Answer:
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 [tex]=\sqrt{npq}[/tex] =[tex]\sqrt{11*0.75*0.25}[/tex] = 1.43614 ≅ 1.44 to two decimal places