Question

a company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. over the years, they have established the following probability distribution. let ( x = ) the number of years a new hire will stay with the company. let ( p(x) = ) the probability that a new hire will stay with the company ( x ) years. on average, how long would you expect a new hire to stay with the company? (enter an exact number as an integer, fraction, or decimal.) yr \\(\

$$\begin{array}{|c|c|} \\hline x & p(x) \\\\ \\hline 0 & 0.13 \\\\ \\hline 1 & 0.17 \\\\ \\hline 2 & 0.30 \\\\ \\hline 3 & 0.15 \\\\ \\hline 4 & \\\\ \\hline 5 & 0.05 \\\\ \\hline 6 & 0.05 \\\\ \\hline \\end{array}$$

\\)

Explanation:

Step1: Find \( P(4) \)

The sum of all probabilities in a probability distribution is 1. So, we calculate \( P(4) \) as:
\( P(4)=1-(0.13 + 0.17+0.30 + 0.15+0.05 + 0.05) \)
\( =1-(0.13+0.17 = 0.3; 0.3+0.30 = 0.6; 0.6+0.15 = 0.75; 0.75+0.05 = 0.8; 0.8+0.05 = 0.85) \)
\( =1 - 0.85=0.15 \)

Step2: Calculate the expected value \( E(X) \)

The formula for the expected value of a discrete random variable is \( E(X)=\sum_{i}x_iP(x_i) \).
So we calculate each term:
- For \( x = 0 \): \( 0\times0.13 = 0 \)
- For \( x = 1 \): \( 1\times0.17 = 0.17 \)
- For \( x = 2 \): \( 2\times0.30 = 0.6 \)
- For \( x = 3 \): \( 3\times0.15 = 0.45 \)
- For \( x = 4 \): \( 4\times0.15 = 0.6 \)
- For \( x = 5 \): \( 5\times0.05 = 0.25 \)
- For \( x = 6 \): \( 6\times0.05 = 0.3 \)

Now sum all these terms:
\( E(X)=0 + 0.17+0.6 + 0.45+0.6 + 0.25+0.3 \)
\( 0.17+0.6 = 0.77; 0.77+0.45 = 1.22; 1.22+0.6 = 1.82; 1.82+0.25 = 2.07; 2.07+0.3 = 2.37 \) Wait, no, wait:
Wait, let's recalculate:
\( 0\times0.13=0 \)
\( 1\times0.17 = 0.17 \)
\( 2\times0.30=0.6 \)
\( 3\times0.15 = 0.45 \)
\( 4\times0.15=0.6 \)
\( 5\times0.05 = 0.25 \)
\( 6\times0.05 = 0.3 \)

Now sum: \( 0+0.17 = 0.17; 0.17+0.6 = 0.77; 0.77+0.45 = 1.22; 1.22+0.6 = 1.82; 1.82+0.25 = 2.07; 2.07+0.3 = 2.37 \)? Wait, no, wait I made a mistake in Step1? Wait no, wait \( P(4) \): Let's recalculate the sum of probabilities:
\( 0.13+0.17 = 0.3; 0.3+0.30 = 0.6; 0.6+0.15 = 0.75; 0.75+0.05 = 0.8; 0.8+0.05 = 0.85 \). Then \( 1 - 0.85 = 0.15 \), that's correct.

Now recalculate \( E(X) \):
\( 0\times0.13 = 0 \)
\( 1\times0.17 = 0.17 \)
\( 2\times0.30 = 0.6 \)
\( 3\times0.15 = 0.45 \)
\( 4\times0.15 = 0.6 \)
\( 5\times0.05 = 0.25 \)
\( 6\times0.05 = 0.3 \)

Now sum all these: \( 0+0.17 = 0.17; 0.17+0.6 = 0.77; 0.77+0.45 = 1.22; 1.22+0.6 = 1.82; 1.82+0.25 = 2.07; 2.07+0.3 = 2.37 \)? Wait, no, wait 0.17+0.6 is 0.77, plus 0.45 is 1.22, plus 0.6 is 1.82, plus 0.25 is 2.07, plus 0.3 is 2.37? Wait, but let's check again:

Wait \( 0.17+0.6 = 0.77 \)

\( 0.77+0.45 = 1.22 \)

\( 1.22+0.6 = 1.82 \)

\( 1.82+0.25 = 2.07 \)

\( 2.07+0.3 = 2.37 \). Wait, but let's check the calculation of \( P(4) \) again. Wait, the sum of the given probabilities: 0.13 (x=0), 0.17 (x=1), 0.30 (x=2), 0.15 (x=3), 0.05 (x=5), 0.05 (x=6). Sum: 0.13+0.17=0.3; +0.30=0.6; +0.15=0.75; +0.05=0.8; +0.05=0.85. So 1 - 0.85 = 0.15 for x=4. That's correct.

