The average credit score in Canada is 650. Assume that credit scores follow a normal distribution with a standard deviation of 80. (Round each to nearest integer)

Question

a) Find the 85th percentile of credit scores in Canada.

b) 65% of Canadians have a credit score higher than what value?

c) Mac’s credit score is 720. In what percentile is his credit score?

Answer 1

With TI-84:

A) inverse norm(mean=650, standard deviation=80, area to left=0.85) = 732.91

B) inverse norm(mean=650, standard deviation=80, area to left=0.65) = 680.83

C) normalcdf(inverse norm(mean=650, standard deviation=80, lower bound -10^99, upperbound=720) =.81

81st percentile

Answer 2

will be using standardized z-score equation to do this problem:

z = (x - mu)/std dev, where mu = 650 and std dev = 80, so z = (x - 650)/80

a) x = credit score that is the 85th percentile

determine z score for 85th percentile from std normal prob table = 1.03

plugging value of z into equation:

1.03 = (x - 650)/80

1.03 * 80 + 650 = x

x = 732.4 ~ 732

b) If 65% have a score higher then we need to find score above what percentile --- the 35th.

z score for 35th percentile = -0.38

plugging value of z into equation:

-0.38 = (x - 650)/80

-0.38 * 80 + 650 = x

x = 619.6 ~ 620

c) here we are given x and want to find z.

plugging value of 720 for x into the equation:

z = (720 - 650)/80

z = 0.875

Look up probability corresponding to 0.875 in standard normal probability table:

0.8089

multiplying by 100 to get percentile: 80.89 ~ 81st percentile