An accounting office has 6 incoming telephone line

Questions: An accounting office has 6 incoming telephone lines. The probability distribution of the number of busy lines, is as follows: x P(x) ------ 0 0.01 1 0.12 2 0.2 3 0.32 4 0.19 5 0.15 6 0.01 Use the above probability distribution table to find the following: a. P(3<X<5)= (Round the answer to 2 decimals) b. P(1<X<4)= (Round the answer to 2 decimals) c. P(X ≤ 5)= (Round the answer to 2 decimals)

Transcript text: An accounting office has 6 incoming telephone lines. The probability distribution of the number of busy lines, is as follows: \begin{tabular}{|c|c|} \hline$x$ & $P(x)$ \\ \hline 0 & 0.01 \\ \hline 1 & 0.12 \\ \hline 2 & 0.2 \\ \hline 3 & 0.32 \\ \hline 4 & 0.19 \\ \hline 5 & 0.15 \\ \hline 6 & 0.01 \\ \hline \end{tabular} Use the above probability distribution table to find the following: a. $P(3

Solution

Solution Steps

To solve the given problems, we will use the provided probability distribution table. We will sum the probabilities for the specified ranges and round the results to two decimal places.

a. For $ P(3 < X < 5) $, we need to sum the probabilities for $ X = 4 $.

b. For $ P(1 < X < 4) $, we need to sum the probabilities for $ X = 2 $ and $ X = 3 $.

c. For $ P(X \leq 5) $, we need to sum the probabilities for $ X = 0, 1, 2, 3, 4, $ and $ 5 $.

Step 1: Calculate $ P(3 < X < 5) $

To find $ P(3 < X < 5) $, we consider the probability for $ X = 4 $: \[ P(3 < X < 5) = P(X = 4) = 0.19 \]

Step 2: Calculate $ P(1 < X < 4) $

To find $ P(1 < X < 4) $, we sum the probabilities for $ X = 2 $ and $ X = 3 $: \[ P(1 < X < 4) = P(X = 2) + P(X = 3) = 0.2 + 0.32 = 0.52 \]

Step 3: Calculate $ P(X \leq 5) $

To find $ P(X \leq 5) $, we sum the probabilities for $ X = 0, 1, 2, 3, 4, $ and $ 5 $: \[ P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.01 + 0.12 + 0.2 + 0.32 + 0.19 + 0.15 = 0.99 \]