Standardization (z-values); inverse normal practice questions.

The random variable X is normally distributed with mean 60 and standard deviation 5. Calculate the z-value corresponding to x = 70.

• 2
• 10
• 0.5
• −2

Answer: 2

z = (x − μ)/σ = (70 − 60)/5 = 2, meaning 70 lies two standard deviations above the mean. Forgetting to divide by σ gives 10.

A normal variable has mean 40 and standard deviation 6. Determine the value of x whose z-value is −1.5.

• 49
• 31
• 39
• 25

Answer: 31

x = μ + zσ = 40 + (−1.5)(6) = 40 − 9 = 31. Getting 49 comes from ignoring the negative sign of z.

The random variable X is normal with mean μ and standard deviation σ. State the distribution of Z = (X − μ)/σ.

• N(μ, 1)
• N(0, σ²)
• N(0, 1)
• N(1, 0)

Answer: N(0, 1)

Standardizing any normal variable produces the standard normal distribution Z ~ N(0, 1): subtracting μ centres it at 0 and dividing by σ scales the spread to 1.

Let Z be the standard normal variable and a > 0. Determine which expression is equal to P(Z < −a).

• P(Z < a)
• 1 − P(Z > a)
• −P(Z < a)
• P(Z > a)

Answer: P(Z > a)

By the symmetry of the standard normal curve about 0, the tail below −a has the same area as the tail above a, so P(Z < −a) = P(Z > a) = 1 − P(Z < a).

The random variable X ~ N(100, 15²) and P(Z < 1.28) ≈ 0.90 for the standard normal variable Z. Determine the value of k such that P(X < k) = 0.90.

• 80.8
• 119.2
• 101.3
• 134.5

Answer: 119.2

Unstandardize: k = μ + zσ = 100 + 1.28 × 15 = 119.2. Subtracting instead of adding gives 80.8; adding z without multiplying by σ gives 101.3.

For any normal distribution, state the approximate proportion of values lying within one standard deviation of the mean.

• 0.68
• 0.50
• 0.95
• 0.997

Answer: 0.68

About 68% of a normal population lies within one standard deviation of the mean; 95% within two and 99.7% within three.