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Subject: probabilities An agricultural cooperative claims 95 percent of the watermelons shipped out are ripe and ready to eat. If 20 watermelons are shipped out , what is the probability that the number of watermelon that are ripe and ready to is (i) exactly 14 (ii) more than 18 (iii) of the 20 watermelons that are shipped what is the expected number of ripe and ready to eat watermelons thank for you assistance
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Hi James, If 95% of the watermelons are ripe and ready to ship and you ship 20 watermelons then the expected number of ripe and ready to eat watermelons you shipped is 95% of 20 which is 0.95 Penny |
20 = 19. The solution to the remainder of this problem comes from seeing it as a binomial distribution. You randomly select a watermelon from those produced by the agricultural cooperative, success is selecting a watermelon that is ripe and ready to eat, p is the probability of success which is then 0.95 and X is the number of successes in 20 trials. |
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