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

zi2lalZ 2021-09-10 Answered

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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timbalemX

Answered 2021-09-11 Author has 108 answers

Step-by-step explanation:

$z = \frac{X - μ_{1}}{σ}$

$μ_{0}$ the mean ( average no. of months that an employee stay in a factory) $σ$ standard deviation

a) $P [\geq 30] = 1 - P [X < 30]$

$P [X < 30]$

We look for z (score)

$z = \frac{X - μ_{1}}{σ} \Rightarrow z = 30 - \frac{36}{6}$

$z = - 1$

From z table we get for -1

$P [X < 30] = 0, 1587$ And $P [X \geq 30] = 1 - P [x < 30] \Rightarrow P [X \geq 30] = 1 - 0, 1587$

$P [X \geq 30] = 0, 8413 or 84, 13 %$

b) $P [x < 24]$

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

z(score) $= - 2$ And from z table we get: $P [X < 24] = 0, 0228 or 2, 28 %$

c) $P [24 < x < 48] i s P [X \leq 48] - P [x \leq 24]$

$P [X < 48]$

s(score) $= 48 - \frac{36}{6} \Rightarrow z = 2$

$P [x < 48] = 0, 9772$

Then $P [24 < X < 48] = 0, 9772 - 0, 0228$

$P [24 < X < 48] = 0, 9544 o r 95, 44 %$

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