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.
 A: 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
$$ P(X \le 40)
= \int_{(-\infty,40]} f_X(x)\,\mathrm{d}x
= \frac{1}{80} \int_{(-\infty,40]} \chi_{[20,80]}(x)\,\mathrm{d}x
= \frac{1}{80} \int_{20}^{40} 1\,\mathrm{d}x
= \frac{20}{80}
= \frac{1}{4}.
$$
A: 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
