Question

Help with last two parts

Answer 1

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Also sum of probabilities from 0 till n = 1, i.eP(0) + P(1) + P(2) +.......+P(n) = 1

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(d) Standard Deviation = SQRT[n * p * (1 - p)]

Here n = 4, p = 0.45 and 1 - p = 0.55

Therefore Standard Deviation = Sqrt(4 * 0.45 *0.55) = 0.99498 $\approx$1.00 (Rounding to 2 decimal places)

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(e) We need to be at least 93% sure, therefore the required minimum probability = 0.93

We also need to interview a minimum of 3 professors.

Let the Total professors to be interviewed = n

Therefore P(3) + P(4) + ------ + P(n) = 0.93

We know that P(0) + P(1) + P(2) + ------+ P(n) = 1

Therefore P(3) + P(4) + ------ + P(n) = 1 - [P(0) + P(1) + P(2)]

1 - [ⁿC₀ * (0.45)⁰ * (0.55)ⁿ + ⁿC₁ * (0.45)¹ * (0.55)^{n -1} + ⁿC₂ * (0.45)² * (0.55)^n-2] = 0.93

By Trial and error, we get n = 11