Answer:
0.048 ; 0.03593 ; (14320, 45680) ; 6000
Step-by-step explanation:
Given a normal distribution :
Mean salary (m) = $30000
Standard deviation (s) = $8000
A.) probability of having a starting salary of atleast 30400
P(x ≥ 30400)
Obtain the standardized value (Z) :
Z = (x - m) / s
Z = (30400 - 30000) / 8000
Z = 400 / 8000 = 0.05
P(Z ≥ 0.05) = 0.48006 (Z probability calculator)
b. Individuals with starting salaries of less than $15,600 receive a low income tax break. What percentage of the graduates will receive the tax break?
P(X < 15600)
Obtain the standardized value (Z) :
Z = (x - m) / s
Z = (15600 - 30000) / 8000
Z = - 14400 / 8000 = - 1.8
P(Z < - 1.8) = 0.03593
c. what are the minimum and the maximum starting salaries of the middle 95% of the LU graduates?
Zscore which corresponds to 0.95 is 1.96
Minimum:
-1.96 = (x - 30000) / 8000
-15680 = x - 30000
x = 14320
Maximum :
1.96 = (x - 30000) / 8000
15680 = x - 30000
x = 45680
d. If 303 of the recent graduates have salaries of at least $43,120, how many students graduated this year from this university?
P(X ≥ 43120)
Z = (43120 - 30000) / 8000
Z = 13120 / 8000
Z = 1.64
P(Z ≥ 1.64) = 0.050503
Hence,
0.050503 = 303
1 = x
x = 303 / 0.05050
= 6000