48.7% of Americans have brown eyes. A convention has 5000 people in attendance.

Question

48.7% of Americans have brown eyes. A convention has 5000 people in attendance. Find the mean (expected value),the standard deviation, and the variance for the number of brown-eyed people at such gatherings of 5000 people.

2.

Assume a procedure yields a binomial distribution with a trial repeated 2 times. Using Table A-1, find the probability of 1 or 2 successes given that the probability of success on a single trial is 99%.

Select one:

a. .100

b. .002

c. 1.000

d. .020

3.

Forty percent of Clinton County residents are uncomfortable using the Internet. Suppose 7 such people are randomly selected. What is the probability exactly 5 are uncomfortable using the Internet?

Select one:

a. .025

b. .077

c. .070

d. .194

Explanation / Answer

1. Mean = np = (5000)(0.487) = 2435

  Variance = np(1 - p) = 1249.155

  Standard Dev. = sqrt(Variance) = 35.343387

2. p(k = 1) = C(n,k)*p^k*(1-p)^(n-k) = C(2,1)*(.99)^1*(0.01)^1 = 0.0198

p(k = 2) = C(2,2)*(.99)^2*(0.01)^0 = 0.9801

probability of 1 or 2 successes = p(k = 1) + p(k = 2) = 0.9999 (rounds up to 1?) answer is c

3. p(k = 5) = C(n,k)*p^k*(1-p)^(n-k) = C(7,5)*(0.4)^5*(0.6)^2 = 0.0774144 (rounds to 0.077) answer is b

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