Question

The average number of words in a romance novel is 64,245 and the standard deviation is 17,316. 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 48,661 and 65,977 words.

c. The 95th 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)

Answer 1

Solution :

Given that ,

mean = = 64245

standard deviation = = 17316

a. What is the distribution of X? X ~ N( 64245 , 17316 )

( b )

P( 48661 < x < 65977 ) = P[(48661 -64254 ) / 17316 ) < (x - ) /  < ( 65977 - 17316 ) / 17316 ) ]

= P( -0.90 < z < 0.10 )

= P( z < 0.10 ) - P( z < -0.90 )

Using z table,

= 0.5398 - 0.1841

= 0.3557

proportion = 0.3557

( c )

The z - distribution of the 90% is

P(Z < z) =90 %

= P(Z < z ) = 0.90

= P(Z < 1.282 ) = 0.90.  

z = 1.282

Using z-score formula,

x = z * +

x = 1.282 * 17316 + 64245

x = 86444.112

x = 86444

( d )

Middle 50%

= 1 - 50%  

= 1 - 0.50 = 0.5

/2 = 0.25

1 - /2 = 1 - 0.25 = 0.75

Z/2 = Z0.25 = -0.674

Z 1- /2 = Z 0.75 = + 0.674

Using z-score formula,

x = z * +

x = - 0.674 * 17316 + 64245

x = 52574.016

x = z * +

x = +0.674 * 17316 + 64245

x = 75915.984

Between 52574 and 75916