Question

The average number of words in a romance novel is 64,290 and the standard deviation is 17,422. 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 57,321 and 71,259 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 = = 64290

standard deviation = = 17422

a) The distribution of X is normal , X ~ N( 64290, 17422 )

b) P( 57321 < x < 71259) = P[(57321 - 64290 ) / 17422) < (x - ) /  < (71259 - 64290) /17422 ) ]

= P(-0.40 < z < 0.40)

= P(z < 0.40) - P(z < -0.40)

Using z table,

= 0.6554 - 0.3446

= 0.3108

c) Using standard normal table,

P(Z < z) = 95%

= P(Z < z) = 0.95  

= P(Z < 1.645) = 0.95

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 * 17422 + 64290

x = 92949

d) Using standard normal table,

P( -z < Z < z) = 50%

= P(Z < z) - P(Z <-z ) = 0.50

= 2P(Z < z) - 1 = 0.50

= 2P(Z < z) = 1 + 0.50

= P(Z < z) = 1.50 / 2

= P(Z < z) = 0.75

= P(Z < 0.6745) = 0.75

= z  ± 0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 17422 + 64290

x = 52539

Using z-score formula,

x = z * +

x = 0.6745 * 17422 + 64290

x = 76041

The middle 50% are from 52539 to 76041