Question

A bank has kept records of the checking balances of its customers and determined that the average daily balance of its customers is $300 with a standard deviation of $50. A random sample of 145 checking accounts is selected.

a) What is the probability that the sample mean will be more than $307.60?

b) What is the probability that the sample mean will be between $300 and $310?

c) What is the probability that the sample mean will be at least $296?

Answer 1

Solution :

Given that ,

mean = = $300

standard deviation = = $50

n =145

= 300

= / n = 50/ 145= 4.1523

P( >$307.60 ) = 1 - P( < 307.60)

= 1 - P[( - ) / < (307.60-300) / 4.1523 ]

= 1 - P(z <1.83 )

Using z table

= 1 - 0.9664

= 0.0336

probability= 0.0336

(B)

P($300<     <$310 ) = P[(300-300) / 4.1523 < ( - ) /   <(310-300) / 4.1523 )]

= P( 0< Z <2.41 )

= P(Z <2.41 ) - P(Z <0)

Using z table

=0.992-0.5

=0.4920

probability=0.4920

(C)

P( >$296 ) = 1 - P( < 296)

= 1 - P[( - ) / < (296-300) / 4.1523 ]

= 1 - P(z <-0.96)

Using z table

= 1 - 0.1685

= 0.8315

probability= 0.8315