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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.

Step by step solution

01

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

02

Calculating the company’s expected profit

Let,

$\begin{array}{rcl}P\left(21â\copyright ½Xâ\copyright ½30\right)& =& P\left(\frac{21-25}{2}â\copyright ½\frac{X-\mathrm{μ}}{\mathrm{σ}}â\copyright ½\frac{30-25}{2}\right)\\ & =& P\left(-2â\copyright ½Zâ\copyright ½2.5\right)\\ & =& P\left(Zâ\copyright ½2.5\right)-P\left(Zâ\copyright ½2\right)\\ & & \end{array}$

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

Now,

$\begin{array}{rcl}P\left(Xâ\copyright ½21\right)& =& P\left(Zâ\copyright ½\frac{21-25}{2}\right)\\ & =& P\left(Zâ\copyright ½-2\right)\\ & =& 0.0228\\ & & \end{array}$

$\begin{array}{rcl}P\left(Xâ\copyright ¾30\right)& =& P\left(Zâ\copyright ½\frac{30-25}{2}\right)\\ & =& P\left(Zâ\copyright ¾2.5\right)\\ & =& 1-P\left(Zâ\copyright ½2.5\right)\\ & & \end{array}$

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

Since,

$E\left(X\right)=\underset{i=1}{\overset{n}{∑}}{X}_i.P\left(X_i\right)$

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

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

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