Questions: The data given below show the number of overtime hours worked in one week per employee. Use the data to complete parts (a) and (b). Overtime hours 0 1 2 3 4 5 6 b 6 12 27 59 46 28 19 Employees (a) Construct a probability distribution. (b) Graph the probability distribution using a histogram and describe its shape. (a) Construct the probability distribution by completing the table below x 0 P(x) 1 2 3 4 5 6 P(x) (Round to three decimal places as needed.)

The data given below show the number of overtime hours worked in one week per employee. Use the data to complete parts (a) and (b). Overtime hours 0 1 2 3 4 5 6 b 6 12 27 59 46 28 19 Employees (a) Construct a probability distribution. (b) Graph the probability distribution using a histogram and describe its shape. (a) Construct the probability distribution by completing the table below x 0 P(x) 1 2 3 4 5 6 P(x) (Round to three decimal places as needed.)
Transcript text: The data given below show the number of overtime hours worked in one week per employee. Use the data to complete parts (a) and (b). Overtime hours \begin{tabular}{rrrrrrrrr} 0 & 1 & 2 & 3 & 4 & 5 & 6 & b \\ 6 & 12 & 27 & 59 & 46 & 28 & 19 & \end{tabular} Employees (a) Construct a probability distribution. (b) Graph the probability distribution using a histogram and describe its shape. (a) Construct the probability distribution by completing the table below \[ \begin{array}{cc} x & 0 \\ \mathrm{P}(\mathrm{x}) & \square \end{array} \] \[ \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ \square & \square & & \square & \end{array} \] 6 $\mathrm{P}(\mathrm{x})$ (Round to three decimal places as needed.)
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Solution

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Solution Steps

Step 1: Calculate the total number of employees

The total number of employees is the sum of the given frequencies: \[ 6 + 12 + 27 + 59 + 46 + 28 + 19 = 197 \]

Step 2: Calculate the probability for each number of overtime hours

The probability for each number of overtime hours is calculated by dividing the frequency of each number of hours by the total number of employees.

\[ \begin{array}{cc} x & \mathrm{P}(x) \\ 0 & \frac{6}{197} \approx 0.0305 \\ 1 & \frac{12}{197} \approx 0.0610 \\ 2 & \frac{27}{197} \approx 0.1371 \\ 3 & \frac{59}{197} \approx 0.2995 \\ 4 & \frac{46}{197} \approx 0.2330 \\ 5 & \frac{28}{197} \approx 0.1421 \\ 6 & \frac{19}{197} \approx 0.0964 \\ \end{array} \]

Final Answer

\[ \begin{array}{cc} x & \mathrm{P}(x) \\ 0 & 0.0305 \\ 1 & 0.0610 \\ 2 & 0.1371 \\ 3 & 0.2995 \\ 4 & 0.2330 \\ 5 & 0.1421 \\ 6 & 0.0964 \\ \end{array} \]

{"axisType": 3, "coordSystem": {"xmin": -1, "xmax": 7, "ymin": 0, "ymax": 0.35}, "commands": ["y = 0.0305", "y = 0.0610", "y = 0.1371", "y = 0.2995", "y = 0.2330", "y = 0.1421", "y = 0.0964"], "latex_expressions": ["$P(x=0) = 0.0305$", "$P(x=1) = 0.0610$", "$P(x=2) = 0.1371$", "$P(x=3) = 0.2995$", "$P(x=4) = 0.2330$", "$P(x=5) = 0.1421$", "$P(x=6) = 0.0964$"]}

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