Question

A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time (minutes) spent with each customer can be modeled by an uniform distribution that spans from 10 to 20 minutes.

a.) What is the variance?

b.) State f(x)

c.) Find P(12 < x < 15.5).

d.) If samples of size 20 are drawn, what is the probability that the sample mean is less than 13?

Answer 1

(a)

Uniform Distribution in (10,20) is given by:

                                   for 10 < X < 20

     = 0, otherwise

                                       between the limits 10 to 20.

Applying limits, we get:
E(X) = 15

between limits 10 to 20.

Applying limits, we get:
E(X²) = 233.3333

Variance = E(X²)-(E(X))² = 233.3333 - 15² = 8.3333

(b)

,

              10 < X < 20

       = 0, OTHERWISE

(c)

between limits 12 to 15.5

Applying limits, we get:

(d)

n = 20

SE = /

    = 2.8868/4.4721 = 0.6455

To find P( < 13):
Z = (13 - 15)/0.6455 = - 3.0984

Table of Area Under Standard Normal Curve gives area = 0.4990

So,

P(<13) = 0.5 - 0.4990 = 0.001