Question

A census study was conducted on base salary of brand managers in Houston, TX as well as in Los Angeles, CA. The census data on base salaries for brand managers in Houston, TX is normally distributed with mean $89,250 and standard deviation $21,325. The census data on base salaries for brand managers in Los Angeles, CA is normally distributed with mean $99,500 and standard deviation $19,675 a) What are the parameter values for the base salary of brand managers in Los Angeles, CA? 1. b) What are the parameter values for the base salary of brand managers in Houston, TX? 2. c) What is the probability that a brand manager in Los Angeles has a base salary in excess of $92,150? 3. d) What is the probability that a brand manager in Houston has a base salary in excess of $89,175 but less than $107,825 4. e) What is the probability that a brand manager in Los Angeles has a base salary of less than $101,750? 5. f) Ms. Jane Roberts is to be hired as a Brand Manager. In order make the offer attractive to her, what should Ms. Jane Roberts salary be so that 92.5% of the brand managers in Los Angeles, CA earn a salary less than what she is to be offered? 6. g) What percentage of brand managers in Houston earn a salary that is higher than what Ms. Jane Roberts is to be offered?

Answer 1

a)
Los Angeles, CA
mean = 99500
sd = 19675

b)
Houston, TX
mean = 89250
sd = 21325

c)
P(X > 92150)
= P(z > (92150 - 99500)/19675)
= P(z > -0.3736)
= 0.6456

d)
P(89175 < X < 107825)
= ((89175 - 89250)/21325 < z < (107825 - 89250)/21325))
= (-0.0035 < z < 0.8710)
= P(z < 0.8710) - P(z < -0.0035)
= 0.8081 - 0.4986
= 0.3095

e)
P(X < 101750)
= P(z < (101750 - 99500)/19675)
= P(z < 0.1144)
= 0.5455

f)
z-value = 1.4395
x = 99500 + 1.4395*19675
x = 127822.16

g)
P(X > 127822.16)
= (z > (127822.16 - 89250)/21325))
= (z > 1.8088)
= 0.0352

i.e. 3.52%