Normal distribution practice questions.

The masses of apples in an orchard are normally distributed with mean 50 g and standard deviation 4 g. Determine the approximate probability that a randomly chosen apple has a mass less than 58 g.

• 84.1%
• 95.4%
• 97.7%
• 99.9%

Answer: 97.7%

58 g is two standard deviations above the mean, so P(X < 58) = 0.5 + 0.477 ≈ 0.977, giving about 97.7%.

Test scores are modelled by X ~ N(60, 5²). Calculate the standardized value (z-score) corresponding to a score of 70.

• 0.5
• 2
• −2
• 10

Answer: 2

z = (x − μ)/σ = (70 − 60)/5 = 2; the score lies two standard deviations above the mean.

X is normally distributed with mean 30. It is known that P(X < 24) = 0.2. Determine P(X > 36).

• 0.2
• 0.3
• 0.5
• 0.8

Answer: 0.2

24 and 36 are the same distance from the mean 30, so by the symmetry of the normal curve P(X > 36) = P(X < 24) = 0.2.

State the transformation that converts a normal random variable X with mean μ and standard deviation σ into the standard normal variable Z.

• Z = (μ − X)/σ
• Z = (X + μ)/σ
• Z = X/σ − μ
• Z = (X − μ)/σ

Answer: Z = (X − μ)/σ

Standardizing subtracts the mean and divides by the standard deviation: Z = (X − μ)/σ, giving Z ~ N(0, 1).

IQ scores follow N(100, 15²). Given that P(Z < 1.28) ≈ 0.90 for the standard normal variable Z, determine the 90th percentile of the IQ scores.

• 80.8
• 101.28
• 119.2
• 128

Answer: 119.2

The 90th percentile is μ + zσ = 100 + 1.28 × 15 = 119.2.

Which of the following is NOT a property of a normal distribution curve?

• It is symmetric about the mean
• The total area under the curve equals 1
• The mean, median and mode are all equal
• It is positively skewed

Answer: It is positively skewed

A normal curve is perfectly symmetric, so it has no skew; skewness would contradict mean = median = mode.