Calculating probability when range of values is given
Assume the number of pounds of peanuts consumed in a year is equally likely to assume any value (including fractions) between 20 and 100 pounds. What is the probability that someone consumes less than 40 pounds of peanuts in a year?
Which method should I use to solve this problem?
If we wish to solve this using normal distribution, mean and standard deviation are unknown. I tried solving this assuming mean = 60 and calculating standard deviation with assumption that 99.7% of the area falls under 3 standard deviations from mean.
Binomial distribution does not seem to fit here. Which other distribution does fit here?
The answer is 0.25.
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$\begingroup$In probability, the phrasing of a question is often really, really important. In this case, the phrasing is a bit imprecise, but I think that the intended meaning is clear: the distribution is uniform ("each quantity is equally likely") and continuous ("the quantity could be any number, including fractions"). Thus the correct probability distribution is a continuous uniform distribution on the interval $[20,100]$. The quick-and-dirty computation is that $$ P(X \le 40) = \frac{|40-20|}{|100-20|} = \frac{20}{80} = \frac{1}{4}. $$ That is, we compare the length of the interval corresponding to the "good" outcomes to the total length of the interval.
Slightly more rigorously, if we let $X$ denote the number of pounds of peanuts consumed, then $X$ is uniform on $[20,100]$, which implies that $X$ has probability density function $$ f_X = C\chi_{[20,100]}, $$ where $C$ is a normalizing constant such that $$ 1 = \int_{\mathbb{R}} f_X(x)\,\mathrm{d}x. $$ This condition is necessary, as $f_X$ is a pdf, and so must integrate to 1 over $\mathbb{R}$. Integrating, we obtain $$ 1 = C \int_{\mathbb{R}} \chi_{[20,100]}(x),\mathrm{d}x = C \int_{20}^{100} 1\,\mathrm{d}x = 80 C \implies C = \frac{1}{80}.$$ Then $$ P(X \le 40) = \int_{(-\infty,40]} f_X(x)\,\mathrm{d}x = \frac{1}{80} \int_{(-\infty,40]} \chi_{[20,80]}(x)\,\mathrm{d}x = \frac{1}{80} \int_{20}^{40} 1\,\mathrm{d}x = \frac{20}{80} = \frac{1}{4}. $$
$\endgroup$ $\begingroup$This interval consists of 4 smaller intervals , 20-40, 40-60,60-80,80-100.
As the probability distribution is uniform, the probability distribution over 20-40 is 1/4 i.e. 0.25
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