# What's the difference between the probability of these weights? [closed]

I have a random x variable of weight(pounds) for a cucumber. It is uniformly distributed between 0.5 and 1.5 (with avg value 1 pound)

Suppose 2 cucumbers are selected at random.

Find the probability that both will weigh less than 0.94 pounds

Find the prob that their avg weight will be less than 0.94 pounds

## closed as off-topic by heropup, Chill2Macht, Daniel W. Farlow, JMP, Claude LeiboviciAug 5 '16 at 5:37

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We can define the probability density function for this problem from the standard uniform p.d.f., namely $$f_{X_i}(x_i)= \frac{1}{1.5-.5} = 1 \ \text{for} \ x\in(.5, 1.5) \ \text{and} \ 0 \ \text{elsewhere}, i=\{1, 2\}$$ Therefore, the cumulative density function will be the integral of this: $$\int_{.5}^{x}f_{Y_i}(y_i)dy_i = \frac{y-.5}{1}$$ Therefore, if we want to find the probability of 1 of the cucumbers being less than .94 pounds, simply plug into c.d.f. to yield $P(X_i<.94)= .44$. Thus the probability that two of the cucumbers will be less than .94, due to independence of cucumbers is $.44^2=0.1936$. Next, the probability that the avg. weight will be less than .94 points will be $.5P(X_1<.94)+.5P(X_2<.94)=.44$.
• The probability that average weight is less than $0.94$ has to do ith the distribution of the sum. That part of the answer needs modification. – André Nicolas Aug 4 '16 at 17:02