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xenn [34]
2 years ago
7

About 25% of the students in Mr. Goodman's class got an Aon their last math test. Of those, 50% are girls. The class has 12 girl

s and 16 boys. Is it
reasonable to say 6 girls got an Aon the test?
A No, because 6 is not 25% of 28.
O
B. No, because 6 girls is 50% of all the girls in the class, and not all of the girls got A's.
O
C. Yes, because 50% of 12 is 6.
D
Yes, because more girls than boys got A's.
Mathematics
1 answer:
bogdanovich [222]2 years ago
3 0
I think the answer is c
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Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. Whe
kherson [118]

Answer:

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

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval 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}=-1.96, z_{1-\alpha/2}=1.96

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)  

And on this case we have that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

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

We don't have a prior estimation for the proportion \hat p so we can use 0.5 as an approximation for this case  

And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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Step-by-step explanation:

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Answer:

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