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Help with statistics due today!

[From: ] [author: ] [Date: 13-02-20] [Hit: ]
56.2% of murders are committed with a firearm.(a) if 200 murdersare randomly selected, how many would we expect to be committed with a firearm?(b) Would it be unusual to observe 134 murders by firearm in a random sample of 200 murders? Why?......
According to a report, 56.2% 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 134 murders by firearm in a random sample of 200 murders? Why?

Help me! it's my last question and i need help.

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a. E(x) = np = 200*0.562 = 112.4

b. z- value with continuity correction
= (value with continuity correction - mean)/sd
= (133.5-112.4)/sqrt(200*0.562*(1-0.562)) = 3.00
so P(≥134) is quite unusual , as by the empirical rule, only 0.15%
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(a) 112 murders (mathematically 112.4, but one cannot have a 0.4 murder)

(b) Depends how your statistics course is defining 'unusual'. This will be very close to 3sigma, but it *is* more than 2sigma (sigma = standard deviation if the term is unfamiliar)
You should have an equation showing you how to calculate standard deviation from the population mean (56.2%), and from that you can use your standard deviation or 'z' tables to determine the probability of this event being this far from the norm.

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200 x 0.562 = 112

134/200 100/1 = 67% , in excess of 112 by 11 % so probably unacceptable but the numbers involved in arriving at 56.2 % would be needed i.e. if 2000 murders were used as data then the first 200 may be higher in their relative frequency .

no expert
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