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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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