(b) {fewer than three lines are in use} (c) (at least three lines are in use} (d) {between two and five lines, inclusive, are in use}

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Q13  Please answer b, c, and d

### Cumulative Distribution Function (CDF) Calculation for a Phone Line Usage Scenario

#### Problem Statement
A mail-order computer business has six telephone lines. Let \( X \) denote the number of lines in use at a specified time. Suppose the probability mass function (pmf) of \( X \) is given in the table below.

| \( x \)   | 0    | 1    | 2    | 3    | 4    | 5    | 6    |
|-----------|------|------|------|------|------|------|------|
| \( p(x) \)| 0.12 | 0.15 | 0.20 | 0.23 | 0.20 | 0.07 | 0.03 |

#### Objective
Calculate the cumulative distribution function \( F(x) \).

#### Cumulative Distribution Function (CDF) Calculation

The cumulative distribution function \( F(x) \) is calculated as follows:

- \( F(0) = P(X \leq 0) = P(X = 0) = 0.12 \)
- \( F(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = 0.12 + 0.15 = 0.27 \)
- \( F(2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.12 + 0.15 + 0.20 = 0.47 \)
- \( F(3) = P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.12 + 0.15 + 0.20 + 0.23 = 0.70 \)
- \( F(4) = P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.12 + 0.15 + 0.20 + 0.23 + 0.20 = 0.90 \)
- \( F(5) = P(X \leq 5
Transcribed Image Text:### Cumulative Distribution Function (CDF) Calculation for a Phone Line Usage Scenario #### Problem Statement A mail-order computer business has six telephone lines. Let \( X \) denote the number of lines in use at a specified time. Suppose the probability mass function (pmf) of \( X \) is given in the table below. | \( x \) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |-----------|------|------|------|------|------|------|------| | \( p(x) \)| 0.12 | 0.15 | 0.20 | 0.23 | 0.20 | 0.07 | 0.03 | #### Objective Calculate the cumulative distribution function \( F(x) \). #### Cumulative Distribution Function (CDF) Calculation The cumulative distribution function \( F(x) \) is calculated as follows: - \( F(0) = P(X \leq 0) = P(X = 0) = 0.12 \) - \( F(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = 0.12 + 0.15 = 0.27 \) - \( F(2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.12 + 0.15 + 0.20 = 0.47 \) - \( F(3) = P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.12 + 0.15 + 0.20 + 0.23 = 0.70 \) - \( F(4) = P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.12 + 0.15 + 0.20 + 0.23 + 0.20 = 0.90 \) - \( F(5) = P(X \leq 5
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