Question

The average number of words in a romance novel is 64,436 and the standard deviation is 17,071. Assume the distribution is normal. Let X be the number of words in a randomly selected romance novel. Round all answers to 4 decimal places where possible.
a. What is the distribution of X? X ~ N(
64436
Correct,
17071
Correct)

b. Find the proportion of all novels that are between 72,972 and 90,043 words.


c. The 75th percentile for novels is
words. (Round to the nearest word)

d. The middle 60% of romance novels have from
words to
words. (Round to the nearest word)

Answer 1

b. You're looking for the probability

Pr [72,972 ≤ X ≤ 90,043]

Transform X to Z ∼ Normal(0, 1) using the rule

X = µ + σ Z

with µ = 64,436 and σ = 17,071. Then the probability is

Pr [(72,972 - µ)/σ ≤ (X - µ)/σ ≤ (90,043 - µ)/σ]

≈ Pr [0.5000 ≤ Z ≤ 1.5000]

You probably have a z-score table available, so you can look up the probabilities to be about

Pr [Z ≤ 0.5000] ≈ 0.6915

Pr [Z ≤ 1.5000] ≈ 0.9332

and then

Pr [0.5000 ≤ Z ≤ 1.5000] ≈ 0.9332 - 0.6915 = 0.2417

c. The 75th percentile word count that separates the lower 75% of the distribution from the upper 25%. In other words, its the count x such that

Pr [X ≤ x] = 0.75

Transforming to Z and looking up the z-score for 0.75, we have

Pr [(X - µ)/σ ≤ (x - µ)/σ] ≈ Pr [Z ≤ 0.6745]

so that

(x - µ)/σ ≈ 0.6745

x ≈ µ + 0.6745σ

x ≈ 75,950

d. Because the normal distribution is symmetric, the middle 60% of novels have word counts between µ - k and µ + k, where k is a constant such that

Pr [µ - k ≤ X ≤ µ + k] = 0.6

Also due to symmetry, we have

Pr [µ - k ≤ X ≤ µ + k] = 2 Pr [µ ≤ X ≤ µ + k]

⇒   Pr [µ ≤ X ≤ µ + k] = 0.3

Transform X to Z :

Pr [(µ - µ)/σ ≤ (X - µ)/σ ≤ (µ + k - µ)/σ] = 0.3

⇒   Pr [0 ≤ Z ≤ k/σ] = 0.3

⇒   Pr [Z ≤ k/σ] - Pr [Z ≤ 0] = 0.3

⇒   Pr [Z ≤ k/σ] - 0.5 = 0.3

⇒   Pr [Z ≤ k/σ] = 0.8

Consult a z-score table:

Pr [Z ≤ k/σ] ≈ Pr [Z ≤ 0.8416]

⇒   k/σ ≈ 0.8416

⇒   k ≈ 14,367

Then the middle 60% of novels have between µ - k = 50,069 and µ + k = 78,803 words.