Currently working on this problem.

Monthly demand of a product has been observed to follow a normal distribution with mean of $50$ pieces and standard deviation of $5$ pieces. Assume each month is independent of other months.

  1. What is the distribution followed by the yearly demand and what is the mean and standard deviation of that distribution?

I think the distribution remains normal and mean is $600$ but how can I calculate the standard deviation?

  1. If the store has $220$ pieces in stock what is the probability it will cover the demand of the next $4$ months?

for that $4$ months mean $= 4*50 =200$; standard deviation $= sqrt(4*25) =10$ ; $P(X<220)= P(Z<220-200/10)$

  1. How many pieces should the store have to cover the full year's demand ($12$ months) with a probability at least equal to $93,7\%$ without having to restock?

And this is where my problem begins. I know I need the standard deviation from 1. but even if I had that I can't think of how to incorporate the minimum probability.Can anyone help me?


1 Answer 1



In your answer to 2, you say for $4$ months "standard deviation $= sqrt(4*25)$" which supposes each month is independent of other months. If that is true, then you could do the same for a year and say the standard deviation for the year ($12$ months) is $\sqrt{12 \times 25}$.

The method for 3 is in a sense the reverse of that used for 2.

  • $\begingroup$ Silly me. For some reason I had it stuck in my head that for the year I needed a different approach than the one I had already used. It didn't even occur to me to do it like that. Also the months are independent and I forgot to add that. Will edit now $\endgroup$
    – apot
    Commented Dec 6, 2015 at 23:46

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