Solution Found!
A 500-page book contains 250 sheets of paper. The
Chapter 4, Problem 2E(choose chapter or problem)
A 500-page book contains 250 sheets of paper. The thickness of the paper used to manufacture the book has mean 0.08 mm and standard deviation 0.01 mm..
a. What is the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers)?
b. What is the 10 th percentile of book thicknesses?
c. Someone wants to know the probability that a randomly chosen page is more than 0.01 mm thick. Is enough information given to compute this probability? If so, compute the probability. If not, explain why not.
Questions & Answers
QUESTION:
A 500-page book contains 250 sheets of paper. The thickness of the paper used to manufacture the book has mean 0.08 mm and standard deviation 0.01 mm..
a. What is the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers)?
b. What is the 10 th percentile of book thicknesses?
c. Someone wants to know the probability that a randomly chosen page is more than 0.01 mm thick. Is enough information given to compute this probability? If so, compute the probability. If not, explain why not.
ANSWER:
Step 1 of 3
(a)
In this question, we are asked to find the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers).
A 500 - page book contains 250 sheets of paper.
Thickness of paper has mean 0.08 mm and standard deviation 0.01 mm .
Let , denote the thickness of the 250 sheets of paper.
We need to find P(X > 20.2).
Now the sample size is n = 250 , which is a large sample.
Then according to Central Limit Theorem,
Let , be a simple random sample from a population with mean and variance
Let = be the total thickness.
Hence can be approximately normal distributed.
CLT specifies that and for sum of the sample items.
Hence we achieve,
Mean
Variance
= 250
=
Standard deviation
Therefore we can write,
Now we will calculate the Z score because our distribution is approximately normal.
thus the Z score of 20.2 mm is
=
=
z = 1.265
From the z table, the area to the left of 1.265 is 0.8962.
Hence the area to the right is
P(X > 20.2) = 1 - the area to the left of 1.265
P(X > 20.2) = 1 - 0.8962
P(X > 20.2) = 0.1038
Hence the probability that a randomly chosen book is more than 20.2 mm thick is 0.1038.