Question

1)

Assume that a sample is used to estimate a population mean μμ. Find the 99.9% confidence interval for a sample of size 585 with a mean of 74.4 and a standard deviation of 16.6. Enter your answers accurate to four decimal places.

Confidence Interval:

2) You measure 20 textbooks' weights, and find they have a mean weight of 74 ounces. Assume the population standard deviation is 10.9 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight.

Keep 4 decimal places of accuracy in any calculations you do. Report your answers to four decimal places.

Confidence Interval:

3)

A population of values has a normal distribution with μ=146.5μ=146.5 and σ=53.4σ=53.4. You intend to draw a random sample of size n=114n=114.

Find the probability that a single randomly selected value is less than 141.
P(X < 141) =

Find the probability that a sample of size n=114n=114 is randomly selected with a mean less than 141.
P(¯xx¯ < 141) =  Enter your answers as numbers accurate to 4 decimal places.

4)

CNNBC recently reported that the mean annual cost of auto insurance is 987 dollars. Assume the standard deviation is 257 dollars. You take a simple random sample of 94 auto insurance policies.

Find the probability that a single randomly selected value is less than 968 dollars.
P(X < 968) =

Find the probability that a sample of size n=94n=94 is randomly selected with a mean less than 968 dollars.
P(¯xx¯ < 968) =

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

1)
Solution
sample mean, xbar = 74.4
sample standard deviation, s = 16.6
sample size, n = 585
degrees of freedom, df = n - 1 = 584

Given CI level is 99.9%, hence α = 1 - 0.999 = 0.001
α/2 = 0.001/2 = 0.0005, tc = t(α/2, df) = 3.307

ME = tc * s/sqrt(n)
ME = 3.307 * 16.6/sqrt(585)
ME = 2.2697

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (74.4 - 3.307 * 16.6/sqrt(585) , 74.4 + 3.307 * 16.6/sqrt(585))
CI = (72.1303 , 76.6697)

2)
sample mean, xbar = 74
sample standard deviation, σ = 10.9
sample size, n = 20

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 10.9/sqrt(20)
ME = 4.8

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (74 - 1.96 * 10.9/sqrt(20) , 74 + 1.96 * 10.9/sqrt(20))
CI = (69.2229 , 78.7771)

3)

Here, μ = 146.5, σ = 53.4 and x = 141. We need to compute P(X <= 141). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (141 - 146.5)/53.4 = -0.1

Therefore,
P(X <= 141) = P(z <= (141 - 146.5)/53.4)
= P(z <= -0.1)
= 0.4602

4)

Here, μ = 987, σ = 26.5075 and x = 968. We need to compute P(X <= 968). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (968 - 987)/26.5075 = -0.72

Therefore,
P(X <= 968) = P(z <= (968 - 987)/26.5075)
= P(z <= -0.72)
= 0.2358