Answer:
a) 280.4; b) E. Yes, because 300 is greater than mu plus 2 sigma .
Step-by-step explanation:
For part a, we multiply the percentage of murders committed by a firearm by the total number of murders:
0.701(400) = 280.4
This will be the average number of murders by a firearm out of a sample of 400.
Since this is a binomial distribution (two outcomes; independent trials; same probability of success for each trial; and fixed number of trials), the standard deviation is given by
√npq
Since n = 0.701, q = 1-0.701 = 0.299.
This gives us
√(400*0.701*0.299) = √83.8396 = 9.16
This means any unusual values will be less than 2 standard deviations below the mean or more than 2 standard deviations above the mean.
2 standard deviations below the mean will be 280.4-2(9.16) = 280.4-18.32 = 262.08. 300 is not below this value.
2 standard deviations above the mean will be 280.4+298.72. 300 is above this, so it is an unusual value.
We would expect 280.4 murders to be committed with a firearm. In turn, it would not be unusual to observe 300 murders by firearm, as long as the margin of error does not exceed 5%.
Given that 70.1% of murders are committed with a firearm, if 400 murders are randomly selected, to determine how many would we expect to be committed with a firearm, and determine if it would be unusual to observe 300 murders by firearm in a random sample of 400 murders, the following calculations must be performed:
Therefore, we would expect 280.4 murders to be committed with a firearm. In turn, it would not be unusual to observe 300 murders by firearm, as long as the margin of error does not exceed 5%.
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