Question

Bank tellers report that when deciding on where to work, career growth, salary and compensation, location and commute, and company culture and values are important factors to them. According to a blog "25 Highest Paying Companies" the mean monthly pay of bank tellers is $4648. Suppose that the monthly pay is normally distributed with a standard deviation of $400. What is the probability that the monthly pay of a bank teller is...

a. less than $4000
b. between $4200 and $4800
c. above $5100
d. Ninety nine percent of the monthly pays are higher than what value?

Answer 1

Answer:

a)

Given,

P(X < 4000) = P((x-u)/(s/sqrt(n)) < (4000 - 4648)/(400/sqrt(25)))

= P(z < -8.1)

= 0 [since from z table]

= 0

b)

P(4200 < X < 4800) = P((4200 - 4648)/(400/sqrt(25)) < (x-u)/(s/sqrt(n)) < (4800 - 4648)/(400/sqrt(25)))

= P(-5.6 < z < 1.9)

= P(z < 1.9) - P(z < - 5.6)

= 0.9712834 - 0 [since from z table]

= 0.9713

c)

P(X > 5100) = P((x-u)/(s/sqrt(n)) > (5100 - 4648)/(400/sqrt(25)))

= P(z > 5.65)

= 0

d)

Here at 99% CI, z value is - 2.33

consider,

z = (x-u)/s

substitute values

- 2.33 = (x - 4648)/80

x - 4648 = - 2.33*80

x = 4648 - 186.4

x = 4461.6