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Consider the production of iron rods having diameter X which is normally distributed with mean 2 in. and standard deviation 0.008 in. Using $\phi$(0.25) = 0.99, where $\phi$ is the cumulative distribution function (c.d.f.) of a standard normal distribution with mean 0 and standard deviation 1, the percentage of defectives we can expect if we set the tolerance limits at 2 ± 0.02 in is  _______ .
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the percentage of defectives we can expect if we set the tolerance limits at 2 ± 0.02 in is

We need to calculate the percentage of defective productions. That means if rods are of length > $2.02$ or < $1.98$

$Z = \frac{X - \lambda }{\sigma}$

$P(1.98 \leq X \leq 2.02) = P(X \leq 2.02) - P( X \leq 1.98)$

$= P(Z \leq \frac{2.02 - 2}{0.008}) - P( Z \leq \frac{1.98 - 2}{0.008})$

$= P(Z \leq 2.5) - P( Z \leq -2.5)$

$= P(Z \leq 2.5) - (1 - P( Z \leq 2.5))$

So the question contains an error "ϕ(0.25) = 0.99"

but $\phi(2.5) = 0.99$

$= P(Z \leq 2.5) - (1 - P( Z \leq 2.5))$

$= 0.99 - (1 -  0.99)$

$P(1.98 \leq X \leq 2.02) = 0.98$

Hence the percentage of defectives we can expect is 100% - 98% which is 2%

 

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