The average rainfall is normally distributed. How many times would you expect there to be less than 52 mm of rainfall?

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It seems as if you are trying to use the Empirical Rule to solve this problem. According to the ER, about 95% of the probability under a normal curve lies between $\mu \pm 2\sigma.$ Then by symmetry about 2.5% of observations lie below $\mu -2\sigma,$ which is $68 - 2(8) = 52$ for your rainfall distribution. So over a 40-year period, one might expect to see $40(.025) = 1$ year with rainfall that low.

If you know how to use printed tables of the standard normal CDF or to use statistical software, you might get a result that is a bit more accurate:

$$P(X < 52) = P\left(Z = \frac{X - \mu}{\sigma} < \frac{52 - 68}{8}\right) = \cdots,$$

where $Z$ is standard normal, and the procedure of subtracting the mean and then dividing by the standard deviation is called 'standardization'.

In R statistical software standardization (and rounding to use tables) can be avoided:

pnorm(52, 68, 8)
## 0.02275013           # 'about' .025
40*pnorm(52, 68, 8)
## 0.9100053            # 'about' 1

Notes: (1) I share @Did's doubts that $3\sigma$ has anything to do with fining your answer. (2) Also, with your guess that the answerbook is rounding to the nearest integer. (3) This is a fine drill problem, but it has to be an approximation that rainfall is normally distributed. If there is a discrepancy between the model and reality it is likely to be in the far 'tails' of the distribution, which is where your problem is focused.

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Updated on August 15, 2022

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  • Mad Banners
    Mad Banners about 1 year

    Full question:

    The average rainfall of Charlesville in August is normally distributed with mean 68 mm and standard deviation 8 mm. Over a 40 year period, how many times would you expect there to be less than 52 mm of rainfall during August in Charlesville?

    I drew a graph of the normal distribution for this data set. Then, finding that $<52$ mm of rain would likely constitute about 2.28% of the data (between $μ-3σ$ and $μ-2σ$), I multiplied $0.0228$ by $40$ to get $0.912$. The textbook answer says that one instance of $<52$ mm of rainfall could be expected over the $40$ year period. Have they just rounded $0.912$ to the nearest whole number, or are they going about it an entirely different way?

    • Did
      Did about 7 years
      They probably solved $$\mathrm{argmax}_n\ P(N_{40}=n)$$ where $N_{40}$ is binomial $(40,P(X<52))$ for $X$ normal $(68,8)$. Unrelated: why "between $μ-3σ$ and $μ-2σ$" instead of "less than $μ-2σ$"?)
    • Mad Banners
      Mad Banners about 7 years
      Oh, that's how they write it in my textbook. Realistically it would be better to write these à la $<μ−2σ$ or $-3σ≤μ≤3σ$, you're right.
    • Did
      Did about 7 years
      Sorry but why is $3\sigma$ even mentioned here?
    • Mad Banners
      Mad Banners about 7 years
      I'm just saying, if you're going to write the measurements in a uniform manner, do the same for each one. Right? @Did
    • Did
      Did about 7 years
      ?? "Less than 52 mm" corresponds to "less than $\mu-2\sigma$", not to "between $\mu-3\sigma$ and $\mu-2\sigma$".
    • Mad Banners
      Mad Banners about 7 years
      That's correct. If it was "between $μ−3σμ−3σ and μ−2σ$" however, would it be valid to write this as $−3σ≤μ≤−2σ$?
    • BruceET
      BruceET about 7 years
      Not really, ... only in the sense that there is almost 0 probability below $\mu - 3\sigma.$
  • Mad Banners
    Mad Banners about 7 years
    I didn't claim that $3σ$ had anything to do with finding the answer, though (spare of what was stated in my original post). @Did's and your answers also seem a tad complicated for this level. On my graphics calculator, using normal CD — with a lower of $2$, upper of $2^{98}$, standard deviation of 8 and mean of 68 — it gives me the result P=1. But why?
  • BruceET
    BruceET about 7 years
    Cure for (sometimes) getting answers at an appropriate level is to follow the site guidelines: tell us the level of the course and explain to us in more detail what you have tried. Also, this is not a commercial hwk-ans site. We don't specialize in giving 'screen-shot ready' answers. I suppose you are near the beginning of your course and hope you'll understand more of this as you move along. I have no idea why your graphics calculator didn't give the correct answer. (Graphics calculators are soooo 20th century.) R software works on Windows, Mac, and Linux machines; free from r-project.org.
  • Did
    Did about 7 years
    @BruceET Wise words. (+1)
  • Mad Banners
    Mad Banners about 7 years
    Asked my teacher. The answer is rounded up to 1; i.e. a probability of 50% or more = certainty. Apologies for the wild goose choose. @Did