📝 Chapter 6: The Normal Distribution

Assessment AS Learning — Practice Quiz

🔄 Not Graded — Unlimited Retakes

Purpose: Self-check your work with z-scores, the empirical rule, the standard normal table, and the normal approximation to the binomial.

Score: 0 / 12

Topic 6.2 — Empirical Rule (68-95-99.7)

Question 1

Heights are normally distributed with \( \mu = 170 \) cm, \( \sigma = 10 \) cm. What percentage of people have heights between 160 and 180 cm?

Solution:

160 = \( \mu - \sigma \), 180 = \( \mu + \sigma \). Within 1 \( \sigma \): 68% by the empirical rule.

Question 2

For \( N(50, 5) \), about what percent of values fall between 35 and 65?

Solution:

35 = \( \mu - 3\sigma \), 65 = \( \mu + 3\sigma \). Within 3 \( \sigma \): 99.7% (or just 99.7).

Topic 6.3 — z-Scores

Question 3

A test has \( \mu = 75 \), \( \sigma = 8 \). A student scores 90. Find their z-score.

Solution:

\( z = \dfrac{x - \mu}{\sigma} = \dfrac{90 - 75}{8} = \dfrac{15}{8} = 1.875 \).

Question 4

A z-score of \( z = -1.5 \) on a test with \( \mu = 70 \), \( \sigma = 6 \) corresponds to what raw score?

Solution:

\( x = \mu + z\sigma = 70 + (-1.5)(6) = 70 - 9 = 61 \).

Question 5

For the standard normal distribution, \( P(Z < 1.5) \) ≈ ? (decimal to 4 places, use standard table)

Solution:

From standard normal table: \( P(Z < 1.5) = 0.9332 \).

Question 6

\( P(Z > 1.0) = ? \) (decimal to 4 places)

Solution:

\( P(Z > 1) = 1 - P(Z < 1) = 1 - 0.8413 = 0.1587 \).

Question 7

\( P(-1 < Z < 1) = ? \) (decimal to 4 places)

Solution:

\( P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 \) (matches the empirical rule's 68%).

Real-World Applications

Question 8

IQ is normally distributed with \( \mu = 100 \), \( \sigma = 15 \). What is \( P(\text{IQ} > 130) \)? (decimal to 4 places)

Solution:

\( z = \dfrac{130 - 100}{15} = 2.0 \). \( P(Z > 2) = 1 - 0.9772 = 0.0228 \).

Question 9

SAT scores: \( \mu = 1000 \), \( \sigma = 200 \). \( P(800 \le X \le 1200) = ? \) (decimal to 4 places)

Solution:

\( z_1 = \dfrac{800 - 1000}{200} = -1 \), \( z_2 = 1 \). \( P(-1 \le Z \le 1) = 0.6826 \).

Question 10

A factory produces bolts with diameter \( N(10\,\mathrm{mm}, 0.05\,\mathrm{mm}) \). A bolt is defective if its diameter is < 9.9 mm or > 10.1 mm. Find \( P(\text{defective}) \). (decimal to 4 places)

Solution:

\( z = \pm\dfrac{0.1}{0.05} = \pm 2 \). \( P(\text{defective}) = 1 - P(-2 \le Z \le 2) = 1 - 0.9544 = 0.0456 \).

Topic 6.4 — Normal Approximation to Binomial

Question 11

For \( X \sim \mathrm{Binomial}(100, 0.5) \), the normal approximation has \( \mu = ? \) and \( \sigma = ? \). State both.

Solution:

\( \mu = np = 100 \cdot 0.5 = 50 \). \( \sigma = \sqrt{np(1-p)} = \sqrt{25} = 5 \).

Question 12

A coin is flipped 100 times. Use the normal approximation to estimate \( P(X \ge 60) \) (continuity correction \( X \ge 59.5 \)). (decimal to 4 places)

Solution:

\( z = \dfrac{59.5 - 50}{5} = 1.9 \). \( P(Z \ge 1.9) = 1 - 0.9713 = 0.0287 \).