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 $48, and is a normally distributed population.  A random sample of 144 checking accounts is selected.

a) What is the probability that the sample mean will be at least $306.60?

b) What is the probability that the sample mean will be less than $308?

c) What is the probability that the sample mean will be between $302 and $308?

Answer 1

Solution :

Given that ,

= 300

= / n = 48 / 144 = 4

a) P( 306.60) = 1 - P( 306.60)

= 1 - P[( - ) / (306.60 - 300) / 4]

= 1 - P(z 1.65)   

= 1 - 0.9505

= 0.0495

b) P( < 308 ) = P(( - ) / < (308 - 300) / 4)

= P(z < 2.00)

Using z table

= 0.9772

c) P(302 < < 308 )  

= P[(302 - 300) / 4 < ( - ) / < (308 - 300) / 4)]

= P( 0.50 < Z < 2.00)

= P(Z < 2.00) - P(Z < 0.50)

Using z table,  

= 0.9772 - 0.6915

= 0.2857