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%

sup

step-by-step explanation:

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )