Ave you ever tried to get out of jury duty? about 25% of those called will find an excuse (work, poor health, travel out of town, etc.) to avoid jury duty.†(a) if 11 people are called for jury duty, what is the probability that all 11 will be available to serve on the jury? (round your answer to three decimal places.) 0.042 correct: your answer is correct. (b) if 11 people are called for jury duty, what is the probability that 5 or more will not be available to serve on the jury? (round your answer to three decimal places.) 0.115 correct: your answer is correct. (c) find the expected number of those available to serve on the jury. what is the standard deviation? (round your answers to two decimal places.)μ = 8 incorrect: your answer is incorrect. peopleσ = 1.436 correct: your answer is correct. people

The answers to the question are;

(a) 0.042 to three decimal places.

(b) 0.115 to three decimal places.

(c) μ = 8.25, σ = 1.44 to two decimal places.

Step-by-step explanation:

To solve the question, we note that this is a binomial distribution problem

(a) The probability that all 11 will be available is given by

P(11) = ₁₁C₁₁ × 0.75¹¹×0.25¹¹⁻¹¹ = 0.0422

(b) Probability of 6 or less success = P(6) + P(5) +P(4) +P(3) + P(2) + P(1) +P(0)

P(6)  = ₁₁C₆ × 0.75⁶×0.25⁵ = 0.080299

P(5) = ₁₁C₅ × 0.75⁵×0.25⁶ = 0.026766

P(4) = ₁₁C₄ × 0.75⁴×0.25⁷ = 0.063729

P(3) = ₁₁C₃ × 0.75³×0.25⁸ = 0.001062

P(2) = ₁₁C₂ × 0.75²×0.25⁹ = 0.0001180

P(1) = ₁₁C₁ × 0.75¹×0.25¹⁰ = 0.000007867

P(0) = ₁₁C₀ × 0.75⁰×0.25¹¹ = 0.00000002384

Therefore P(6) + P(5) +P(4) +P(3) + P(2) + P(1) +P(0) = 0.11462

Which is 0.115 to three decimal places

(c) The expected number of those available to serve on the jury =

Probability of success = n·p = 11×0.75 = 8.25

μ = 8.25

The standard deviation,σ = = 1.43614 ≅ 1.44 to two decimal places

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