Question

The average number of words in a romance novel is 64,317 and the standard deviation is 17,458. 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( 64,317 , 17,458 ) b. Find the proportion of all novels that are between 50,351 and 59,080 words. c. The 75th percentile for novels is words. (Round to the nearest word) d. The middle 50% of romance novels have from words to words. (Round to the nearest word) License Points possible: 1 This is attempt 1 of 1.

Answer 1

Page Solution! Given that, M = 64317 6- 17 458 9 The distribution of - XNN [64317, 17458] 5035 1<x<59080 =P 5035-64317 CX-4 <50080-64317 - 17458 6 17 17458 ep - 13966 . 2 . -5237 17458 17458) =P [-0.80 <2 <-0.30] =p [z(-0.30] -p [z(-0.80 = 0.3821 – 0.2119 = 0.1702 P[s0351 LXC sa oso)=0-1702 proportion = 0.1702

c) The 2-distribution of the 75% is P(Z <2) = 75 % P(Z <2) = 0.75 P(Z <0.674)= 0.75 z = 0.674 Using a-score formula x=26t4 x= 0.674 X 174587 64317 x= 76083.692 X= 76084 The 75 do percentile for nevels is 176084 words

uge Middle 50 % 2=1- so lo x=1-0.50 -0.5 x = 0.5 -0.25 1-2/2 - 1-0.25 = 0.75 2 2x12 - 20.25 = -0.674 21-2/2 = 20.75 = 0.674 Using Z-score formula x= 26tu x=-0.674 X 17458 + 64317 x = 52sso. 308 - Zote x = 0.674 X 17458 x = 76083.692 & 64317 Between 52550 and 76084 The middle 50% novels have of romance from 152550 words to [760845