Let X1, X2, … , X24 be independent observations from a random sample of size 24 from the uniform distribution
𝑈
(
0
,
1
)
with pdf
𝑓
(
𝑥
)
=
1
, and support
0

Question

Let X1, X2, … , X24 be independent observations from a random sample of size 24 from the uniform distribution 
𝑈
(
0
,
1
)
with pdf 
𝑓
(
𝑥
)
=
1
, and support 
0
<
𝑥
<
1
.

(a) The (1) mean, 
𝜇
, of the uniform distribution, (2) variance, 
𝜎
2
, of the uniform distribution, (3) mean, 
𝜇
𝑋
¯
, of the sampling distribution, and (4) variance, 
𝜎
𝑋
¯
2
, of the sampling distribution are: (1) 1/2, (2) 1/12, (3) 1/2, (4) 1/288

(b) Using a normal approximation to find 
𝑃
(
1
/
2
≤
𝑋
¯
≤
5
/
8
)
, the answer is: 0.483

BlackTom AI · Accepted Answer

Mean of a Uniform(a,b) is (a+b)/2, so for U(0,1) μ = (0+1)/2 = 1/2. Variance of U(a,b) is (b−a)^2/12, so for U(0,1) σ^2 = (1−0)^2/12 = 1/12. The sampling distribution of the sample mean X̄ has mean equal to the population mean, so μ_X̄ = μ = 1/2. Its variance is σ^2/n = (1/12)/24 = 1/(12·24) = 1/288. This verifies part (a): (1) 1/2, (2) 1/12, (3) 1/2, (4) 1/288.

For part (b) use the Central Limit Theorem: for n = 24 the distribution of X̄ is approximately Normal(μ_X̄, σ^2_X̄) = Normal(1/2, 1/288). We want P(1/2 ≤ X̄ ≤ 5/8). Compute the standard deviation of X̄:
σ_X̄ = sqrt(1/288) = 1/√288 = 1/(12√2) ≈ 0.0589256.
Convert endpoints to standard normal z-scores:
z_lower = (1/2 − 1/2)/σ_X̄ = 0/0.0589256 = 0.
z_upper = (5/8 − 1/2)/σ_X̄ = (0.625 − 0.5)/0.0589256 = 0.125/0.0589256 ≈ 2.1213203.
Thus P(1/2 ≤ X̄ ≤ 5/8) = Φ(z_upper) − Φ(z_lower) = Φ(2.1213203) − Φ(0).
From standard normal tables or calculator Φ(2.1213) ≈ 0.9830 and Φ(0) = 0.5, so the probability ≈ 0.9830 − 0.5 = 0.4830 ≈ 0.483. This confirms part (b).

Why the choices above are correct and typical mistakes: the key is using the correct population variance 1/12 and then dividing by n (not by √n) to get the variance of X̄. A common error would be using σ/n instead of σ/√n when standardizing, which would produce wrong z-values. Using CLT is appropriate here because the underlying distribution is continuous and n=24 is moderate; for exact probabilities one could use the Irwin–Hall distribution for sums of uniforms, but the normal approximation gives the stated 0.483.

Final answer confirmed: (a) (1) 1/2, (2) 1/12, (3) 1/2, (4) 1/288; (b) 0.483.

Similar Questions

"Why is the Central Limit Theorem so important anyway?" asks Quackers, somewhat irritated. "It seems too complex and abstract to be of any relevance to anything we actually care about." What would you tell him?

"Why is the Central Limit Theorem so important anyway?" asks Quackers, somewhat irritated. "It seems too complex and abstract to be of any relevance to anything we actually care about." What would you tell him?

The distribution of gestational times of zebra fish is skewed left with a mean of 6.5 days and standard deviation of 0.8 days. If we take repeated samples of size 50, we know the distribution of the sample mean is approximately normal here because of

Which of the following best describes the Central Limit Theorem in the context of a sampling distribution of the sample mean?

For which of the following situations would the Central Limit Theorem for Sample Means APPLY?Select ALL correct answers:

The central limit theorem says that when a simple random sample of size n is drawn from any population with mean μ and standard deviation σ, then when n is sufficiently largeSelect ALL correct answers

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler