Question

The average number of words in a romance novel is 64,174 and the standard deviation is 17,215. 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 59,009 and 71,060 words.

c. The 80th percentile for novels is ... words. (Round to the nearest word but try one word up or down if needed.)

d. The middle 60% of romance novels have from ..... words to ..... words. (Round to the nearest word but try one word up or down if needed.)

Answer 1

a) Average number of words = 64174 = mean number of words

Standard deviation = 17215

X ~ N (64174 , 17215)

B) P(59009< X <71060) = P(59009- 64174 /17215 < X-64174/17215 < 71060-64174/17215)

X -64174 / 17215 ~ N(0,1)

P(-0.300 < Z < 0.4) = P(Z<0.4 ) - (1-P(Z<-0.3))

0.655- 0 3829 = 0.2721

C) the 80th percentile for novel is

P(Z<x- 64174/17215) =0.8

X-64174/17215 = 0.842

X = 78669 = 80th percentile

D) . Because normal distributions are symmetrical, you will get the top 20% above the z-score of +0.84 and the bottom 20% below the z-score of -0.84. So you will get the middle 60% between -0.84 and +0.84.

0.84 = x-64174/17215

X = 78635

-0.84 = x -64174/17215 =49713

So the middle 60% of population lies between

49713 to 78635