Answer:
a.0.8664
b. 0.23753
c. 0.15866
Step-by-step explanation:
The comptroller takes a random sample of 36 of the account balances and calculates the standard deviation to be N42.00. If the actual mean (1) of the account balances is N175.00, what is the probability that the sample mean would be between
a. N164.50 and N185.50?
b. greater than N180.00?
c. less than N168.00?
We solve the above question using z score formula
z = (x-μ)/σ/√n where
x is the raw score,
μ is the population mean = N175
σ is the population standard deviation = N42
n is random number of sample = 36
a. Between N164.50 and N185.50?
For x = N 164.50
z = 164.50 - 175/42 /√36
z = -1.5
Probability value from Z-Table:
P(x = 164.50) = 0.066807
For x = N185.50
z = 185.50 - 175/42 /√36
z =1.5
Probability value from Z-Table:
P(x=185.50) = 0.93319
Hence:
P(x = 185.50) - P(x =164.50)
= 0.93319 - 0.066807
= 0.866383
Approximately = 0.8664
b. greater than N180.00?
x > N 180
Hence:
z = 180 - 175/42 /√36
z = 5/42/6
z = 5/7
= 0.71429
Probability value from Z-Table:
P(x<180) = 0.76247
P(x>180) = 1 - P(x<180) = 0.23753
c. less than N168.00?
x < N168.
z = 168 - 175/42 /√36
z = -7/42/6
z = -7/7
z = -1
Probability value from Z-Table:
P(x<168) = 0.15866
