7. In a forest, the probability that a randomly sampled tree is oak is 0.1, pine 0.3, maple 0.35 and 0.25 is the probability that a randomly sampled tree is of any other type. If 8 trees are sampled randomly from a forest (Hint: You may also think of the problem as probability of a tree being oak is 0.1 and non-oak is 0.9).a) (6 pts) What is the probability that 2 of the 8 trees sampled is oak?P(Y=8)= 8!/(2!(8-2)!(.1)^2(1-.1)^6=.1488Binomial with n = 8 and p = 0.1x P( X = x )1 0.3826382 0.1488033 0.0330674 0.0045935 0.0004086 0.0000237 0.0000018 0.000000b) (6 pts) What is the probability that at least one of the 8 trees sampled is either an oak or a pine?c) (5 pts) Let Y be a random variable denoting the number of maple trees in the sample of 8. What is the expected value of Y (mean value of Y)? If this value is not a whole number, will you round it up to a whole number? d) (3 pts) What is the standard deviation of Y?
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(a) P(x = 2) = 0.1488
(b) P(At least one is Oak or Pine) = 1 - P(None of the trees is either Oak or Pine)
= 1 - [8!/0! (8 - 0)! 0.4^0 0.68]
= 1 - 0.0168
(c) p = 0.35
Expected value of Y = np = 8 * 0.35 = 2.8
The rounded up answer is 3
(d) Standard deviation of Y = √(np(1 - p))
= √(8 * 0.35 * (1 - 0.35))
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