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Stats without Tears
Solutions for Chapter 7

Updated 1 Jan 2016 (What’s New?)
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1
2 sketch to accompany problem solution 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:
3 sketch to accompany problem solution “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 sketch to accompany problem solution 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 sketch to accompany problem solution 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 sketch to accompany problem solution (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 normal probability plot as described in text 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 sketch to accompany problem solution 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 sketch to accompany problem solution 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 sketch to accompany problem solution z0.01 = invNorm(1−0.01, 0, 1) = 2.326347877 → z0.01 = 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

sketch to accompany problem solution 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.

12 normal probability plot as described in text 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.

sketch to accompany problem solution

xm1 = invNorm(.05, 69.3, 2.92) = 64.49702741

xm2 = invNorm(1−.05, 69.3, 2.92) = 74.10297259

xf1 = invNorm(.05, 64.1, 2.75) = 59.57665253

xf2 = 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.

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