Question

The average number of words in a romance novel is 64,357 and the standard deviation is 17,202. 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 66,077 and 76,398 words.  

c. The 90th 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

Solution(a)
Distribution of X is X~ N(mean , Standard deviation)
Distribution of X is X~ N(64357 , 17202)
Solution(b)
We need to find
P(66077<X<76398) = P(X<76398) - P(X<66077)
Z = (76398-64357)/17202 = 0.7
Z = (66077 - 64357)/17202 = 0.1
From Z table we found p-value
P(66077<X<76398) = P(X<76398) - P(X<66077) = 0.7580 - 0.5398 = 0.2182
Solution(C)
P-value = 0.9
Z -score from Z table is
Z-Score = 1.28
1.28 = (X-64357)/17202
X = 22018.56 + 64357 = 86375.56
Solution(d)
P(-c<X<c) = 0.4
alpha = 0.6, P-value = 0.3 & 0.7
Z = -0.525 and 0.525
0.525 = (X-64357)/17202
X = 9031.05 + 64357 = 73388
-0.525 = (X-64357)/17202
X = 64357 - 9031.05 = 55326