Three businesswomen are trying to convene in Northwest Arkansas for a business meeting. The first (Woman 1) is arriving on a flight from Atlanta, the second (Woman 2) is arriving on a flight from Dallas, and the third (Woman 3) is arriving on a flight from Chicago. Historical data suggests that the Atlanta flight is "on time" 90% of the time, the Dallas flight is on time 95%of the time, and the Chicago flight is on time 80% of the time. Furthermore, historical data suggests that the three flights are independent with respect to on time behavior. Let X denote the number of women who arrive on time.

What is the probability mass function function of X?

What is the cumulative distribution function of X?

What is the probability that at least two businesswomen arrive on time?

What is the expected value of X?

What is the standard deviation of X?

a) The probability mass function of X is then presented in the table below.

X | probability P(X=x) or p

0 | 0.001

1 | 0.032

2 | 0.283

3 | 0.684

b) The cumulative distribution function of X

Cdf = Σ pdf = P(X=0) + P(X=1) + P(X=2) + P(X=3)

= 1.000

c) The probability that at least two businesswomen arrive on time

P(X ≥ 2) = P(X=2) + P(X=3) = 0.967

d) Expected value of X = E(X) = 2.65

e) Standard deviation = 0.545

Step-by-step explanation:

The probability that the woman coming from Atlanta arrives on time = P(A) = 0.90

The probability that the woman coming from Atlanta DOES NOT arrive on time = P(A') = 1 - 0.90 = 0.10

The probability that the woman coming from Dallas arrives on time = P(B) = 0.95

The probability that the woman coming from Dallas DOES NOT arrive on time = P(B') = 1 - 0.95 = 0.05

The probability that the woman coming from Chicago arrives on time = P(C) = 0.80

The probability that the woman coming from Chicago DOES NOT arrive on time = P(C') = 1 - 0.80 = 0.20

Since X is the random variable that represents how many women arrive on time,

To evaluate the probability function, we will first obtain the probability that the number of women that arrive in time = 0, 1, 2, and 3.

First probability; that no woman arrives on time. X = 0

P(X=0) = P(A') × P(B') × P(C')

= 0.10 × 0.05 × 0.20

P(X=0) = 0.001

Second probability; that only one of the women arrive on time. X = 1

P(X=1) = [P(A) × P(B') × P(C')] + [P(A') × P(B) × P(C')] + [P(A') × P(B') × P(C)]

= [0.90 × 0.05 × 0.20] + [0.10 × 0.95 × 0.20] + [0.10 × 0.05 × 0.80]

= 0.009 + 0.019 + 0.004

P(X=1) = 0.032

Third probability; that only two women arrive on time. X = 2

P(X=2) = [P(A) × P(B) × P(C')] + [P(A) × P(B') × P(C)] + [P(A') × P(B) × P(C)]

= [0.90 × 0.95 × 0.20] + [0.90 × 0.05 × 0.80] + [0.10 × 0.95 × 0.80]

= 0.171 + 0.036 + 0.076

P(X=2) = 0.283

Fourth probability; that all 3 women arrive on time. X = 3

P(X=3) = P(A) × P(B) × P(C)

= 0.90 × 0.95 × 0.8

P(X=3) = 0.684

The probability mass function of X is then presented in the table below.

X | probability P(X=x) or p

0 | 0.001

1 | 0.032

2 | 0.283

3 | 0.684

b) The cumulative distribution function of X

Cdf = Σ pdf = P(X=0) + P(X=1) + P(X=2) + P(X=3)

= 0.001 + 0.032 + 0.283 + 0.684 = 1.000

c) The probability that at least two businesswomen arrive on time

P(X ≥ 2) = P(X=2) + P(X=3) = 0.283 + 0.684 = 0.967

d) Expected value of X

Expected value is given as

E(X) = Σ xᵢpᵢ

E(X) = (0)(0.001) + (1)(0.032) + (2)(0.283) + (3)(0.684) = 0 + 0.032 + 0.566 + 2.052 = 2.65

e) What is the standard deviation of X?

Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = E(X) = 2.65

Σx²p = (0²)(0.001) + (1²)(0.032) + (2²)(0.283) + (3²)(0.684)

= (0)(0.001) + (1)(0.032) + (4)(0.283) + (9)(0.684)

= 0 + 0.032 + 1.132 + 6.156

= 7.32

Variance = Var(X) = 7.32 - 2.65² = 7.32 - 7.0225

Var(X) = 0.2975

Standard deviation = √(variance) = √0.2975

Standard deviation = 0.545

Hope this Helps!!!

146 days

step-by-step explanation:

1. (365*2)/5 =

730/5 =

146 days

2. 365/5=

73*2   = 146

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