# A report revealed that the average number of months that an employee stays in a factory is 36 months.

A report revealed that the average number of months that an employee stays in a factory is 36 months. Assuming that the number of months of an employee tenure in the factory is normally distributed with a standard deviation of 6 months, find the probability that a certain employee will stay. a. More than 30 months b. Less than 24 months c. Between 24 to 48 months
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Step-by-step explanation:

$z=\frac{X-{\mu }_{1}}{\sigma }$

${\mu }_{0}$ the mean ( average no. of months that an employee stay in a factory) $\sigma$ standard deviation

a) $P\left[\ge 30\right]=1-P\left[X<30\right]\phantom{\rule{0ex}{0ex}}$

$P\left[X<30\right]$

We look for z (score)

$z=\frac{X-{\mu }_{1}}{\sigma }⇒z=30-\frac{36}{6}$

$z=-1$

From z table we get for -1

$P\left[X<30\right]=0,1587$ And $P\left[X\ge 30\right]=1-P\left[x<30\right]⇒P\left[X\ge 30\right]=1-0,1587$

b) $P\left[x<24\right]$

z(score)$=\frac{24-36}{6}$

z(score)$=-2$ And from z table we get:

c)

$\phantom{\rule{0ex}{0ex}}P\left[X<48\right]$

s(score)$=48-\frac{36}{6}⇒z=2$

$P\left[x<48\right]=0,9772$

Then $P\left[24