Question

3. Recall that the distribution of the lengths of calls coming into a Boston, Massachusetts, call center each month is strongly skewed to the right. The mean call length is µ = 90 seconds and the standard deviation is σ= 120 seconds.

a. Let x be the sample mean from 10 randomly selected calls. What is the mean and standard deviation of xbar? What, if anything, can you say about the shape of the distribution of xbar? Explain.

b. Let x be the sample mean from 100 randomly selected calls. What is the mean and standard deviation of xbar? What, if anything, can you say about the shape of the distribution of xbar? Explain.

c. In a random sample of 100 calls from the call center, what is the probability that the average length of these calls will be over 2 minutes?

Answer 1

Given,

= 90, = 120

a)

For given n = 10,

= = 90

= / sqrt(n)

= 120 / sqrt(10)

= 37.9473

Since sample size is small and original distribution is skewed, the shape of distribution with

sample size n = 10 is skewed.

b)

For given n = 100,

= = 90

= / sqrt(n)

= 120 / sqrt(100)

= 12

Since sample size is Sufficinetly large (That is greater than 30) , the shape of distribution with

sample size n = 100 is approximately normal.

c)

We have to calculate P( > 2 minutes) = P( > 120 seconds) =?

The central limit thoerem states that

P( < x) = P( Z < x - / )

So,

P( > 120) = P( Z > 120 - 90 / 12)

= P( Z > 2.5)

= 1 - P( Z < 2.5)

= 1 - 0.9938

= 0.0062