Question

The average number of words in a romance novel is 64,419 and the standard deviation is 17,160. 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 71,283 and 83,295 words.

c. The 85th 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

Solution :

Given that ,

mean = = 64419

standard deviation = = 17160

a.

X N (64419 , 17160)

b.

P(71283 < x < 83295) = P[(71283 - 64419)/ 17160) < (x - ) /  < (83295 - 64419) / 17160) ]

= P(0.4 < z < 1.1)

= P(z < 1.1) - P(z < 0.4)

= 0.8643 - 0.6554

= 0.2089

proportion = 0.2089

c.

Using standard normal table ,

P(Z < z) = 85%

P(Z < 1.04) = 0.85

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 17160 + 64419 = 82265

85th percentile = 82265

d.

Middle 60% as the to z values are -0.8416 and 0.8416

Using z-score formula,

x = z * +

x = -0.8416 * 17160 + 64419 = 49977

and

x = 0.8416 * 17160 + 64419 = 78861

The middle 60% of romance novels have from 49977 words to 78861 words