Alright posting this question with an answer myself because the person who keeps answering it keeps giving the wrong answer. So I finally figured it out myself and will answer it myself for the sake of the CHEGG community. CHEGG you need to get better people doing your answers!!!!

Let's say you are ordering food for your wedding reception, and you sent out 150 invitations. Everyone RSVP'd, with120 people saying they would attend. You know however that not everyone who RSVP'd will actually show up; there are always, always, always some no-shows (at least from the question-writers experience). Thus you expect to need 100 meals. But just to be safe, you plan to order 110 meals, estimating this is a 90% service level (accepting a 10% probability you'll run out).

What is the implied standard deviation in the number of meals demanded?

(please include the detailed steps to getting the answer)

A) 110

B) 12.8

C) 7.8 (correct Answer) ( Looking at the graph a 90% service level is roughly a z value of 1.3 plugging into the equation we find: 110= 100 (1.3 * x) which equals 10=1.3x so x equals roughly 7.7 which is roughly 7.8. When calculating based of the graph your answer may be very slightly off so you may have to choose a number very similar to the one you estimated)

You can also use the NORMSINV(0.9) (.9 is our service level) to use this excel equation to get the Z score. This will equal 1.28. We can then do 110 = 100(1.28 * x) and do simple algebra to get our answer of 7.8.

D) 10

Use this left graph if Co is greater than Cu Use this right graph if Cu is greater than Co Find Co/Cu on this top scale (note it's a "log" scale), Find Cu/Co on this top scale (note it's a "log" scale), then on down vertically to find 7 on the lower crale then on down vertically to find z on the lower scale Zv no. devi: (sub fri me