Question

The Bank of Connecticut issues Visa and MasterCard credit cards. It is estimated that the balances on all Visa credit cards issued by the Bank of Connecticut have a mean of $845 and a standard deviation of $260 . Assume that the balances on all these Visa cards follow a normal distribution. a. What is the probability that a randomly selected Visa card issued by this bank has a balance between $1100 and $1420? Round your answer to three decimal places. b. What percentage of the Visa cards issued by this bank have a balance $725 of or more? Round your answer to two decimal places.

Answer 1

Solution:

Given that,

mean = = 845

standard deviation = = 260

a ) p (1100 < x < 1450 )

= p ( 1100 - 845 / 260 ) < ( x -  / ) < ( 1450 - 845 / 260)

= p (255 / 260< z < 605 / 260 )

= p (0.98 < z < 2.33)

= p ( z < 2.33) - p ( z < 0.98 )

Using z table

= 0.9901 - 0.8365

= 0.154

Probability = 0.154

c ) p ( x > 725 )

= 1 - p (x < 725 )

= 1 - p ( x -  / ) < ( 725 - 845 / 260)

= 1 - p ( z < - 120 / 260 )

= 1 - p ( z < -0.46)

Using z table

= 1 - 0.3228

= 0.6742

Probability = 0.6772 = 67.72%

The Visa cards issued by this is  67.72%