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.
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