Question

At an airport, 79% of recent flights have arrived on time. A sample of 7flights is studied. a.Compute the mean of this probability distribution. Round to two decimal places, if needed.b.Compute the standard deviation of this probability distribution. Round to two decimal places, if needed.c.Find the probability that exactly4of the flights were on time. Round to three decimal places.d.Find the probability that lessthan 4of the flights were on time. Round to three decimal places.e.Find the probability that more than 5of the flights were on time. Round to three decimal places. f.Find the probability that at least5of the flights were on time. Round to three decimal places.g.Find the probability that no more than 5of the flights were on time. Round to three decimal places

Answer 1

Answer:

a. 5.53

b. 1.078

c. 0.126

d. 0.109

e. 0.549

f. 0.834

g.  0.451

Step-by-step explanation:

The percentage of the flights that arrive on time, P(x) = 79%

The number of flights in the sample, n = 7 flights

a. The mean of the probability distribution, μ = ∑x·P(x)

Therefore, we have; μₓ = n·p

μₓ = 7 × 79/100 = 5.53

b. The standard deviation, σₓ = √(n·p·(1 - p))

∴ σₓ = √(7 × 0.79 × (1 - 0.79)) ≈ 1.078

c. We have;

p = 0.79

q = 1 - p = 1 - 0.79 = 0.21

By binomial probability distribution formula, we have;

The probability of exactly four, P(Exactly 4) = ₇C₄·p⁴·q³

P(Exactly 4) = 35 × 0.79⁴×0.21³ ≈ 0.12625

d. The probability of less than 4 is given as follows;

P(Less than 4) = ₇C₀·p⁰·q⁷ + ₇C₁·p¹·q⁶ + ₇C₂·p²·q⁵ + ₇C₃·p³·q⁴

∴ P(Less than 4) = 1×0.79^0 * 0.21^7 + 7 * 0.79^1 × 0.21^6 + 21*0.79^2*0.29^5+ 85×0.79^3*0.21^4 ≈ 0.109

The probability of less than 4 is ≈ 0.109

e. The probability that more than 5 is given as follows;

P(More than 5) = ₇C₆·p⁶·q¹ + ₇C₇·p⁷·q⁰

7×0.79^6 * 0.21 + 1 * 0.79^7 × 0.21^0 ≈ 0.549

f. The probability that at least 5 of the flight were on time is given as follows;

P(At least 5) = ₇C₅·p⁵·q² + ₇C₆·p⁶·q¹ + ₇C₇·p⁷·q⁰

∴ P(At least 5) = 21×0.79^5 * 0.21^2 + 7×0.79^6 * 0.21 + 1 * 0.79^7 × 0.21^0 ≈ 0.834

g.  For the probability that no more than 5 of the flights were on time, e have;

P(At most 5) = 1 - P(More than 5)

∴ P(At most 5) = 1 - 0.549 ≈ 0.451.