Question

Suppose that the ages of PSU (undergraduate) students is independently and normally distributed witha mean of 19 and a standard deviation of 2. Suppose a first sample of four PSU students and a second sample of four PSU students are selected at random 1. What is the distribution of the sample mean, X, of the first four students? 2. What is the distribution of the sample mean, Y, of the other four students? 3. What is the distribution of 2X-Y? Please show the details. Your answer should include the name of distribution and the values of parameters.

Answer 1

Solution

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ²,

then, X bar ~ N(µ, σ²/n). ……..……………………………………………………………............…….(1),

where X bar is average of a sample of size n from population of X.

If X ~ N(µ₁, σ₁²), Y ~ N(µ₂, σ₂²), then

(aX + bY) ~ N(µ, σ²) µ = aµ₁ + bµ₂, and σ² = a²σ₁² + b²σ₂² + 2abCov(x, y) …………………… (2a)

(aX - bY) ~ N(µ, σ²) µ = aµ₁ - bµ₂, and σ² = a²σ₁² + b²σ₂² - 2abCov(x, y) ……...……………… (2b)

If X and Yare independent, then

(aX + bY) ~ N(µ, σ²) µ = aµ₁ + bµ₂, and σ² = a²σ₁² + b²σ₂² …………………………....……..… (2c)

(aX - bY) ~ N(µ, σ²) µ = aµ₁ - bµ₂, and σ² = a²σ₁² + b²σ₂²…………………………....……..… (2d)

Now to work out the solution,

Given ages are Normally distributed with mean, µ = 19 and standard deviation, σ = 2, i.e., N(19, 2²)

Part (a)

Vide (1), X bar ~ N(19, 1²) Answer

[Note: n = 4]

Part (b)

By the same theory, Y bar ~ N(19, 1²)   Answer

Part (c)

Since the samples are independent, vide (2d),

(2Xbar - Ybar) ~ N[(2 x 19) – 19, (4 x 1) + (1 x 1)]

i.e., N(19, 5) Answer [Note: 5 is the variance]

[Here, a = 2, µ₁ = µ₂,b = 1, σ₁² = σ₂² = 1]

DONE