The average rainfall is normally distributed. How many times would you expect there to be less than 52 mm of rainfall?
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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Mad Banners
Updated on August 15, 2022Comments
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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?
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Did about 7 yearsThey 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 $μ-2σ$"?)
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Mad Banners about 7 yearsOh, 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.
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Did about 7 yearsSorry but why is $3\sigma$ even mentioned here?
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Mad Banners about 7 yearsI'm just saying, if you're going to write the measurements in a uniform manner, do the same for each one. Right? @Did
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Did about 7 years?? "Less than 52 mm" corresponds to "less than $\mu-2\sigma$", not to "between $\mu-3\sigma$ and $\mu-2\sigma$".
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Mad Banners about 7 yearsThat's correct. If it was "between $μ−3σμ−3σ and μ−2σ$" however, would it be valid to write this as $−3σ≤μ≤−2σ$?
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BruceET about 7 yearsNot really, ... only in the sense that there is almost 0 probability below $\mu - 3\sigma.$
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Mad Banners about 7 yearsI 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?
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BruceET about 7 yearsCure 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
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Did about 7 years@BruceET Wise words. (+1)
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Mad Banners about 7 yearsAsked 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