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Irina-Kira [14]
3 years ago
6

The table gives the probability distribution of the number of books sold in a day at a bookstore. What is the probability of 16

or more books being sold on a given day?
number of books sold in a day

0-5

6-10

11-15

16-20

21-25

probability

0.110

0.206

0.464

0.201

?
Mathematics
1 answer:
kakasveta [241]3 years ago
4 0
We are to find the probability of 16 or more books being sold on a given day. So all the categories listing the sale of more than 16 books will be counted for this problem.

From the given table we can see that there are 2 such categories:

1) 16 - 20
2) 21 - 25

To find the probability of 16 or more books sold , we need to add the probability of both these categories. Probability of selling 16 - 20 books is 0.201 and probability of selling 21 - 25 books is unknown. The given distribution is a probability distribution. So sum of probabilities of all categories must be equal to 1. So, the probability of selling 21-25 books is 0.019 only then the probabilities sum up to 1.

Thus, 
The probability that 16 or more books are sold on a give day = 0.201 + 0.019 = 0.220
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n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Solution to the problem

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

If solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

The margin of error desired for this case is ME= \pm 0.02 equivalent to 2% points

For this case we need to assume a confidence level, let's assume 95%. And since we don't have prior estimation for the population proportion of interest the best value to do an approximation is \hat p =0.5

In order to find the critical value we need to take in count that we are finding the margin of error for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:  

z_{\alpha/2}=\pm 1.96  

Now we have all the values needed and if we replace into equation (b) we got:

n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

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I think it would be 16.95 Hope this help you in someway.

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