Question #320797

According to a​ report, 67.4​% of murders are committed with a firearm. ​(a) If 200 murders are randomly​ selected, how many would we expect to be committed with a​ firearm? ​(b) Would it be unusual to observe 147 murders by firearm in a random sample of 200 ​murders? Why?





1
Expert's answer
2022-03-31T05:21:06-0400

a:2000.674=134.8Thenumberwillbearound135b:P(p^147200)=P(np^pp(1p)n147200pp(1p))==P(Z2001472000.6740.674(10.674))=P(Z1.8404)==Φ(1.8404)=0.033Theprobabilitytoobtain147ormoremurdersbyfirearmof200islessthan4%.Thusthisisunusual.a:\\200\cdot 0.674=134.8\\The\,\,number\,\,will\,\,be\,\,around\,\,135\\b:\\P\left( \hat{p}\geqslant \frac{147}{200} \right) =P\left( \sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}\geqslant \sqrt{n}\frac{\frac{147}{200}-p}{\sqrt{p\left( 1-p \right)}} \right) =\\=P\left( Z\geqslant \sqrt{200}\frac{\frac{147}{200}-0.674}{\sqrt{0.674\left( 1-0.674 \right)}} \right) =P\left( Z\geqslant 1.8404 \right) =\\=\varPhi \left( -1.8404 \right) =0.033\\The\,\,probability\,\,to\,\,obtain\,\,147 or\,\,more\,\,murders\,\,by\,\,firearm\,\,of\,\,200 is\,\,less\,\,than\,\,4\%. \\Thus\,\,this\,\,is\,\,unusual.


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