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.

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