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 Leibovici Aug 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 $.5P(X_1<.94)+.5P(X_2<.94)=.44$.

  • $\begingroup$ How do you justify your solution to the second question? $\endgroup$ – Landon Carter Aug 4 '16 at 16:58
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Aug 4 '16 at 17:02

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