Stats without Tears
Solutions for Chapter 7

1

• On any given trip, there’s a 9% chance that Chantal’s commute will be less than 17 minutes.
• 9% of Chantal’s commutes are shorter than 17 minutes.

2 P(x ≥ 76.5) = normalcdf(76.5, 10^99, 69.3, 2.92) = .0068362782.
Here are the two interpretations, from Interpreting Probability Statements in Chapter 5:

• The probability that a randomly selected man is 76.5″ or taller is 0.0068 or 0.68%.
• Only 0.68% of men are 76.5″ tall or taller.

3 “Have boundaries, find probability.”

P(64 ≤ x ≤ 67) = normalcdf(64, 67, 64.1, 2.75) = 0.3686871988 → 0.3687

36.87% of women are 64″ to 67″ tall.

4 5% probability in the two tails means 2.5% or 0.025 in each tail.

x₁ = invNorm(.025, 69.3, 2.92) = 63.57690516

x₂ = invNorm(1−.025, 69.3, 2.92) = 75.02309484

Heights under 63.6″ or over 75.0″ would be considered unusual.

5 The area to left is given as 15% or 0.15, and you need the boundary.

P15 = invNorm(.15, 69.3, 2.92) = 66.27361453

You must be at least 66″ or 5′6″ tall. Also acceptable: at least 66¼ inches, or at least 66.3 inches.

6 (a) By the definition of percentile, the number of the desired percentile is also the area to left.

P25 = invNorm(.25, 64.1, 2.75) = 62.24515319 → P25 = 62.2″

P75 = invNorm(.75, 64.1, 2.75) = 65.95484681 → P75 = 66.0″

(b) Q3 is P75 and Q1 is P25, so the IQR is P75−P25 = 65.95484681−62.24515319 = 3.70969362 → IQR = 3.7″.

(c) 1.35σ = 1.35×2.75 = 3.7125 → 3.7″, matching the IQR as expected. (The match isn’t perfect, because 1.35 is a rounded number.)

7 Use MATH200A Program part 4. The screens are shown at right. The points fall reasonably close to a line. r = 0.9595 and crit = 0.9383. r > crit, and therefore you can say that the normal model is a good fit to the data.

8 The percentile is the percent of the population that scored ≤735.

P(x ≤ 735) = normalcdf(−10^99, 735, 500, 100) = 0.9906.

A score of 735 is at the 99th percentile.

9 2% or 0.02 is area to right, but invNorm needs area to left, so you subtract from 1.

x₁ = invNorm(1−.02, 1500, 300) = 2116.124673

You must score at least 2117. (If you round to 2116, you get a number that is a bit less than the computed minimum. While rounding usually makes sense, there are situations where you have to round up, or round down, instead of following the usual rule.)

10 z_0.01 = invNorm(1−0.01, 0, 1) = 2.326347877 → z_0.01 = 2.33

11 P(x < 60) = normalcdf(−10^99, 60, 69.3, 2.92) = 7.240062385E−4
P(x < 60) = 7.24×10^-4 or (better) 0.0007

Common mistake: The probability is not 7.24! That’s not just wrong, it’s very wrong — probabilities are never greater than 1. “E−4” on your calculator comes at the end of the number, but it’s critical info. It means “times 10 to the minus 4th power”, so the probability is 7×10⁻⁴ or 0.0007.

• The probability that a randomly selected man is under 60″ tall is 0.0007 or 0.07%.
• 0.07% of men are under 60″ tall.

12 The plot is pretty clearly not a straight line — there’s a sharp bend around the second and third data points. The numbers confirm this: r = .8363, crit = .9121, r < crit, and therefore the normal model is not a good fit for this data set.

13 The middle 90% leaves 10% in the two tails, or 5% in each tail.

x_m1 = invNorm(.05, 69.3, 2.92) = 64.49702741

x_m2 = invNorm(1−.05, 69.3, 2.92) = 74.10297259

x_f1 = invNorm(.05, 64.1, 2.75) = 59.57665253

x_f2 = invNorm(1−.05, 64.1, 2.75) = 68.62334747

Men must be 64.5 to 74.1 inches tall; women must be 59.6 to 68.6 inches tall.