Question

The average number of words in a romance novel is 64,464 and the standard deviation is 17,262. 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(64,464,17262)

b. Find the proportion of all novels that are between 67,916 and 81,726 words.

c. The 80th percentile for novels is words. (Round to the nearest word)

d. The middle 40% of romance novels have from words to words. (Round to the nearest word)

Answer 1

Using the normal distribution, it is found that:

a) X ~ N(64,464,17262)

b) 0.262 = 26.2% of all novels that are between 67,916 and 81,726 words.

c) The 80th percentile for novels is 78,964 words.

d) The middle 40% of romance novels have from 55,401 words to 73,527 words.

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Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.
• After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

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In this problem:

• Mean of 64,464 words, thus $\mu = 64464$.
• Standard deviation of 17,262 words, thus $\sigma = 17262$.
• This also means that the answer for item a is X ~ N(64,464,17262).

Item b:

This proportion is the p-value of Z when X = 81,726 subtracted by the p-value of Z when X = 67,916, thus:

X = 81,726:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{81726 - 64464}{17262}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 67,916:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67916 - 64464}{17262}$

$Z = 0.2$

$Z = 0.2$ has a p-value of 0.5793.

0.8413 - 0.5793 = 0.262.

0.262 = 26.2% of all novels that are between 67,916 and 81,726 words.

Item c:

• The 80th percentile is X when Z has a p-value of 0.8, so X when Z = 0.84.

$Z = \frac{X - \mu}{\sigma}$

$0.84 = \frac{X - 64464}{17262}$

$X - 64464 = 0.84(17262)$

$X = 78964$

The 80th percentile for novels is 78,964 words.

Item d:

• The normal distribution is symmetric, which means that the middle 40% is between the 30th percentile and the 70th percentile.
• The 30th percentile is X when Z has a p-value of 0.3, so X when Z = -0.525.

• The 70th percentile is X when Z has a p-value of 0.7, so X when Z = 0.525.

$Z = \frac{X - \mu}{\sigma}$

$-0.525 = \frac{X - 64464}{17262}$

$X - 64464 = -0.525(17262)$

$X = 55401$

$Z = \frac{X - \mu}{\sigma}$

$0.525 = \frac{X - 64464}{17262}$

$X - 64464 = 0.525(17262)$

$X = 73527$

The middle 40% of romance novels have from 55,401 words to 73,527 words.

A similar problem is given at https://brainly.com/question/24850210