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kdenormandie3122
18.10.2020 •
Mathematics
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
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Ответ:
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%
Ответ:
538,757
Step-by-step explanation:
Population=(current population)*(1+rate)^(year)
Population=(300000)*(1+5/100)^12=538,756.899 or 538,757. Love from Gauthmath