Question

The average number of words in a romance novel is 64,151 and the standard deviation is 17,400. 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(,)

b. Find the proportion of all novels that are between 65,891 and 79,811 words.

c. The 90th percentile for novels is  words. (Round to the nearest word)

d. The middle 70% of romance novels have from  words to  words. (Round to the nearest word)

Answer 1

a)
X ~ N(64151, 17400^2)

b)
z = (x - μ)/σ
z1 = (65891 - 64151)/17400 = 0.1
z2 = (79811 - 64151)/17400 = 0.9

Therefore, we get
P(65891 <= X <= 79811) = P((79811 - 64151)/17400) <= z <= (79811 - 64151)/17400)
= P(0.1 <= z <= 0.9) = P(z <= 0.9) - P(z <= 0.1)
= 0.8159 - 0.5398
= 0.2761

c)
z-value = 1.28
x = 64151 + 1.28*17400 = 86423

d)
z-value = 1.04

64151 - 1.04*17400 = 46055
64151 + 1.04*17400 = 82247

The middle 70% of romance novels have from 46055 words to 82247 words