The average teacher’s salary is $45,000. Assume a normal distribution with standard deviation equal to $10,000. a) What is the probability that a randomly selected teacher makes greater than $65,000 a year? b) If we sample 100 teachers’ salaries, what is the probability that the sample mean will be greater than $65,000?
a)
Given,
= 45000 ,
=
10000
We convert this to standard normal as
P( X < x) = P( Z < x - /
)
So,
P( X > 65000) = P( Z > 65000 - 45000 / 10000)
= P( Z > 2)
= 0.0228 (Probability calculated from Z table)
b)
Using central limit theorem,
P( < x) = P
(Z < x -
/
/ sqrt(n)
)
So,
P( > 65000)
= P( Z > 65000 - 45000 / 10000 / sqrt(100) )
= P( Z > 20)
= 0
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