Question

1. According RCA 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. What is the probability that a randomly selected tax return refund from the 2011 tax year will be

(a) more than $2,000?

(b) between $3,200 and $4,000?

Answer 1

Answer:

$\mu$= 2913, $\sigma$ = 950

a)

We want to find P(x > 2000)

formula for z-score is

τ – μ

2000-2913 950

z =−0.9611

P(x > 2000) = 1 - P(Z < -0.96)

find P(Z < -0.96) using normal z table we get

P(Z < -0.96) = 0.1683

P(x > 2000) = 1 -  0.1683

P(x > 2000) = 0.8317

Probability = 0.8317

b)

We want to find P( 3200 < x < 4000 )

P( 3200 < x < 4000 ) = P(x < 4000) - P(x < 3200)

first find P(x < 4000)

formula for z-score is

τ – μ

4000 - 2913 950

z =1.1442

P(x < 4000) = P(Z < 1.14)

find P(Z < 1.14) using normal z table we get

P(Z < 1.14) = 0.8737

P(x < 4000) = 0.8737

now find P(x < 3200)

formula for z-score is

τ – μ

3200-2913 950

z =0.3021

P(x < 3200) = P(Z < 0.30)

find P(Z < 0.30) using normal z table we get

P(Z < 0.30) = 0.6187

P(x < 3200) = 0.6187

P( 3200 < x < 4000 ) = P(x < 4000) - P(x < 3200)

P( 3200 < x < 4000 ) = 0.8737−0.6187

P( 3200 < x < 4000 ) = 0.2550

Probability = 0.2550