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Statistics Problems

In: Other Topics

Submitted By joe2315
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Statistical Analysis 1
MA260

1. Compute the mean and variance of the following discrete probability distribution. X P(X) 2 .50 8 .30 10 .20

Mean 2*.50+8*.30+10*.20 = 5.4 Variance 2^2*.50+8^2*.30+10^2*.20-5.40^2 = 2.00+19.20+20.00-29.16 = 12.04

2a. The total number of ways to choose 3 members out of 8 is C(8, 3) = 8!/(3! 5!) = (8*7*6*5*4*3*2*1) / ((3*2*1) * (5*4*3*2*1)) = 56. The total number of ways to choose 3 tenured people from 6 is C(6,3) = 6!/(3! 3!) = (6*5*4*3*2*1) / ((3*2*1) * (3*2*1)) = 20. So the probability is 20/56 = 5/14.

2b. Probability (at least one member is not tenured) = 1 – Probability (All three are tenured) = 1- 5/14 = 9/14.

3. P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = e^(-2) (1 + 2 + 2 + 8/6 + 16/24) = 7 e^(-2) = 0.9473 = 94.73%.

They have not lived up to their internal goal of 95%.

4a. P(X = 0) = (5/9)^3 = 0.1715. P(X = 1) = 3 * (4/9) * (5/9)^2 = 0.4115. P(X = 2) = 3 * (4/9)^2 * (5/9) = 0.3292. P(X = 3) = (4/9)^3 = 0.0878

4b. P(X ≥1) = 1 - P(X = 0) = 1 - 0.1715 = 0.8285

References
Bennett, J., Briggs , W., & Triola, M. (2014). Statical Reasoning for Everyday Life. Boston, MA: Pearson.…...

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