Question

In: Statistics and Probability

wildlife officials tagged 80 deer in an area that had approximately 120 deer a) if they...

wildlife officials tagged 80 deer in an area that had approximately 120 deer

a) if they later took a sample of 25 deer, how many would they expect to have been tagged?

b) should the officials be surprised if the sample has fewer than 13 tagged deer?

Jamal is successful on basketball free throws 80% of the time.

a) how likely is he to be successful on eight of 10 free-throw attempts?

b) how likely is he to be successful on at least eight of 10 free-throws attempts?

Expert Solution

Q1) a) The expected number of tagged deer in the sample is computed here as:
= Sample size * Proportion of tagged deer in the population

= 25*(80/120)

= 16.67

Therefore 16.67 deer is the expected number here.

b) The probability to get fewer than 13 tagged deer is computed here as:

Note that this is a case of a hypergeometric distribution with parameters:
N = 120 as the population size,
n = 25 as the sample size,
K = 80 as the number of successes (that is tagged deer) in the population

and the random variable X here is the number of tagged deer in the sample taken.

The probability required here is:
P(X < 13) = P(X <= 12)

This is computed in EXCEL as:
=HYPGEOM.DIST(12,25,80,120,TRUE)

0.0252 is the output here.

As the probability is 0.0252 < 0.05, therefore yes the event is unusual and official should be surprised on the basis of this theoritical probability.

Q2) We are given here,
p = P(success) = 0.8

a) Probability to be successful on 8/10 attempts is computed using the binomial probability function as:

Therefore 0.3020 is the required probability here.

b) Probability of being successful on at least 8 of the 10 attempts is computed here as:

= P(X = 8) + P(X = 9) + P(x = 10)

Therefore 0.6778 is the required probability here.

orchestra answered 2 years ago

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