Answer to Question #233882 in Statistics and Probability for samuel

Question #233882

The average demand on a factory store for a certain electricmotoris8 per week. When the storeman places an order for these motors, delivery takes one week. If the demand for motors has a Poisson distribution, how low can the storeman allow his stock to fall before ordering a new supply if he wants to be at least 95%sureofmeetingallrequirementswhilewaitingforhisnewsupplytoarrive? [6marks] b) A bank has 175 000 credit card holders. During one month the average amount spent by each card holder totalled $192,50 with a standard deviation of $60,20. Assuming a normal distribution, determine the number of card holders who spent more than $250. 


1
Expert's answer
2021-09-13T16:45:51-0400

a) Mean λ=8

"P(X=k) = \\frac{\u03bb^k \\times e^{-\u03bb}}{k!} \\\\\n\nP(X\u2264n)\u22650.95"

Find for value of “n”

"\\frac{8^0 \\times e^{-8}}{0!} + \\frac{8^1 \\times e^{-8}}{1!}+\\frac{8^2 \\times e^{-8}}{2!}+...+\\frac{8^n \\times e^{-8}}{n!}\u22650.95 \\\\\n\ne^{-8}(\\frac{8^0}{0!} + \\frac{8^1}{1!}+...+\\frac{8^n}{n!} ) \u22650.95 \\\\\n\n\\frac{8^0}{0!} + \\frac{8^1}{1!}+...+\\frac{8^n}{n!} \u22652831.91009"

From the above series, minimum n=13 satisfied to meet 95% sure of meeting all requirements while waiting for his new supply to arrive. Therefore, storeman can allow minimum n=13.


b) Mean

"\\mu =192.50"

Standard deviation

"\\sigma =60.20 \\\\\n\nP(X>250)= 1- P(X\u2264250) \\\\\n\n= 1-P(Z\u2264 \\frac{250-192.50}{60.20}) \\\\\n\n= 1 -P(Z\u2264 0.9551) \\\\\n\n= 1-0.8302 \\\\\n\n= 0.1698"

Therefore, Required number of card holders who spent more than $250 "= 175000 \\times 0.1698= 29715"

Answer= 29715


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