# Thread: In Need of Math Help

1. ## In Need of Math Help

Basically, I fail at life when it comes to all sorts of math. Right now, that math is Statistics.

I was wondering if anyone could point me in the right direction with the proper calculations/or anything for this:

Assume that you own a small factory. A critical piece of machinery in your factory will need to be replaced in 180 days. If the machinery does not show up on time, you will need to shut down until it arrives. This might cause you to permanately lose customers.When you order the part you wil need to pay the \$500,000 in advance. That is a lot of money for your small business. If you keep the money in the bank, it will earn interest each month. If you spend the money now, it will leave you with very litte money on hand, and you might have to borrow money to make payroll. You know from past experience that the delivery times are normally distributed with a mean of 45 days and a standard deviation of 15 days. When should you order the part? It is your company, but you must write up an explanation for your actions that convinces your investors that your actions are best. (Unfortunately, the investors cannot afford \$500,000 at this time.)

Be sure to include

1) Explain why 135 days from now is the date to order it if you want to be 50% sure of getting the delivery on time.

2) Explain to the investors why you cannot be 100% sure that the part arrives in time.

3) How sure do you want to be that the machinery arrives on time? Explain your answer clearly, and explain why you chose that figure. Remember, you need to confince your investors.
(the answer here cannot be 50%)

4) Based on how sure you want to be, calculate when you need to order the part. Explain all calculations clearly, because your investors need to understand what you did. They are bright and understand math, but you will have to explain the statistics to them. This needs to include z-scores and a calculation of x based on z.
I'm honestly only aware that-
mean/mu = 45 standard deviation/sigma = 15 & that I want to be 75% sure, which I'm assuming would be .67 in terms of z-score, unless I'm looking that up incorrectly.
From there, I have no real clue. (so basically, I'm lost as all hell)
Any help whatsoever would be appreciated.

2. Well statistics is the one thing I never had in my University lectures, but I'ld say:

1) Because on day 135th it's 15 days after and 15 days before a possible delivery
2) If I could I wouldnt need statistics.
3) Well to be mathematically 99,9 % sure you'ld need to order it 60 days in advance, so on day 120. From now on it depends on how much you want to poker. You cant get more than 50 % if you order it any day after the 135th. Because after that it still can be 15 days too late until you have it. So the magic day would be between day 120 - 135, with day 127,5 being the day of 75%.

I dunno what's the right course then, I'ld prolly order 60 days in advance, if you want to poker go with 75% days.

4) well Z = x - mu / sigma , with mu = 120 and sigma = 97.7 % and Z = 2
That is x = 121,954 (that would be the day to be 97.7% sure you get the thing)

I have no clue tho what kind of answers they'ld like to hear. Do they want you to play it safe or to go for maximum danger ? Looks like a trick question or am I only too dumb ?

The part must be replaced in 180 days, mean is 45 days, std dev is 15 days.

The mean is 45 days, there is a 50% chance of being delivered at that date and before, but, naturally, also a 50% chance of being delivered after.

Originally Posted by empirical rule
Approximately two-thirds of a normal curve is within 1 standard deviation of the mean, 95% is within 2 standard deviations, and almost all is within 3 standard deviations.
So, in your case, two standard deviations would mean that between 15 and 75 days delivery time comes 95% of the time. That 95% obviously includes the part coming a month after the mean. One standard deviation would put the delivery time between 30 and 60 days, and has a 68% of happening. Thats not a whole lot better than the mean (50%), but it is better.

You cannot be 100% sure of any range of delivery time, theoretically, no matter how far in either direction the curve extends, it never meets the x axis—but it gets increasingly closer. (Area underneath still equals 100, can be proved with the theory of calculus.)

It becomes a judgment/risk assessment problem from here on for you to select your percentage, find z values, etc., since it doesn't really help you if the part comes early. I hope this helps somewhat

4. You guys do smart people stuff

5. 1) because 50% of a normal curve's area is enclosed between its starting point (minus infinity) and the mean. Area of a normal curve up to X = probability that the represented event occurs up to X; mean = 45; 135+45 = 180, thus there will be a 50% probability for the piece to be delivered in 45 days.
2) we cannot be 100% sure because if a normal curve is the right probability representation, we would need an infinite amount of time (associated to the X axis) to obtain an area of 1 (equivalent to 100% probability).
3) this part might be tricky. A relatively uncomplicated answer is similar to what entropie has provided. If you want to write a more in-depth analysis (which requires assumptions though!), you'd have to:
- check out on the internet what's the (daily) average interest rate a firm can earn on its bank account
- assume how much money can you lose by shutting down the firm for one day
- compute the probability of delivery one day past 180, two days past 180...for some time
- set up a maximization problem (with T as decisional variable) such as:

where the first term represents what you would earn by leaving 500'000 in your bank account for T days, while the second would represent what you'd potentially lose if the piece were to be delivered T days late.
I understand this sounds complicated...take it as an idea on how to motivate your choice, maybe. A straightforward example:
- suppose daily interest rates are 0.002
- suppose you can lose 10'000 dollars a day
- computing the probability of not receiving the piece for 1 day is straightforward: set x=45+T and x1=45+(T-1), then what you're looking for is CDF(z)-CDF(z1) with z=(x-mu)/sigma and z1=(x1-mu)/sigma, so for 181 it'd be 0.0265; then for day 1 you'd have:

[(500000*e^0.002)-(10'000*0.02659)]-500'000

then you can compute this opportunity-cost function for some days and see what the results are...

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