A report revealed that the average no. of months that an employee stays in a factory is 36 months. Assuming that the no. 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

a) P [X ≥ 30 ] =  0,8413      or    84,13%

b)  P [X < 24]  = 0,0228      or    2,28 %

c) P [ 24 < X < 48 ]   =  0,9544     or   95,44%

Step-by-step explanation:

z = ( X - μ₀ )/σ

μ₀ the mean ( average no. of months that an employee stay in a factory)

σ standard deviation

a) P [X ≥ 30 ] = 1 -  P [X < 30 ]

P [X < 30 ]

We look for z (score)

z = ( X - μ₀ )/σ      ⇒  z =  30 - 36 / 6

z = - 1

From z table we get for -1  

P [X < 30 ] = 0,1587

And

P [X ≥ 30 ] = 1 -  P [X < 30 ]  ⇒        P [X ≥ 30 ] = 1 - 0,1587

P [X ≥ 30 ] =  0,8413      or    84,13%

b) P [X < 24]

z (score) = ( 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 ]    is  P[X ≤ 48] -  P[X ≤ 24]

P [X < 48]

s (score) = 48 - 36 / 6      ⇒   z = 2

P [X < 48] = 0,9772

Then

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

P [ 24 < X < 48 ]   =  0,9544     or   95,44%

answer: -2+√x,-2-√x

i hope this , i don't know if you're looking for a pair or not.