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Trying to calculate the odds on something and was getting myself confused. I'll try to summarize into a simple problem with made up numbers.

Say a cannon fires projectiles with a population mean of 100 m/s and a standard deviation of 10 m/s, represented by a normal distribution.

I wanted to calculate the odds of firing off 15 rounds in a row that would have a standard deviation between 0 m/s and 2 m/s.

I basically calculated two z-scores:

Z1 = (101-100)/10 and Z2 = (99-100)/10.

Then assumed the probability of getting one round within that range was (using table for standardized z-scores):

P = P(X < Z1) - P(X < Z2)

To fire 15 rounds within that range, then I said P_15 = P^15.

Although, I feel more like I am calculating the odds of my sample to have more more like 3+ sigma (of 2 m/s), since with 1-sigma all the rounds from the sample don't necessarily have to fall within the +/- 1 m/s range, just ~68% of them. But, I really would like the sample to have a 1-sigma between 0m/s and 2 m/s.

Question: what is the correct way to formulate this problem and what are the details of the calculation?

Thanks.

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  • $\begingroup$ Your question has no unique answer for the information given. If you assume a specific distribution for the speeds, then the question can be answered, so please edit it to include that information. [There exist analytic answers for some distributions, such as a Normal distribution. For it, a known multiple of the sample standard deviation has a chi distribution. Note that the chance that the sample SD is exactly $2$ will be zero: you need to ask for the chance that the SD lies within some given range, such as $0-2$. For a Normal distribution, that chance is one in 48 million.] $\endgroup$
    – whuber ♦
    Commented May 23, 2017 at 3:42
  • $\begingroup$ Sorry, I assumed it was obvious it was a normal distribution. Although I guess that brings up another question: Is that assumption not as obvious as I thought? Are there reasons to think cannon (or any type of gun) muzzle velocities would not be represented by a normal distribution around some mean velocity? However, I guess for the sake of this problem, let's assume both the population and samples can be represented by normal distributions. $\endgroup$
    – EthanT
    Commented May 23, 2017 at 3:48
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    $\begingroup$ There's obvious reasons to think it wouldn't be normal -- muzzle velocity is necessarily non-negative, and presuming you'll only consider projectiles that exit the barrel, actually necessarily positive. As a result you'd tend to expect somewhat skewed distributions (one can at least consider a possibility of a cannonball exiting with a bit more than double the mean velocity but it's impossible to be that below the mean, since that would be negative). It might be reasonable in some situations to use a normal approximation, but it's by no means obvious that one should automatically do so. $\endgroup$
    – Glen_b
    Commented May 23, 2017 at 10:53
  • $\begingroup$ Hello Glen_b, I have access to large databases of the particular "gun" in question. I would have to say it is a near impossibility to ever see a muzzle velocity at double the mean for this particular "gun". Physically, where is that energy going to come from? All the prop charges are within 1% of each other in terms of weight, coming from a relatively consistent manufacturing process. In addition, plots of the muzzle velocity sure look like a normal distribution, with a tight standard deviation around the mean that is fractions of the mean velocity. $\endgroup$
    – EthanT
    Commented May 23, 2017 at 15:46
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    $\begingroup$ Ethan, concerning the distribution: Suppose, for the sake of imagining what might go on, that the dimensions of the explosive charge in the cannon have a joint Normal distribution. That implies (from geometry alone) that the volume of the charge itself varies like the cube of a Normal distribution. Assuming muzzle speed is proportional to the mass of the charge and the mass if proportional to its volume, that would yield a positively skewed, non-Normal distribution of speeds. Regardless, the answer might not depend too strongly on the shape of the distribution. $\endgroup$
    – whuber ♦
    Commented May 23, 2017 at 16:01

1 Answer 1

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We know for a normal distribution that $${ \left( n-1 \right) s^2 \over \sigma^2} \sim \chi^2_{n-1} $$

Equivalently (since a chi-squared distribution is a gamma distribution), we can say that $${ \left( n-1 \right) s^2 \over \sigma^2} \sim \textrm{Gamma} \left( \frac{n-1}{2},2 \right) .$$

Now if $X \sim \textrm{Gamma} \left( \alpha, \beta \right) ,$ then $kX \sim \textrm{Gamma} \left( \alpha, k \beta \right) $

Therefore $$ s^2 \sim \textrm{Gamma} \left( \frac{n-1}{2},\frac{2 \sigma^2}{n-1} \right)$$

For the values you have of $n$ and $\sigma^2,$ that means $$s^2 \sim \textrm{Gamma} \left(7,\frac{100}{7} \right) $$

Now checking the CDF for this gamma random variable, $$P[s^2 < 4]=2.096444E-08,$$ which is about 1 in 47,699,809

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  • $\begingroup$ Thanks for the reply soakley. I'm a bit confused going from eq 1 to 2. Is that a Chi^2 on the RHS of your first equation? How does that jive with a gamma distribution in the second equation. Also, can you provide a reference/link where I can see the justification (or even a derivation) for the first equation? (Or, if you could provide any details, that would be great too). $\endgroup$
    – EthanT
    Commented May 25, 2017 at 20:04
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    $\begingroup$ That is a chi-squared distribution and a $\chi^2_p$ distribution is a special case of the Gamma Distribution, where $\alpha = p/2$ and $\beta=2$, using the pdf characterization described in Casella and Berger, 2nd ed., 2002. $\endgroup$
    – StatsStudent
    Commented May 25, 2017 at 20:12
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    $\begingroup$ onlinecourses.science.psu.edu/stat414/node/174 has a derivation of statement 1 $\endgroup$
    – Max S.
    Commented May 25, 2017 at 20:17
  • $\begingroup$ Where does it help the OP understand how to get to a Gamma from a chi-squared distribution on that page, @MaxS.? $\endgroup$
    – StatsStudent
    Commented May 25, 2017 at 20:32
  • $\begingroup$ It doesn't. OP asked "Also, can you provide a reference/link where I can see the justification (or even a derivation) for the first equation?" @Analyst1 $\endgroup$
    – Max S.
    Commented May 25, 2017 at 20:37

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