(1 point) You are interested in finding out the mean number of customers enterin
ID: 3313802 • Letter: #
Question
(1 point) You are interested in finding out the mean number of customers entering a 24-hour convenience store every 10-minutes. You suspect this can be modeled by the Poisson distribution with a a mean of = 3.87 customers. You are to randomly pick n = 51 10- minute time frames, and observe the number of customers who enter the convenience store in each. After which, you are to average the 51 counts you have. That is, compute the value of X (a) What can you expect the value of X to be? Enter your answer to one decimal. l (use at least two decimals in your answer) (b) Find the value of the standard deviation of X. Enter your answer using at least two decimals. (c) Find the probability that mean of the sample of 51 is between 1.1 and 4. Use at least four decimals in your answer. Ensure you use three decimals in your z-values. P(1.1 X 4) = (d) 94% of the time, the observed value of X will exceed what value? (use at least two decimals in your answer)Explanation / Answer
(A) Here the expected value of x = 3.87 customers
(B) Standard deviation of x = (/n) = (3.87/51) = 0.2755
(C) As n > 30, we have done normal approximation with the help of central limit theorm.
P(1.1 < x < 4) = (Z2 ) - (Z1)
Z1= (1.1 - 3.87)/ 0.2755 = -10.056
Z2 = (4 - 3.87) / 0.2755 = 0.472
Pr(1.1 < x < 4) = (0.472) - (-10.056) = 0.6815
(d) Let say that time would be X
so Pr( x > x ) = NORM ( x > x ; 3.87; 0.2755) = 0.94
NORM ( x < x ; 3.87; 0.2755) = 1 - 0.94 = 0.06
Z- value for the given p - value
Z = -1.555 = (x - 3.87)/ 0.2755
z = 3.87 - 1.555 * 0.2755 = 3.44
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