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iren2701 [21]
3 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]3 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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xxMikexx [17]

Answer:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      32   18    50

Medium                                 68     7    75

Large                                     89    11    100

_____________________________________

Total                                     189    36   225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1:  NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*189}{225}=42

E_{2} =\frac{50*36}{225}=8

E_{3} =\frac{75*189}{225}=63

E_{4} =\frac{75*36}{225}=12

E_{5} =\frac{100*189}{225}=84

E_{6} =\frac{100*36}{225}=16

And the expected values are given by:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      42    8    50

Medium                                 63     12    75

Large                                     84    16    100

_____________________________________

Total                                     189    36   225

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\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

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