(1 point) On a typical day at a local popular mall, the number of shoplifters ca

Question

(1 point) On a typical day at a local popular mall, the number of shoplifters caught by mall security fluctuates from day to day, with an average of 7.2 caught per day. Suppose you are to count the number of Part (a) What is the probability that mall security apprehended 9 shoplifters? shoplifters apprehended at this mall today(Assume today is a typical day at the mall.) l (use four decimals in your answer) Part (b) What is the probability that mall security will apprehend at least 6 shoplifters. Enter your answer to four decimals. Part (c) Find the probability that between 5 and 12 shoplifters will be apprehended. Enter your answer to four decimals IEE Part (d) Think about the distribution of the number of shoplifters mall security apprehends on a typical day. What can you say about this distribution? Select the most appropriate reason below A. The distribution of values is skewed to the right, with a mean of 7.2 shoplifters and a standard deviation of 2.68328157299975 shoplifters. B. The distribution of values is roughly symmetrical, with a mean of 7.2 shoplifters and a standard deviation of 2.68328157299975 shoplifters. C. The number of shoplifters apprehended each day is 7.2 D. The distribution of values is roughly symmetrical, with a mean of 7.2 shoplifters and a variance of 2.68328157299975 shoplifters E. The distribution of values is skewed to the right, with a mean of 7.2 shoplifters and a variance of 2.68328157299975 shoplifters Part (e) On a typical day, the mall is open from 10:00am to 9:00pm, a total of 11 hours. You are interested in the number of shoplifters apprehended in any given hour How many shoplifters would you expect mall security to catch in any given hour? Provide the variance of this count as well. (use three decimals in your answer) (use three decimals in your answer

Explanation / Answer

Par(A) THe given distribution is poission distribution with parameter = 7.2 shoplyters caught per day

If X is the number of shoplyfters caught per day is X then

Pr(X = 9) = POISSON(X = 9 ; 7.2) = e-7.2 7.29/9! = 0.1070

Part(B) Pr(X 6) = 1 - POISSOn (X < 6 ; 7.2) = 1 - 0.2759 = 0.7241

Part(C) Pr(12 X 5) = Pr(X 12) - Pr(X < 5) = 0.9673 - 0.1555 = 0.8118

part(D)

The distribtuoon is roughly symmetrical, with a mean of 7.2 shoplifters and a standard deviation of 2.68328 shoplifters Option B is right.

Part(e)

Here as total of 11 hours, so

Expected number of shoplifters caught by mall security in any given hour = 7.2/11 = 0.655

Varaince of this count = 0.655

It is because the distribution will be poisson again.

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