Question

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 $48. A random sample of 144 checking accounts is selected. a. What is the probability that the sample mean will be more than $306.60? b. What is the probability that the sample mean will be less than $308? c. What is the probability that the sample mean will be between $302 and $308? d. What is the probability that the sample mean will be at least $296?

Answer 1

a)

Here, μ = 300, σ = 4 and x = 306.6. We need to compute P(X >= 306.6). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (306.6 - 300)/4 = 1.65

Therefore,
P(X >= 306.6) = P(z <= (306.6 - 300)/4)
= P(z >= 1.65)
= 1 - 0.9505 = 0.0495

b)

Here, μ = 300, σ = 4 and x = 308. We need to compute P(X <= 308). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (308 - 300)/4 = 2

Therefore,
P(X <= 308) = P(z <= (308 - 300)/4)
= P(z <= 2)
= 0.9772

c)

Here, μ = 300, σ = 4, x1 = 302 and x2 = 308. We need to compute P(302<= X <= 308). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (302 - 300)/4 = 0.5
z2 = (308 - 300)/4 = 2

Therefore, we get
P(302 <= X <= 308) = P((308 - 300)/4) <= z <= (308 - 300)/4)
= P(0.5 <= z <= 2) = P(z <= 2) - P(z <= 0.5)
= 0.9772 - 0.6915
= 0.2857

d)

Here, μ = 300, σ = 4 and x = 296. We need to compute P(X >= 296). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (296 - 300)/4 = -1

Therefore,
P(X >= 296) = P(z <= (296 - 300)/4)
= P(z >= -1)
= 1 - 0.1587 = 0.8413