Question

According to the Internal Revenue Service, the mean tax refund for the year 2014 was $2,800. Assume the standard deviation is $450 and that the amounts refunded follow a normal probability distribution. What percent of the refunds are more than $3,100

Answer 1

Approximately 25.14% of the tax refunds are more than $3,100, calculated using the normal distribution and the z-score for a $3,100 refund.

Step-by-step explanation:

To determine what percent of tax refunds are more than $3,100 when the mean tax refund is $2,800 with a standard deviation of $450, we'll use the normal distribution properties.

First, we need to find the z-score for $3,100, which is calculated by subtracting the mean from the value and then dividing by the standard deviation. So, the z-score for a $3,100 refund is:

Z = ($3,100 - $2,800) / $450 ≈ 0.67

Now, using a z-table or normal distribution calculator, we can find the probability that a tax refund is less than $3,100, and then subtract this value from 1 to find the probability that a refund is more than $3,100.

The z-table gives us a probability of approximately 0.7486 that a refund is less than $3,100, so the probability that a refund is greater is:

1 - 0.7486 = 0.2514

This means approximately 25.14 percent of the tax refunds are more than $3,100.