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Use the probability distribution for the random variable \(x\) to answer the questions. $$\begin{array}{l|lllll}x & 0 & 1 & 2 & 3 & 4 \\\\\hline p(x) & .1 & .3 & .3 & ? & .1\end{array}$$ Calculate the population mean, variance, and standard deviation.

Short Answer

Expert verified
& .1\end{array}$$ Answer: The population mean is 1.9, the population variance is 1.29, and the standard deviation is approximately 1.14.

Step by step solution

01

Find the missing probability value

To find the missing probability value, we have to make sure the sum of all probabilities is equal to 1. The sum of given probabilities is 0.1+0.3+0.3+0.1 = 0.8. Therefore, the missing probability (when \(x=3\)) is: $$ p(3) = 1 - 0.8 = 0.2 $$ Now, the updated probability distribution table is: $$\begin{array}{l|lllll}x & 0 & 1 & 2 & 3 & 4 \\\\\hline p(x) & .1 & .3 & .3 & .2 & .1\end{array}$$
02

Calculate the population mean

The population mean (µ) can be calculated with the following formula: $$ \mu = \sum_{i=1}^n x_i p(x_i) $$ Thus, the population mean is: $$ \mu = (0 \cdot 0.1) + (1 \cdot 0.3) + (2 \cdot 0.3) + (3 \cdot 0.2) + (4 \cdot 0.1) = 1.9 $$
03

Calculate the population variance

The population variance (σ²) can be calculated with the following formula: $$ \sigma^2 = \sum_{i=1}^n (x_i - \mu)^2 p(x_i) $$ Thus, the population variance is: $$ \sigma^2 = (0-1.9)^2 \cdot 0.1 + (1-1.9)^2 \cdot 0.3 + (2-1.9)^2 \cdot 0.3 + (3-1.9)^2 \cdot 0.2 + (4-1.9)^2 \cdot 0.1 = 1.29 $$
04

Calculate the population standard deviation

The population standard deviation (σ) is the square root of the population variance: $$ \sigma = \sqrt{\sigma^2} = \sqrt{1.29} \approx 1.14 $$ So, the population mean is 1.9, the population variance is 1.29, and the population standard deviation is approximately 1.14.

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Key Concepts

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

Understanding Population Mean
The population mean, denoted as \( \mu \), is a fundamental concept in statistics, representing the average value in a population. In the context of probability distributions, calculating the mean involves multiplying each possible outcome by its corresponding probability and then summing all these products.

To visualize this, imagine you have a set of weighted numbers, where each number (\(x\)) represents a possible outcome, and the weight (\(p(x)\)) is the probability of that outcome occurring. The population mean is like the balance point of a scale that has these weighted numbers on it. In the exercise provided, the mean is calculated as \( \mu = (0 \cdot 0.1) + (1 \cdot 0.3) + (2 \cdot 0.3) + (3 \cdot 0.2) + (4 \cdot 0.1) = 1.9 \).

  • Tip 1: Always make sure that the probabilities add up to 1; otherwise, the calculations will not reflect the true mean of the distribution.
  • Tip 2: Remember, the mean is a measure of central tendency, indicating where the center of the data distribution lies.
Deciphering Population Variance
Population variance, denoted as \( \sigma^2 \), measures the spread of a set of data points in a population. To put it simply, it tells us how much the values in a distribution deviate from the mean, on average. This measure of dispersion is particularly useful because it gives us an idea of the diversity or variability within our data set.

The variance is calculated by taking the squared differences between each outcome and the mean, then multiplying these by their respective probabilities, and finally, adding them all together. Using the improved probability distribution from the exercise, we get \( \sigma^2 = (0-1.9)^2 \cdot 0.1 + (1-1.9)^2 \cdot 0.3 + (2-1.9)^2 \cdot 0.3 + (3-1.9)^2 \cdot 0.2 + (4-1.9)^2 \cdot 0.1 = 1.29 \).

  • Tip 1: Computing variance requires prior knowledge of the mean, emphasizing the mean's foundational role in statistical analysis.
  • Tip 2: A smaller variance indicates that the data points are closer to the mean, while a larger variance shows greater spread.
Population Standard Deviation Made Simple
The population standard deviation, represented by the Greek letter \( \sigma \), is the square root of the population variance. It's a powerful statistic offering insights into the amount of variation or dispersion from the population mean, in the same units as the data itself. A lower standard deviation means the data points are closer to the mean, and a higher value indicates a wider spread of data.

In the exercise at hand, by taking the square root of the variance \( \sigma^2 \), we find the standard deviation: \( \sigma = \sqrt{1.29} \approx 1.14 \).

  • Tip 1: Standard deviation provides an intuitive sense of the variability because it's in the same units as the original data, which isn't the case with variance.
  • Tip 2: It's easier to compare the standard deviation to the mean than the variance, which can offer a quick gauge of relative dispersion in a data set.

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Most popular questions from this chapter

In Exercises 34-37, explain why \(x\) is or is not a binomial random variable. (Hint: compare the characteristics of this experiment with those of a binomial experiment given in this section.) If the experiment is binomial, give the value of \(n\) and \(p\), if possible. Two balls are randomly selected without replacement from a jar that contains three red and two white balls. The number \(x\) of red balls is recorded.

For the random variables described, find and graph the probability distribution for \(x .\) Then calculate the mean, variance, and standard deviation. A piece of electronic equipment contains 6 computer chips, two of which are defective. Three chips are randomly selected and inspected, and \(x\), the number of defective chips in the selection is recorded.

Car color preferences change over the years and according to the particular model that the customer selects. In a recent year, suppose that \(10 \%\) of all luxury cars sold were black. If 25 cars of that year and type are randomly selected, find the following probabilities: a. At least five cars are black. b. At most six cars are black. c. More than four cars are black. d. Exactly four cars are black. e. Between three and five cars (inclusive) are black. f. More than 20 cars are not black.

A candy dish contains five brown and three red M&Ms. A child selects three M&Ms without checking the colors. Use this information to answer the questions in Exercises \(18-21 .\) Write down \(p(x)\), the probability distribution for \(x\), the number of red M&Ms in the selection for \(x=0,1,2,3\)

Use the probability distribution for the random variable \(x\) to answer the questions in Exercises 12-16. $$\begin{array}{l|rrrrrr}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ $$ \text { Find } \mu, \sigma^{2}, \text { and } \sigma \text { . } $$

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