Question

After deducting grants based on need, the average cost to attend the University of Southern California (USC) is $27,175. Assume the population standard deviation is $7,400. Suppose that a random sample of 69 USC students will be taken from this population.

(a)

What is the value of the standard error of the mean? (Round your answer to the nearest whole number.)

$

(b)

What is the probability that the sample mean will be more than $27,175?

(c)

What is the probability that the sample mean will be within $1,000 of the population mean? (Round your answer to four decimal places.)

(d)

What is the probability that the sample mean will be within $1,000 of the population mean if the sample size were increased to 100? (Round your answer to four decimal places.)

Answer 1

a)

std.errro = 7400/sqrt(69) = 890.8553

b)

Here, μ = 27175, σ = 890.8553 and x = 27175. We need to compute P(X >= 27175). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (27175 - 27175)/890.8553 = 0

Therefore,
P(X >= 27175) = P(z <= (27175 - 27175)/890.8553)
= P(z >= 0)
= 1 - 0.5 = 0.5

c)

Here, μ = 27175, σ = 890.8553, x1 = 26175 and x2 = 28175. We need to compute P(26175<= X <= 28175). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (26175 - 27175)/890.8553 = -1.12
z2 = (28175 - 27175)/890.8553 = 1.12

Therefore, we get
P(26175 <= X <= 28175) = P((28175 - 27175)/890.8553) <= z <= (28175 - 27175)/890.8553)
= P(-1.12 <= z <= 1.12) = P(z <= 1.12) - P(z <= -1.12)
= 0.8686 - 0.1314
= 0.7372

d)

Here, μ = 27175, σ = 740, x1 = 26175 and x2 = 28175. We need to compute P(26175<= X <= 28175). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (26175 - 27175)/740 = -1.35
z2 = (28175 - 27175)/740 = 1.35

Therefore, we get
P(26175 <= X <= 28175) = P((28175 - 27175)/740) <= z <= (28175 - 27175)/740)
= P(-1.35 <= z <= 1.35) = P(z <= 1.35) - P(z <= -1.35)
= 0.9115 - 0.0885
= 0.8230