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The average number of words in a romance novel is 64,464 and the standard deviation is 17,262. 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(64,464,17262)

b. Find the proportion of all novels that are between 67,916 and 81,726 words.

c. The 80th 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)

Sagot :

Using the normal distribution, it is found that:

a) X ~ N(64,464,17262)

b) 0.262 = 26.2% of all novels that are between 67,916 and 81,726 words.

c) The 80th percentile for novels is 78,964 words.

d) The middle 40% of romance novels have from 55,401 words to 73,527 words.

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Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.
  • After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

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In this problem:

  • Mean of 64,464 words, thus [tex]\mu = 64464[/tex].
  • Standard deviation of 17,262 words, thus [tex]\sigma = 17262[/tex].
  • This also means that the answer for item a is X ~ N(64,464,17262).

Item b:

This proportion is the p-value of Z when X = 81,726 subtracted by the p-value of Z when X = 67,916, thus:

X = 81,726:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{81726 - 64464}{17262}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a p-value of 0.8413.

X = 67,916:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{67916 - 64464}{17262}[/tex]

[tex]Z = 0.2[/tex]

[tex]Z = 0.2[/tex] has a p-value of 0.5793.

0.8413 - 0.5793 = 0.262.

0.262 = 26.2% of all novels that are between 67,916 and 81,726 words.

Item c:

  • The 80th percentile is X when Z has a p-value of 0.8, so X when Z = 0.84.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.84 = \frac{X - 64464}{17262}[/tex]

[tex]X - 64464 = 0.84(17262)[/tex]

[tex]X = 78964[/tex]

The 80th percentile for novels is 78,964 words.

Item d:

  • The normal distribution is symmetric, which means that the middle 40% is between the 30th percentile and the 70th percentile.
  • The 30th percentile is X when Z has a p-value of 0.3, so X when Z = -0.525.
  • The 70th percentile is X when Z has a p-value of 0.7, so X when Z = 0.525.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.525 = \frac{X - 64464}{17262}[/tex]

[tex]X - 64464 = -0.525(17262)[/tex]

[tex]X = 55401[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.525 = \frac{X - 64464}{17262}[/tex]

[tex]X - 64464 = 0.525(17262)[/tex]

[tex]X = 73527[/tex]

The middle 40% of romance novels have from 55,401 words to 73,527 words.

A similar problem is given at https://brainly.com/question/24850210

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