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The life of army shoes is normally distributed with mean $8$ months and standard deviation $2$ months. If $5000$ pairs are issued, how many pairs would be expected to need replacement within $12$ months.

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Toolbox:
  • Standard normal distribution:
  • In a standard normal distribution $\mu=0,\sigma ^2=1$
  • The random variable $X$ can be converted to the standard normal variable $Z$ by the transformation
  • $Z=\large\frac{X-\mu}{\sigma}$
Step 1:
Let $X$ be the random variable denoting the life of a pair of army shoes.
$X\sim N(8,2^2)$
Step 2:
A pair of shoes would require replacement within 12 months if its life is less than 12 months.
$\therefore$ we have to find $P(X < 12)$
Step 3:
Let $Z$ be the standard normal variate.
$Z=\large\frac{X-\mu}{\sigma}$
$\;\;\;=\large\frac{X-8}{2}$
When $X=12$
$\;\;\;=\large\frac{12-8}{2}$
$\;\;\;=\large\frac{4}{2}$
$\;\;\;=2$
Step 4:
$\therefore P(X < 12)=P(Z < 2)$
$\qquad\qquad\;\;\;\;=0.5+P(0 < Z < 2)$
$\qquad\qquad\;\;\;\;=0.5+0.4772$
$\qquad\qquad\;\;\;\;=0.9772$
Step 5:
Out of 5000 pairs (N=5000) the number of pairs that are expected to be replaced within 12 months=$5000\times 0.9772$
$\Rightarrow 4886$
answered Sep 19, 2013 by sreemathi.v
 

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