Sums of normal variables; central limit theorem practice questions.

X ~ N(50, 8²) and Y ~ N(30, 6²) are independent. The distribution of X + Y is normal. Determine its mean and variance.

• mean 80, variance 100
• mean 80, variance 14
• mean 20, variance 100
• mean 80, variance 196

Answer: mean 80, variance 100

Means add: 50 + 30 = 80. Variances add: 8² + 6² = 64 + 36 = 100. The sum of independent normals is normal.

X ~ N(100, 25) is the population. For samples of size n = 4, determine the distribution of the sample mean X̄.

• N(100, 25)
• N(100, 6.25)
• N(25, 6.25)
• N(100, 100)

Answer: N(100, 6.25)

X̄ ~ N(μ, σ²/n) = N(100, 25/4) = N(100, 6.25); the mean is unchanged but the variance is divided by n.

The weight of a component has mean 200 g and standard deviation 10 g (any distribution). For a random sample of 100 components, determine the standard deviation (standard error) of the sample mean weight.

• 10 g
• 0.1 g
• 1 g
• 0.01 g

Answer: 1 g

Standard error = σ/√n = 10/√100 = 10/10 = 1 g.

The Central Limit Theorem allows the sample mean to be treated as approximately normal. Which condition makes this approximation reliable when the population is not normal?

• The sample size n is large (n ≥ 30)
• The population variance is exactly 1
• The population mean is 0
• The sample size n is less than 5

Answer: The sample size n is large (n ≥ 30)

The CLT states that for a sufficiently large sample size (rule of thumb n ≥ 30), the distribution of X̄ is approximately normal regardless of the population distribution.

X ~ N(60, 16) and Y ~ N(45, 9) are independent. Determine the distribution of D = X − Y.

• N(15, 7)
• N(15, 25)
• N(105, 25)
• N(15, 5)

Answer: N(15, 25)

Mean = 60 − 45 = 15; variance = 16 + 9 = 25 (variances add even for a difference). So D ~ N(15, 25).

The heights of adult men are N(178, 7²) cm, independently. Two men are chosen at random. Determine the distribution of the difference in their heights, D = H₁ − H₂.

• N(0, 49)
• N(0, 98)
• N(0, 14)
• N(0, 7)

Answer: N(0, 98)

Mean = 178 − 178 = 0; variance = 7² + 7² = 49 + 49 = 98. So D ~ N(0, 98).