In: Statistics and Probability
A bank has kept records of the checking balances of its 4000 customers and determined that the average daily balance of its customers is $300 with a standard deviation of $48. A random sample of 144 checking accounts is selected. Please answer the following questions.
(a) What is the probability that the sample mean will be more than $309?
(b) What is the probability that the sample mean will be between $291 and $307?
(c) Suppose the number of customers decreases from 4000 to 2000. What is the probability that the sample mean will be between $296 and $306?
a)
0.05N = 4000*0.05=200
n<0.05N, so, population finite correction factor will not be used here,
µ = 300
σ = 48
n= 144
X = 309
Z = (X - µ )/(σ/√n) = ( 309
- 300 ) / ( 48 /
√ 144 ) =
2.250
P(X ≥ 309 ) = P(Z ≥
2.25 ) = P ( Z <
-2.250 ) = 0.0122
(answer)
b)
we need to calculate probability for ,
291 ≤ X ≤ 307
X1 = 291 , X2 =
307
Z1 = (X1 - µ )/(σ/√n) = ( 291 -
300 ) / ( 48 / √
144 ) = -2.25
Z2 = (X2 - µ )/(σ/√n) = ( 307
- 300 ) / ( 48 /
√ 144 ) =
1.75
P ( 293 < X <
307 ) = P ( -2.25< Z <
1.75 )
= P ( Z < 1.75 ) - P ( Z
< -1.75 ) = 0.988 -
0.040 = 0.948
(answer)
c)
we need to calculate probability for ,
P ( 296 < X < 306
)
=P( (296-300)/( 48 / √144 )< Z <
(306-300)/( 48 / √144)
P ( -1 < Z < 1.5 )
= 0.933 - 0.159
= 0.774
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