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iren2701 [21]
2 years ago
13

Suppose that the thickness of one typical page of a book printed by a certain publisher is a random variable with mean 0.1 mm an

d a standard deviation of 0.002 mm. A new book will be printed on 500 sheets of this paper. Approximate the probability that the
Mathematics
1 answer:
marusya05 [52]2 years ago
6 0

Answer:

The probability that the thicknesses at the entire book will be between 49.9 mm and 50.1 mm is 0.97.

Step-by-step explanation:

The complete question is:

Suppose that the thickness of one typical page of a book printed by a certain publisher is a random variable with mean 0.1 mm and a standard deviation of 0.002 mm Anew book will be printed on 500 sheets of this paper. Approximate the probability that the thicknesses at the entire book (excluding the cover pages) will be between 49.9 mm and 50.1 mm. Note: total thickness of the book is the sum of the individual thicknesses of the pages Do not round your numbers until rounding up to two. Round your final answer to the nearest hundredth, or two digits after decimal point.

Solution:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e S, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{S}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{S}=\sqrt{n}\sigma

The information provided is:

n=500\\\mu=0.1\\\sigma=0.002

As <em>n</em> = 500 > 30, the central limit theorem can be used to approximate the total thickness of the book.

So, the total thickness of the book (S) will follow N (50, 0.045²).

Compute the probability that the thicknesses at the entire book will be between 49.9 mm and 50.1 mm as follows:

P(49.9

                               =P(-2.22

Thus, the probability that the thicknesses at the entire book will be between 49.9 mm and 50.1 mm is 0.97.

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Step-by-step explanation:

Hello, please consider the following.

\sqrt{(x^2-1)}-x\\\\=\sqrt{x^2(1-\dfrac{1}{x^2})}-x\\\\=x\left( \sqrt{1-\frac{1}{x^2}}-1\right)

For x close to 0, we can write

\sqrt{1+x}=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+o(x^2)\\\\\ \text{x tends to } +\infty \text{ means }\dfrac{1}{x} \text{ tends to 0}\\\\\text{So, when }\dfrac{1}{x}\text{  is close to 0, we can write.}\\\\\sqrt{1-\dfrac{1}{x^2}}=1-\dfrac{1}{2}\dfrac{1}{x^2}-\dfrac{1}{8}\dfrac{1}{x^4}+o(\dfrac{1}{x^4})

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x\left( \sqrt{1-\frac{1}{x^2}}-1\right)\\\\=x(1-\dfrac{1}{2}\dfrac{1}{x^2}+o(\dfrac{1}{x^2})-1)\\\\=-\dfrac{1}{2x}+o(\dfrac{1}{x})

It means that

\displaystyle \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}\\\\=\lim_{x \rightarrow +\infty} {-\dfrac{x}{2x}}=-\dfrac{1}{2}

Thank you

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