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.
(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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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.
(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
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