(b) A department store has 7,000 charge accounts. 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 N164.50 and N185.50?
b. greater than N180.00?
less than N168.00?
a.
C.

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

chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.

the answer is c

hope this !