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 $840 and a standard deviation of $275. 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 $950 and $1450? Round your answer to three decimal places.

P = ______

b. What percentage of the Visa cards issued by this bank have a balance of $735 or more? Round your answer to two decimal places.

P= ___________%

Answer 1

Solution :

Given that ,

mean = = $840

standard deviation = = $275

P($950< x <$1450 ) = P[(950-840) /275 < (x - ) / < (1450 -840) / 275)]

= P(0.4<Z < 2.22)

= P(Z <2.22 ) - P(Z <0.4 )

Using z table   

= 0.9868-0.6554

probability= 0.3314

(b)

P(x >$735 ) = 1 - P(x<735 )

= 1 - P[(x -) / < (735-840) /275) ]

= 1 - P(z < -0.38)

Using z table

= 1 - 0.352

= 0.6480

answer= 64%80