In: Statistics and Probability
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?
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