Respuesta :
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%
z = ( X - μ₀ )/σ
μ₀ the mean ( average no. of months that an employee stays 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%
To learn more about normal distribution:
https://brainly.com/question/18443127
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The right question is:
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
