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.
