Homework Sourse2 The recent average starting salary for new college graduat
2- The recent average starting salary for new college graduates in computer information systems is $47,500. Assume salaries are normally distributed with a standard deviation of $4,500.
What is the probability of a new graduate receiving a salary between $45,000 and $50,000?
What is the probability of a new graduate getting a starting salary at least $55,000?
What percent of starting salaries that are no more than $42,250?
Solution
a)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 45000
x2 = upper bound = 50000
u = mean = 47500
s = standard deviation = 4500
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.555555556
z2 = upper z score = (x2 - u) / s = 0.555555556
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.289257361
P(z < z2) = 0.710742639
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.421485278 [ANSWER]
*********************
b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 55000
u = mean = 47500
s = standard deviation = 4500
Thus,
z = (x - u) / s = 1.666666667
Thus, using a table/technology, the right tailed area of this is
P(z > 1.666666667 ) = 0.047790352 [ANSWER]
*************************
c)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 42250
u = mean = 47500
s = standard deviation = 4500
Thus,
z = (x - u) / s = -1.166666667
Thus, using a table/technology, the left tailed area of this is
P(z < -1.166666667 ) = 0.121672505 [ANSWER]
