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Probability that a washer is defective

$\begin{array}{l}=1-\mathrm{Probability}\mathrm{that}\mathrm{a}\mathrm{washer}\mathrm{is}\mathrm{not}\mathrm{defective}\hfill \\ =1-0.7698\hfill \\ =0.2302\hfill \end{array}$

Therefore the percentage of defective washers=23.02.

Example 4.93: The mean of a normal distribution is 60 and 6% of the values are greater than 70. Find the SD of the distribution?

Solution: Let X denote the normal random variable

We have $P\left(X>70\right)=\frac{6}{100}=0.6$ (given)

Therefore

$P\left(\frac{x-\mu }{\sigma }>\frac{70-\mu }{\sigma }\right)=0.6$

i.e.,

$P\left(Z>\frac{70-60}{\sigma }\right)=0.6$

or

$P\left(Z>\frac{10}{\sigma }\right)=0.6$

Let ${Z}_{1}=10/$

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