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miv72 [106K]
3 years ago
9

A Gallup poll of 1236 adults showed that​ 12% of the respondents believe that it is bad luck to walk under a ladder. Consider th

e probability that among 30 randomly selected people from the 1236 who were​ polled, there are at least 2 who have that belief. Given that the subjects surveyed were selected without​ replacement, the events are not independent. Can the probability be found by using the binomial probability​ formula? Why or why​ not?
A)No. The selections are not​ independent, and the​ 5% guideline is not met.B) No. The selections are not independentC) Yes. Although the selections are not​independent, they can be treated as being independent by applying the​5% guideline.D) Yes. There are a fixed number of selections that are​independent, can be classified into two​categories, and the probability of success remains the same.
Mathematics
1 answer:
SOVA2 [1]3 years ago
8 0

Answer:

c

Step-by-step explanation:

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uysha [10]

Answer:

1) 0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219    

b) ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (2)

Replacing we got:

n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=0.0826052 represent the sample mean

\mu population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size  

Part 1

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom, given by:

df=n-1=5-1=4

The Confidence level is 0.95 or 95%, and the significance would be \alpha=0.05 and \alpha/2 =0.025, the critical value would be using the t distribution with 4 degrees of freedom: t_{\alpha/2}=2.776

Now we have everything in order to replace into formula (1):

0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219    

Part 2

The original margin of error is given by:

ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (2)

Replacing we got:

n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12

So the answer for this case would be n=12 rounded up to the nearest integer

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4) .   1, 2, 5, 6

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