Answer:
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!!!
. The first term reduces to 
, the second to 
, and the third to 3. Putting them all together with the signs in between we have 