Question

A research company reports that the average monthly household cell phone bill in Indiana is $120 with a standard deviation of $14. Assuming the distribution of household cell phone bills is normally distributed find the following probabilities:

The probability that a randomly selected household in Indiana has a cell phone bill of more than $100.

The probability that a randomly selected household in Indiana has a cell phone bill of $110 to $150.

Suppose you are asked to take a random sample of 100 households in Indiana. What is the probability you get an average monthly household cell phone bill of less than $115?

Answer 1

Given,

= 120 , = 14

We convert this to standard normal as

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

a)

P( X > 100) P( Z > 100 - 120 / 14)

= P( Z > -1.4286)

= P( Z < 1.4286)

= 0.9234

b)

P(110 < X < 150) = P( X < 150) - P( X < 110)

= P( Z < 150 - 120 / 14) - P( Z < 110 - 120 / 14)

= P( Z < 2.1429) - P (Z < -0.7143)

= P( Z < 2.1429) - ( 1 - P( Z < 0.7143) )

= 0.9839 - ( 1 - 0.7625)

= 0.7464

c)

Using central limit theorem,

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

So,

P( < 115) = P( Z < 115 - 120 / (14 / sqrt(100) ) )

= P( Z < -3.57)

= ( 1 - P( Z< 3.57) )

= 1 - 0.9998

= 0.0002