"A day care program has an average daily expense of $75.00. The standard deviation is $5.00. The owner takes a sample of 64 bill
2 answers:
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Answer:
0.1855
Step-by-step explanation:
Assuming normal distribution, Sampling mean = $190
Sampling standard deviation = = $2
a) Probability that the average annual cost of a random sample of 100 such checking accounts at U.S. community banks is less than $187 = P(X < 187)
= P(Z < (187 - 190)/2)
= P(Z < -1.5)
= 0.0668
b) Probability that the average annual cost of a random sample of 100 such checking accounts at U.S. community banks is more than $193.5
= 1 - P(X < 193.5)
= 1 - P(Z < (193.5 - 190)/2)
= 1 - P(Z < 1.75)
= 1- 0.9599
= 0.0401
c) Probability that the average annual cost of a random sample of 100 such checking accounts at U.S. community banks is between $191.70 to 194.5 = P(X < 194.5) - P(X <191.7)
= P(Z < (194.5 - 190)/2) - P(Z < (191.7-190)/2)
= P(Z < 2.25) - P(Z < 0.85)
= 0.9878 - 0.8023
= 0.1855