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I am currently studying about Sampling distribution of Sample means, and came across below example here.

Question:
The average male drinks 2L of water when active outdoors with a standard deviation of .7L. You're planning a full day nature trip for 50 men and will bring 110L of water. What's the probability you'll run out?

Given Answer:(transcript taken from here which is same as khan's)
The probability of running out of water is the probability of using more than 110L of water. This is the same as the probability of the average water use is greater than 2.2L (110L divided by 50 men) per man

P(average water use > 2.2L per man)

$\mu_\bar{x}$ = $\mu$ = 2L
$\sigma_\bar{x}^2 = \dfrac {\sigma^2}{n}$
$\sigma_\bar{x} = \dfrac {\sigma}{\sqrt{n}} = \dfrac {0.7}{\sqrt{50}} = 0.099$

We just need to figure out how many standard deviations 2.2L is away from the mean (known as the z-score)

$\dfrac {2.2-\mu}{\sigma} = \dfrac {2.2-2}{0.099} = 2.02$

The probability that average water us > 2.2L per man is the same as probability that the sample mean will be more than 2.02 standard deviations above the mean. Now you can use a z-table to figure out that probability.

0.9783 is the probability that we're less than 2.02 standard deviations above the mean

P(running out of water) = 1 - .9783 = .0217

My questions:

  1. Broadly, what is the inference getting a 2.17%, how is different from 1% or 3% or even 5% practically? What realistic action or usefulness there could be out of this inference?

  2. Average male indicates huge or even entire population of male. And sampling distribution with just 50 men isn't too small to consider as a normal distribution? (given that, we do not know about population distribution in question. Unless one assumes that also as normal)

  3. Even if 50 is normal, it is just 1 sample (of size 50 men). Shouldn't we get a normal distribution only when we repeat this N-trials or N-number of times, to have the normal distribution effect to take place?

  4. How is it not Sampling distribution of sample proportion?

  5. Isn't it counter intuitive that our sampling distribution has lower SD ( so higher certainty), makes one wonder if sampling distribution is better than population distribution? How latter is more beneficial?

Kindly clarify.

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  • $\begingroup$ The "Disturbing" thing about the question is that it is using mathematics to justify carrying the least theoretical amount of water needed by humans to stay in their comfort zone. Take twice the recommended average. $\endgroup$ – Weather Vane Aug 9 '18 at 18:11
  • $\begingroup$ An engineer would take three times the recommended average. $\endgroup$ – Weather Vane Aug 9 '18 at 18:27
  • $\begingroup$ Normal distributions are often useful, especially for modeling probabilities near the average. But there is an adage among those who use normal approximations in real life: "Never trust normal tails." As a practical matter with a supply of 110 L, your guesses of 1% and 3% may be about as good as 2.17% for the probability of running out. $\endgroup$ – BruceET Aug 9 '18 at 18:44
  • $\begingroup$ Even more disturbing is using theoretical people for mathematics, as if they are just numbers, and not real people. Which century does this come from? $\endgroup$ – Weather Vane Aug 9 '18 at 18:57
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The main problem with this question, I think, is that we don't know that your $50$ men are a random sample from the population of all males, or how the activity they will be doing will compare to whatever those average males were doing. If the weather is like what it is in much of North America and Europe these days, they'll need a lot more water!

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My comments are from a practical aspect of this analysis.

  1. The $2.17\%$ probability is a metric used to compare against a predetermined risk factor. In this particular case, the consequences of running out of water would have to be assessed for an acceptable level of risk. Is someone going to die or become dehydrated or just a little thirsty? Normally a predetermined risk factor would be $.05$. If lives are at stake then $.01$ or even smaller may be more appropriate.

  2. Ask yourself if the $50$ men on a nature trip are a typical cross section of the male population. They probably weren't a random sample so are likely to be biased or non representative in some way. If they are older or different in some way that would effect their hydration then this needs to be taken into account.

  3. You are mistaking or taking this sample as a simple random sample which it probably isn't. Even so, what aspects of normal are you looking at to decide if the sample is normal? If the population is normal, and the sample is random and large enough, then the criteria for a valid test have been met.

  4. Please Clarify this point.

  5. No, it is a recognized phenomenon that generally a random sample will have a smaller SD than the population it came from. Hence the calculation of sample SD versus population SD by division by $n-1$ versus $n$ to correct for this anomaly. The goal is to obtain a valid and accurate test by applying criteria that have been determined to work best for a majority of cases. It isn't an exact method and is prone to error. However, in a well designed test, measures can be taken to limit these errors as much as possible

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  • $\begingroup$ For 4, the bigger picture is I am currently trying to understand "Sampling distribution of Sample proportions" and "Sampling distribution of Sample means", and unable to distinguish when to use what. The example in question came in latter, so wondering how its sample means. In both sample proportions and sample means, afaik we take samples, take average and plot Sample distribution graph. I am sorry if my clarification is incomplete, but it just represents my current understanding so kindly enlighten on that as well if possible $\endgroup$ – Parthiban Rajendran Aug 9 '18 at 18:57
  • $\begingroup$ For 3, it is assumed in answer it is normal, and I counter that, how could it be as sample size seem too less (1 sample of 50 men or 50 samples of 1 men) and we do not know population distribution (if its normal or not) from question. $\endgroup$ – Parthiban Rajendran Aug 9 '18 at 18:58
  • $\begingroup$ This case is applicable to sample mean whereby the mean amount of water consumed by the average male is used to determine a quantity to take and then the probability of running out whereby the water is shared as needed. A sample proportion would be more applicable to a scenario to determine what proportion of the sample are likely to run out of water if each has 2.2 L? You seem to be hung up on the sample being normally distributed. The analysis determines the probability of a skewed distribution and different mean from a normal population and hence the probability of running out of water. $\endgroup$ – Phil H Aug 9 '18 at 20:09
  • $\begingroup$ could not understand how to fit "proportion of the sample are likely to run out of water if each has 2.2 L" in the problem, can you kindly elaborate that in your answer? $\endgroup$ – Parthiban Rajendran Aug 10 '18 at 5:38
  • $\begingroup$ The probability of a single person running out of water, mean 2L, SD = 0.7 with a quantity of 2.2L is p = .3875. So we would expect 39 out of 100 people to run out of water. A one proportion z test type question would be very different and something like, if the probability of a person running out of water is 0.39, what is the probability that no more than 30 out of a 100 people will run out of water? A one proportion z test gives p = .0325 with a z = -1.845. Neither of these apply as the probability of the whole group running out of water is based on the mean consumption of the whole group. $\endgroup$ – Phil H Aug 10 '18 at 16:26

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