lab each 1 4. On Exam 1, you had the following situation: The 6 people in your b

Question

lab each 1 4. On Exam 1, you had the following situation: The 6 people in your botany go on separate trips into a wild area; and complete an assignment identify one of the 20 rare species known to live in that area. Assume equally easy to identify. to each correctly the 20 species are time we will ask a different question: On average, how many diferent species do a (Spts.) Let X, be 1 if the ith species is identified, and O if none of the 6 people b) (5 pts.) IfX species identified, find E(X). This your 6 people identify? We wil use the method of indicators: 0). , 20). Find the P(x, identify that species (so i , 2,

Explanation / Answer

Probability of correctly identifying a species by a student = 1/20

Hence, Probability of not correctly identifying a species by a student = 1-1/20=19/20

a) P(Xi=0) = Probability that ith species is not identified by any one of six students =

(19/20)(19/20)(19/20)(19/20)(19/20)(19/20) = (19/20)^6

Therefore, P(Xi=1) = 1- (19/20)^6

b) X is number of species identified

So, X = X1 + X2 + ... + X20

E[X1]=E[X2] = ...= E[X20] = (0)(19/20)^6 + 1(1-(19/20)^6) = 1- (19/20)^6

E[X] = E[X1 + X2 + ... + X20] = E{X1] + E[X2] + ... + E[X20] = 20E[X1] = 20(1-(19/20)^6) = 20(0.265) = 5.39(approx. 5)

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