Question

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
employees 8 14 34 57 41 30 15
(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 1 2 3 4 5 6
p(x)
(round to three decimal places as needed.)

Explanation:

Step1: Calculate total number of employees

First, we sum up the number of employees for each overtime hour. The number of employees for 0,1,2,3,4,5,6 hours are 8,14,34,57,41,30,15 respectively. So total number of employees \( N=8 + 14+34 + 57+41+30+15\). Let's calculate that: \(8+14 = 22\), \(22+34=56\), \(56 + 57=113\), \(113+41 = 154\), \(154+30=184\), \(184 + 15=199\). Wait, wait, let's recalculate: 8+14=22; 22+34=56; 56+57=113; 113+41=154; 154+30=184; 184+15=199? Wait, no, 8+14=22, 22+34=56, 56+57=113, 113+41=154, 154+30=184, 184+15=199. Wait, maybe I made a mistake. Let's add again: 8 + 14 = 22; 22 + 34 = 56; 56 + 57 = 113; 113 + 41 = 154; 154 + 30 = 184; 184 + 15 = 199. So total \(N = 199\).

Step2: Calculate probabilities for each x

The probability \(P(x)=\frac{\text{Number of employees with }x\text{ overtime hours}}{N}\)

For \(x = 0\): \(P(0)=\frac{8}{199}\approx0.040\) (rounded to three decimal places)

For \(x = 1\): \(P(1)=\frac{14}{199}\approx0.070\)

For \(x = 2\): \(P(2)=\frac{34}{199}\approx0.171\)

For \(x = 3\): \(P(3)=\frac{57}{199}\approx0.286\)

For \(x = 4\): \(P(4)=\frac{41}{199}\approx0.206\)

For \(x = 5\): \(P(5)=\frac{30}{199}\approx0.151\)

For \(x = 6\): \(P(6)=\frac{15}{199}\approx0.075\)

Wait, let's check the sum: 0.040+0.070=0.11; +0.171=0.281; +0.286=0.567; +0.206=0.773; +0.151=0.924; +0.075=0.999, which is approximately 1 (due to rounding). So that's correct.

Answer:

For \(x = 0\), \(P(x)\approx0.040\); \(x = 1\), \(P(x)\approx0.070\); \(x = 2\), \(P(x)\approx0.171\); \(x = 3\), \(P(x)\approx0.286\); \(x = 4\), \(P(x)\approx0.206\); \(x = 5\), \(P(x)\approx0.151\); \(x = 6\), \(P(x)\approx0.075\)

So the table is:

x	0	1	2	3	4	5	6