Question

According to the Internal Revenue Service, the average income tax refund for the 2011 tax year was $2,913. Assume the refund per person follows the normal probability distribution with a standard deviation of $950. a. What is the probability that a randomly selected tax return refund from the 2011 tax year will be 1. more than $ 2000 2. between $1,600 and $2,500? 3. between $3,200 and $4,000? b. Confirm the answers to part a using Excel or PH Stat.

Answer 1

Let X ~ Normal distribution with mean ( ) = 2913 and standard deviation ( ) = 950

1. Here we want to find P( X > 2000) = 1 - P( X < 2000) ......( 1 )

Let's find z score by using following formula.

From z table, the probability corresponding to the z score -0.96 ( because we need to round z score up to two decimal places)  is as follow:

P(Z < -0.96 ) = P( X < 2000) = 0.1686

Plug this in equation ( 1 ), we get

P( X > 2000) = 1 - 0.1686 = 0.8314

Let's use excel:

P( X > 2000) = "=1-NORMDIST(2000,2913,950,1)" = 0.831737

There is small difference between the table values and the software values, because when we use excel, we use exact values and not rounded values.

2. between $1,600 and $2,500?

Here we want to find P( 1600 < X < 2500 ) = P( X < 2500 ) - P( X < 1600) ...( 2)

Let's find z scores for x1 = 2500 and x2 = 1600

So that P( Z < -0.43) = 0.3336

So that P( Z < -1.38) = 0.0838

Plug these values in equation ( 2 ), we get

P( 1600 < X < 2500 ) = 0.3336 - 0.0838 = 0.2498

Let's use excel:

P( 1600 < X < 2500 ) = "=NORMDIST(2500,2913,950,1)-NORMDIST(1600,2913,950,1)" = 0.248407

3. between $3,200 and $4,000?

Here we want to find P( 3200 < X < 4000) = P( X < 4000) - P( X < 3200)

Let's use excel:

P( 3200 < X < 4000) = "=NORMDIST(4000,2913,950,1)-NORMDIST(3200,2913,950,1)" = 0.255018