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Lubov Fominskaja [6]
3 years ago
14

According to a​ report, 67.5​% of murders are committed with a firearm. ​(a) If 200 murders are randomly​ selected, how many wou

ld we expect to be committed with a​ firearm? ​(b) Would it be unusual to observe 153 murders by firearm in a random sample of 200 ​murders? Why?
Mathematics
1 answer:
kolezko [41]3 years ago
8 0

Answer: The answer is 135 murders.

Step-by-step explanation: The report tells us that statistically 67.5% of murders are committed using a firearm. It follows therefore that in a sample of 200 randomly selected murders, one would expect that 67.5% of those would be by a firearm. \frac{67.5}{100} * 200 = 135.

It would certainly be higher that the expected value based on previous data collected but it would not be unusual because one sample may have a higher than "normal" amount of murders by firearm. Statistics aren't going to be exact for every sample.

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And for this case the 95% confidence interval is given by (2.13; 2.37)

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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".

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

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

And for this case the 95% confidence interval is given by (2.13; 2.37)

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And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

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