac{30-25}{2}\right)\\ & =& P\left(Z⩾2.5\right)\\ & =& 1-P\left(Z⩽2.5\right)\\ & & \end{array}$

$$\begin{gathered} =1-0.9938 \\ =0.0062 \end{gathered}$$

Since,

$E\left(X\right)=\sum _{i=1}^n{X}_i.P\left(X_i\right)$

$\begin{array}{rcl}E\left(X\right)& =& 0.P\left(X=0\right)+1.P\left(X=1\right)+2.P\left(X=2\right)+3.P\left(X=3\right)\\ & =& \left(0.0228*\left(-2\right)\right)+\left(0.9710*10\right)+\left(0.0062*\left(-1\right)\right)\\ & =& 9.6582\\ & & \end{array}$

Therefore, the company’s expected profit per part is $9.66.