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

25%

Explanation:

GIVEN THE FOLLOWING ;

mean tax refund(m) = $2800

Standard deviation(sd) = $450

Assuming a normal distribution;

Percentage of refunds greater than $3100.

Taking the value of X as $3100 and finding the standardized score(Z-score)

Z-score = (X - m) ÷ sd

Z-score = (3100 - 2800) ÷ 450

Z-score = 300 ÷ 450 = 0.67

Therefore,

P(X > 3100)=P(Z-score > 0.67)

Similarly written as;

1 - P(Z-score < 0.67):

Locating the intersection 0.6 on the y-axis and .07 on the x-axis of the Z-table. This is equal to 0.7486.

Therefore,

1 - P(Z-score < 0.67) = 1 - 0.7486 = 0.2514 (approximately 25%)

= 25%