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Chapter 8: Problem 4

The probability distribution of a random variable \(X\) is given. Compute the mean, variance, and standard deviation of \(X\). $$\begin{array}{lllllll}\hline \boldsymbol{x} & 10 & 11 & 12 & 13 & 14 & 15 \\\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & 1 / 8 & 2 / 8 & 1 / 8 & 2 / 8 & 1 / 8 & 1 / 8 \\\\\hline\end{array}$$

Short Answer

Expert verified
The mean (Expected Value) of the random variable \(X\) is \(E[X] = 12\). The expected value of the square of the random variable \(X\), i.e., \(E[X^2] = 146\). Using these values, the variance of the random variable \(X\) is \(Var(X) = 2\). Finally, the standard deviation of the random variable \(X\) is \(SD(X) = \sqrt{2}\).

Step by step solution

01

Compute the mean (Expected Value)

Calculate the mean of the random variable \(X\) by multiplying the values of \(x\) with their respective probabilities and summing up the products. Using the given probability distribution, the mean is computed as: \(E[X] = (10 \times \frac{1}{8}) + (11 \times \frac{2}{8}) + (12 \times \frac{1}{8}) + (13 \times \frac{2}{8}) + (14 \times \frac{1}{8}) + (15 \times \frac{1}{8})\)
02

Compute the value of \(E[X^2]\)

Now, we need to calculate the expected value of the square of the random variable \(X\), i.e., \(E[X^2]\). This can be done using the same process as in step 1, but with the square of \(x\) instead: \(E[X^2] = (10^2 \times \frac{1}{8}) + (11^2 \times \frac{2}{8}) + (12^2 \times \frac{1}{8}) + (13^2 \times \frac{2}{8}) + (14^2 \times \frac{1}{8}) + (15^2 \times \frac{1}{8})\)
03

Compute the Variance

With \(E[X]\) and \(E[X^2]\) calculated, we can now compute the variance using the formula mentioned in the analysis: \(Var(X) = E[X^2] - (E[X])^2\)
04

Compute the Standard Deviation

Finally, the standard deviation can be calculated by taking the square root of the variance: \(SD(X) = \sqrt{Var(X)}\) After completing each step, you will have the mean, variance, and standard deviation of the random variable \(X\).

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Key Concepts

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

Mean
The mean, also known as the expected value, is a measure of the central tendency of a probability distribution. It provides a single number that describes the average outcome you would expect if you repeated an experiment many times.
In probability distributions, calculating the mean involves multiplying each possible value (\(x\)) by its probability (\(P(X = x)\)) and then summing these products.
  • Formula: \(E[X] = \sum (x \cdot P(X = x))\)
This formula tells us that for each value of \(x\), we find its contribution to the mean by weighing it according to how often it occurs. In our example, the values 10, 11, 12, 13, 14, and 15 are considered, each multiplied by their respective probabilities, and then added together to find \(E[X]\).
Thus, the mean is a powerful tool to give a quick snapshot of where the bulk of the probability mass is located in the distribution.
Variance
Variance is a measure that tells us how much the values of a random variable differ from the mean. It's a crucial concept for understanding the spread of the distribution.To calculate variance, you first determine the expected value of the squared deviations from the mean. This involves two main steps:

1. Calculate the expected value of the square of the random variable, \(E[X^2]\).
  • Formula: \(E[X^2] = \sum (x^2 \cdot P(X = x))\)
2. Use the variance formula:
  • \(Var(X) = E[X^2] - (E[X])^2\)
By squaring each value of the random variable before multiplying it by its probability, you give more weight to larger deviations. Once you subtract the square of the mean, you get the variance. This value can help you understand how spread out the values of the random variable are in relation to the mean.
Variance is always non-negative and a larger variance indicates more spread in the distribution.
Standard Deviation
Standard deviation is another measure of spread in a distribution, but it is more intuitive than variance. It indicates how much individual data points typically deviate from the mean.The standard deviation is simply the square root of the variance. This transformation brings the units of variance back to the same units as data, making it easier to interpret.
  • Formula: \(SD(X) = \sqrt{Var(X)}\)
Understanding standard deviation is crucial because it gives a real-sense feeling of variation in the data set. In our example, once the variance is computed, taking its square root provides the standard deviation, offering a straightforward depiction of how data disperses from the average across the distribution.
Generally, a small standard deviation implies the values are close to the mean, while a large standard deviation signals a wide spread of values. This knowledge is instrumental for statisticians and researchers when analyzing data trends and variability.

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Most popular questions from this chapter

In a four-child family, what is the expected number of boys? (Assume that the probability of a boy being born is the same as the probability of a girl being born.)

Records kept by the chief dietitian at the university cafeteria over a 30-wk period show the following weekly consumption of milk (in gallons). $$\begin{array}{lccccc}\hline \text { Milk } & 200 & 205 & 210 & 215 & 220 \\\\\hline \text { Weeks } & 3 & 4 & 6 & 5 & 4 \\\\\hline\end{array}$$ $$\begin{array}{lcccc}\hline \text { Milk } & 225 & 230 & 235 & 240 \\\\\hline \text { Weeks } & 3 & 2 & 2 & 1 \\ \hline\end{array}$$ a. Find the average number of gallons of milk consumed per week in the cafeteria. b. Let the random variable \(X\) denote the number of gallons of milk consumed in a week at the cafeteria. Find the probability distribution of the random variable \(X\) and compute \(E(X)\), the expected value of \(X\).

A man wishes to purchase a 5 -yr term-life insurance policy that will pay the beneficiary $$\$ 20,000$$ in the event that the man's death occurs during the next 5 yr. Using life insurance tables, he determines that the probability that he will live another \(5 \mathrm{yr}\) is \(.96\). What is the minimum amount that he can expect to pay for his premium?

The accompanying data were obtained in a study conducted by the manager of SavMore Supermarket. In this study, the number of customers waiting in line at the express checkout at the beginning of each 3 -min interval between 9 a.m. and 12 noon on Saturday was observed. $$\begin{array}{llllll}\hline \text { Customers } & 0 & 1 & 2 & 3 & 4 \\ \hline \text { Frequency of } & & & & & \\ \text { Occurrence } & 1 & 4 & 2 & 7 & 14 \\ \hline\end{array}$$ $$\begin{array}{lllllll}\hline \text { Customers } & 5 & 6 & 7 & 8 & 9 & 10 \\\\\hline \text { Frequency of } & & & & & & \\\\\text { Occurrence } & 8 & 10 & 6 & 3 & 4 & 1 \\\\\hline\end{array}$$ a. Find the probability distribution of the random variable \(X\), where \(X\) denotes the number of customers observed waiting in line. b. Draw the histogram representing the probability distribution.

A survey was conducted by the market research department of the National Real Estate Company among 500 prospective buyers in a large metropolitan area to determine the maximum price a prospective buyer would be willing to pay for a house. From the data collected, the distribution that follows was obtained. Compute the mean, variance, and standard deviation of the maximum price \(x\) (in thousands of dollars) that these buyers were willing to pay for a house. $$ \begin{array}{ll} \hline \text { Maximum Price } & \\ \text { Considered, } x & P(X=x) \\ \hline 280 & \frac{10}{500} \\ \hline 290 & \frac{20}{500} \\ \hline 300 & \frac{75}{500} \\ \hline 310 & \frac{85}{506} \\ \hline 320 & \frac{70}{500} \\ \hline 350 & \frac{90}{500} \\ \hline 380 & \frac{90}{500} \\ \hline 400 & \frac{55}{500} \\ \hline 450 & \frac{5}{500} \\ \hline\end{array}$$
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