Wait, but let's recalculate the expected value:

\( E(X)=0\times0.13 + 1\times0.17+2\times0.30 + 3\times0.15+4\times0.15+5\times0.05 + 6\times0.05 \)

\( = 0 + 0.17+0.6 + 0.45+0.6 + 0.25+0.3 \)

Now add 0.17 + 0.6 = 0.77; 0.77 + 0.45 = 1.22; 1.22 + 0.6 = 1.82; 1.82 + 0.25 = 2.07; 2.07 + 0.3 = 2.37. Wait, but let's check with another approach. Wait, maybe I made a mistake in \( P(4) \). Wait, let's sum all probabilities again:

0.13 (x=0) + 0.17 (x=1) = 0.3

+0.30 (x=2) = 0.6

+0.15 (x=3) = 0.75

+P(4) +0.05 (x=5) +0.05 (x=6) = 1

So 0.75 + P(4) + 0.05 + 0.05 = 0.75 + P(4)+0.10 = 0.85 + P(4) = 1 => P(4)=0.15. Correct.

Now, let's recalculate the terms:

x=0: 0*0.13=0

x=1:1*0.17=0.17

x=2:2*0.30=0.6

x=3:3*0.15=0.45

x=4:4*0.15=0.6

x=5:5*0.05=0.25

x=6:6*0.05=0.3

Now sum all these: 0 + 0.17 = 0.17; 0.17 + 0.6 = 0.77; 0.77 + 0.45 = 1.22; 1.22 + 0.6 = 1.82; 1.82 + 0.25 = 2.07; 2.07 + 0.3 = 2.37. Wait, but let's check with a calculator:

0.17 + 0.6 = 0.77

0.77 + 0.45 = 1.22

1.22 + 0.6 = 1.82

1.82 + 0.25 = 2.07

2.07 + 0.3 = 2.37. So the expected value is 2.37? Wait, but let's check again. Wait, maybe I miscalculated the sum of the terms. Wait, 0.17 (x=1) + 0.6 (x=2) = 0.77; +0.45 (x=3) = 1.22; +0.6 (x=4) = 1.82; +0.25 (x=5) = 2.07; +0.3 (x=6) = 2.37. And x=0 term is 0. So total is 2.37. Wait, but let's check if the problem has a typo or if I made a mistake. Wait, the probabilities: 0.13, 0.17, 0.30, 0.15, 0.15, 0.05, 0.05. Sum: 0.13+0.17=0.3; +0.30=0.6; +0.15=0.75; +0.15=0.9; +0.05=0.95; +0.05=1.0. Oh! Wait a minute! I made a mistake in calculating P(4). Wait, the sum of the given probabilities (excluding P(4)) is 0.13 + 0.17+0.30 + 0.15+0.05 + 0.05 = 0.13+0.17=0.3; +0.30=0.6; +0.15=0.75; +0.05=0.8; +0.05=0.85. But wait, no: 0.13 (x=0) + 0.17 (x=1) = 0.3; +0.30 (x=2) = 0.6; +0.15 (x=3) = 0.75; +0.05 (x=5) = 0.8; +0.05 (x=6) = 0.85. So P(4) = 1 - 0.85 = 0.15. But when we sum all probabilities: 0.13+0.17+0.30+0.15+0.15+0.05+0.05 = 0.13+0.17=0.3; +0.30=0.6; +0.15=0.75; +0.15=0.9; +0.05=0.95; +0.05=1.0. Ah! I see my mistake. The sum of the given probabilities (excluding P(4)) is 0.13 + 0.17+0.30 + 0.15+0.05 + 0.05 = 0.13+0.17=0.3; 0.3+0.30=0.6; 0.6+0.15=0.75; 0.75+0.05=0.8; 0.8+0.05=0.85. But 0.13+0.17+0.30+0.15+0.05+0.05 = 0.13+0.17=0.3; +0.30=0.6; +0.15=0.75; +0.05=0.8; +0.05=0.85. So P(4) = 1 - 0.85 = 0.15. Then when we sum all probabilities: 0.13+0.17+0.30+0.15+0.15+0.05+0.05 = 0.13+0.17=0.3; +0.30=0.6; +0.15=0.75; +0.15=0.9; +0.05=0.95; +0.05=1.0. So that's correct. Then why when I calculated the expected value, I got 2.37? Wait, let's recalculate the expected value:

x=0: 0 * 0.13 = 0

x=1: 1 * 0.17 = 0.17

x=2: 2 * 0.30 = 0.6

x=3: 3 * 0.15 = 0.45

x=4: 4 * 0.15 = 0.6

x=5: 5 * 0.05 = 0.25

x=6: 6 * 0.05 = 0.3

Now sum these: 0 + 0.17 = 0.17; 0.17 + 0.6 = 0.77; 0.77 + 0.45 = 1.22; 1.22 + 0.6 = 1.82; 1.82 + 0.25 = 2.07; 2.07 + 0.3 = 2.37. Wait, but let's check with another approach. Let's list all terms:

0*0.13 = 0

1*0.17 = 0.17

2*0.30 = 0.6

3*0.15 = 0.45

4*0.15 = 0.6

5*0.05 = 0.25

6*0.05 = 0.3

Now add them step by step:

0 + 0.17 = 0.17

0.17 + 0.6 = 0.77

0.77 + 0.45 = 1.22

1.22 + 0.6 = 1.82

1.82 + 0.25 = 2.07

2.07 + 0.3 = 2.37

Yes, so the expected value is 2.37.

Answer:

2.37