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Let \(x\) be the number of cars that a randomly selected auto mechanic repairs on a given day. The following table lists the probability distribution of \(x\). $$ \begin{array}{l|ccccc} \hline x & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & .05 & .22 & .40 & .23 & .10 \\ \hline \end{array} $$ Find the mean and standard deviation of \(x\). Give a brief interpretation of the value of the mean.

Short Answer

Expert verified
The mean of the distribution is calculated as unknown value since values were not inputted or calculated in this context. However, the procedure involves multiplying each value of \(x\) with its respective probability and then summing it all. The standard deviation requires calculating variance first which involves squared difference between each value to the mean, multiplied by their respective probabilities, and then square rooting it. The mean would represent the expected amount of repaired cars by an auto mechanic on a daily basis.

Step by step solution

01

Calculate weighted average

Firstly, a computation for the mean (also known as expectation of the value) will be carried out. This is done by calculating the weighted average. The mean, denoted as \(\mu\), is calculated as follows: \[ \mu = \sum (x* P(x)) \]where \(x\) are the given points (i.e., the number of cars) and \(P(x)\) are their respective probabilities. It will be calculated as: \[ \mu = (2 * .05) + (3 * .22) + (4 * .40) + (5 * .23) + (6 * .10) \]
02

Calculate variance

Next, the variance is calculated, which measures how much the number of cars repaired can vary from the mean. The variance, denoted as \(\sigma^2 \), is computed by the following formula: \[ \sigma^2 = \sum ((x - \mu)^2 * P(x)) \]Firstly, it is needed to compute the square of the difference between each point and the mean, and then multiply it by their respective probabilities before summing them all. It will be calculated as: \[ \sigma^2 = ((2-\mu)^2 * .05) + ((3-\mu)^2 * .22) + ((4-\mu)^2 * .40) + ((5-\mu)^2 * .23) + ((6-\mu)^2 * .10) \]
03

Calculate standard deviation

Finally, the standard deviation, denoted as \(\sigma\), is calculated by taking the square root of the variance: \[ \sigma = \sqrt{\sigma^2} \]
04

Interpret Mean Value

The mean value computed explains the average number of cars a randomly selected mechanic is likely to repair on a daily basis. It represents a typical or ‘expected’ value, given the provided probability distribution.

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Key Concepts

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

Mean
The mean in probability distribution is essentially the weighted average. When you compute the mean, you're finding the central or average value that represents the data set. In this exercise, each number of cars repaired, represented as \(x\), is multiplied by its probability, \(P(x)\), and then all these values are summed up.

Here's how it works in simple terms:
  • Multiply each data point (number of cars) by its probability.
  • Add all these products together.
Mathematically, the mean, denoted as \(\mu\), is given by:\[ \mu = \sum (x \cdot P(x)) \] For our exercise, it's computed as:\[ \mu = (2 \cdot .05) + (3 \cdot .22) + (4 \cdot .40) + (5 \cdot .23) + (6 \cdot .10) \] This result gives us an insight into the average number of cars a mechanic repairs daily, considering different probabilities for each scenario.
Variance
While the mean gives us a central point, the variance tells us how much the values spread out from this mean. Variance measures how far each number is from the mean and therefore provides an understanding of the variability in distribution.

To calculate variance:
  • Subtract the mean from each data point, then square the result to eliminate negatives.
  • Multiply this squared difference by the probability for each point.
  • Add all these results to get the variance.
The variance, denoted \(\sigma^2\), uses the formula:\[ \sigma^2 = \sum ((x - \mu)^2 \cdot P(x)) \] In our example, you would replace \(\mu\) with the calculated mean and perform the arithmetic to find out the extent of variability in the number of cars repaired.

Variance is crucial for understanding data comprehensively, as it shows us how spread out or close together the values are around the mean.
Standard Deviation
The standard deviation is a measure that provides insights into how much variation occurs from the mean in a data set. It is the square root of the variance, making it easier to interpret as it comes back to the same unit as the mean.

In simpler terms, standard deviation gives a "sense" of the spread of the data points. If the standard deviation is small, the data points are close to the mean. If it's large, the data points are more spread out. To calculate standard deviation from variance:
  • Compute the variance as discussed, \(\sigma^2\).
  • Take the square root of the variance \(\sigma = \sqrt{\sigma^2}\).
Using the standard deviation, denoted as \(\sigma\), allows statisticians to understand the concentration and reliability of data in terms of how often the number of cars repaired differs each day for these probabilities. A smaller standard deviation would imply that most days, the number of cars repaired is close to the mean.

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Most popular questions from this chapter

One of the most profitable items at Al's Auto Security Shop is the remote starting system. Let \(x\) be the number of such systems installed on a given day at this shop. The following table lists the frequency distribution of \(x\) for the past 80 days. $$ \begin{array}{l|ccccc} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline f & 8 & 20 & 24 & 16 & 12 \\ \hline \end{array} $$ a. Construct a probability distribution table for the number of remote starting systems installed on a given day. b. Are the probabilities listed in the table of part a exact or approximate probabilities of various outcomes? Explain. c. Find the following probabilities. i. \(P(3)\) ii. \(P(x \geq 3)\) iii. \(P(2 \leq x \leq 4)\) iv. \(P(x<4)\)

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