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Chapter 6: Problem 37

The vice president of human resources. at Lowes is studying the number of on- the-job accidents over a period of one month. He developed the following probability distribution. Compute the mean, variance, and standard deviation of the number of accidents in a month. $$\begin{array}{|cc|}\hline \text { Number of } & \\\\\text { Accidents } & \text { Probability } \\\\\hline 0 & .40 \\\1 & .20 \\\2 & .20 \\\3 & .10 \\\4 & .10 \\\\\hline\end{array}$$

Short Answer

Expert verified
Mean: 1.30, Variance: 1.81, Standard Deviation: 1.345.

Step by step solution

01

Identify Random Variable and Probabilities

The random variable, \(X\), represents the number of on-the-job accidents in a month. The probabilities for each number of accidents are:\(P(X=0) = 0.40\), \(P(X=1) = 0.20\), \(P(X=2) = 0.20\), \(P(X=3) = 0.10\), \(P(X=4) = 0.10\).
02

Calculate the Mean (Expected Value)

The mean of a probability distribution is calculated as \(\mu = \sum \left[ x_i \times P(x_i) \right]\).Here, \(\mu = (0 \times 0.40) + (1 \times 0.20) + (2 \times 0.20) + (3 \times 0.10) + (4 \times 0.10)\).Calculate each term: \(0 = 0\), \(0.20 = 0.20\), \(0.40 = 0.40\), \(0.30 = 0.30\), \(0.40 = 0.40\).Sum them up: \(\mu = 0 + 0.20 + 0.40 + 0.30 + 0.40 = 1.30\).
03

Calculate the Variance

The variance of a probability distribution is calculated using \(\sigma^2 = \sum [(x_i - \mu)^2 \times P(x_i)]\).For each \(x_i\), calculate \((x_i - \mu)^2\):- \((0 - 1.3)^2 = 1.69\)- \((1 - 1.3)^2 = 0.09\)- \((2 - 1.3)^2 = 0.49\)- \((3 - 1.3)^2 = 2.89\)- \((4 - 1.3)^2 = 7.29\).Then, \(\sigma^2 = (1.69 \times 0.40) + (0.09 \times 0.20) + (0.49 \times 0.20) + (2.89 \times 0.10) + (7.29 \times 0.10)\).Calculate each product: \(0.676\), \(0.018\), \(0.098\), \(0.289\), \(0.729\).Add them: \(\sigma^2 = 0.676 + 0.018 + 0.098 + 0.289 + 0.729 = 1.81\).
04

Calculate the Standard Deviation

The standard deviation is the square root of the variance. Calculate \(\sigma = \sqrt{1.81}\), which approximately equals \(1.345\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation
The mean of a probability distribution, often called the expected value, represents the long-term average or central value of a set of outcomes. In simple terms, it tells us what we can "expect" to happen most of the time. For this exercise, the number of on-the-job accidents in a month is the random variable, denoted as \(X\). We are given the probability of each number of accidents. To compute the mean (\(\mu\)), follow this formula:
\[\mu = \sum (x_i \times P(x_i))\]
Here, each \(x_i\) is a possible number of accidents, and \(P(x_i)\) is its corresponding probability.

  • Calculate each term: Multiply each accident number by its probability.
  • Add them: Sum the products to find the mean.

For instance, with the provided probabilities, the mean \(\mu\) is:
\[ 0 \times 0.40 + 1 \times 0.20 + 2 \times 0.20 + 3 \times 0.10 + 4 \times 0.10 = 1.30 \]
This value, 1.30, indicates the average number of accidents expected in one month.
Variance Calculation
Variance provides insight into how spread out a distribution is. It measures the average degree to which each number differs from the mean. The higher the variance, the more spread out the numbers are. The variance of a probability distribution (denoted by \(\sigma^2\)) is calculated using:
\[\sigma^2 = \sum ((x_i - \mu)^2 \times P(x_i))\]
Here's a breakdown of the calculation steps:
  • Subtract the mean (\(\mu\)) from each possible number of accidents to find \(x_i - \mu\).
  • Square these results. This step amplifies larger deviations.
  • Multiply each squared deviation by its corresponding probability.
  • Sum all these products to find the variance.
Using the given distribution, the variance is:
\[ (0 - 1.3)^2 \times 0.40 + (1 - 1.3)^2 \times 0.20 + (2 - 1.3)^2 \times 0.20 + (3 - 1.3)^2 \times 0.10 + (4 - 1.3)^2 \times 0.10 = 1.81 \]
This result, 1.81, represents the spread of accident occurrences around the average.
Standard Deviation Calculation
The standard deviation is a crucial measure in statistics that offers a more understandable gauge of spread than variance. It is simply the square root of the variance and it shares the same units as the mean, thus making it easier to interpret. To calculate the standard deviation (denoted \(\sigma\)), use the formula:
\[\sigma = \sqrt{\sigma^2}\]

  • Find the square root of the variance.
This exercise shows that the variance was calculated as 1.81. Taking the square root of this value, we find:
\[\sigma = \sqrt{1.81} \approx 1.345\]
This indicates that, on average, the number of accidents will vary about 1.345 around the mean. The smaller the standard deviation, the closer the numbers are to the mean, indicating less variability in monthly accident occurrences.

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