The weekly salaries of elementary school teachers are normally distributed with

Question

The weekly salaries of elementary school teachers are normally distributed with a mean of $490 and a standard deviation of $45 (i.e., not a sample).

a. What is the probability that a randomly selected school teacher earns MORE than $415 a week (to 2 decimal places)?

b. What is the 30th percentile of the weekly salary of elementary school teachers (to ONE decimal place)?

c. What is the probability that the mean salary of a random sample of 30 school teachers is less than $500 a week (to 2 decimal places)?

d. In the previous question, you calculated the probability that the mean salary of a random sample of 30 school teachers is less than $500 a week. Consider using a sample of 300 school teachers instead of 30. How would we expect the probability we calculated (that the mean salary is less than $500 a week) to change?

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 490
standard Deviation ( sd )= 45
a.
GREATER THAN
P(X > 415) = (415-490)/45
= -75/45 = -1.6667
= P ( Z >-1.6667) From Standard Normal Table
= 0.9522
b.
P ( Z < x ) = 0.3
Value of z to the cumulative probability of 0.3 from normal table is -0.524401
P( x-u/s.d < x - 490/45 ) = 0.3
That is, ( x - 490/45 ) = -0.524401
--> x = -0.524401 * 45 + 490 = 466.401977
c.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 490
standard Deviation ( sd )= 45/ Sqrt ( 30 ) =8.2158
sample size (n) = 30
P(X < 500) = (500-490)/45/ Sqrt ( 30 )
= 10/8.2158= 1.2172
= P ( Z <1.2172) From Standard NOrmal Table
= 0.89
d.
P(X < 500) = (500-490)/45/ Sqrt ( 300 )
= 10/2.5981= 3.849
= P ( Z <3.849) From Standard NOrmal Table
= 1

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