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Ch. 4 - Discrete Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 4, Problem 4.1.41

Linear Transformation of a Random Variable In Exercises 41 and 42, use this information about linear transformations. For a random variable x, a new random variable y can be created by applying a linear transformation , where a and b are constants. If the random variable x has mean and standard deviation , then the mean, variance, and standard deviation of y are given by the formulas


The mean annual salary of employees at an office is originally \$46,000. Each employee receives an annual bonus of \$600 and a 3% raise (based on salary). What is the new mean annual salary (including the bonus and raise)?

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Step 1: Understand the linear transformation formula. A new random variable y can be created using the formula y = a + b * x, where 'a' represents a constant addition (bonus) and 'b' represents a scaling factor (raise percentage).
Step 2: Identify the given values. The mean annual salary of employees (mean of x) is \$46,000. The bonus (a) is \$600, and the raise percentage (b) is 3%, which can be expressed as 0.03 in decimal form.
Step 3: Apply the formula for the mean of the transformed random variable y. The formula for the mean of y is given by: mean(y) = a + b * mean(x). Substitute the values of a, b, and mean(x) into this formula.
Step 4: Perform the substitution. Replace 'a' with 600, 'b' with 0.03, and 'mean(x)' with 46,000 in the formula mean(y) = a + b * mean(x).
Step 5: Simplify the expression to find the new mean annual salary. Combine the terms to calculate the new mean, which includes the bonus and raise.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Linear Transformation

A linear transformation involves modifying a random variable using a linear equation of the form y = ax + b, where 'a' and 'b' are constants. This transformation affects the mean and standard deviation of the original variable. Specifically, the mean of the new variable y is calculated as E(y) = aE(x) + b, while the standard deviation is adjusted by the absolute value of 'a', given by SD(y) = |a|SD(x).
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Mean

The mean, or expected value, of a random variable is a measure of central tendency that represents the average outcome. It is calculated by summing all possible values of the variable, each weighted by its probability. In the context of salary, the mean provides a baseline for understanding the average earnings of employees before any adjustments like bonuses or raises are applied.
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Variance and Standard Deviation

Variance measures the spread of a set of values around the mean, indicating how much the values differ from the average. The standard deviation is the square root of the variance and provides a more interpretable measure of spread in the same units as the original data. In salary calculations, understanding variance and standard deviation helps assess the consistency of salaries among employees.
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Related Practice
Textbook Question

Unusual Events In Exercise 19, would it be unusual for a household to have no HD televisions? Explain your reasoning.

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Textbook Question

In Exercises 5–8, find the indicated probability using the Poisson distribution.


P(3) when μ = 6

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Textbook Question

Using a Distribution to Find Probabilities In Exercises 11–26, find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine whether the events are unusual. If convenient, use a table or technology to find the probabilities.


Immigration The mean number of people who immigrated to the United States per hour was about 5.5 in April 2021. Find the probability that the number of people who immigrate to the U.S. in a given hour in April 2021 was (a) zero, (b) exactly five, and (c) exactly eight. (Source: U.S. Census Bureau)

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Textbook Question

Independent and Dependent Random Variables Two random variables x and y are independent when the value of x does not affect the value of y. When the variables are not independent, they are dependent. A new random variable can be formed by finding the sum or difference of random variables. If a random variable x has mean and a random variable y has mean , then the means of the sum and difference of the variables are given by . If random variables are independent, then the variance and standard deviation of the sum or difference of the random variables can be found. So, if a random variable x has variance and a random variable y has variance , then the variances of the sum and difference of the variables are given by In Exercises 43 and 44, the distribution of SAT mathematics scores for college-bound male seniors in 2020 has a mean of 531 and a standard deviation of 121. The distribution of SAT mathematics scores for college-bound female seniors in 2020 has a mean of 516 and a standard deviation of 112. One male and one female are randomly selected. Assume their scores are independent. (Adapted from College Board)


Find the mean and standard deviation of the sum of their scores.

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Textbook Question

Using a Distribution to Find Probabilities In Exercises 11–26, find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine whether the events are unusual. If convenient, use a table or technology to find the probabilities.


Pass Completions NFL player Aaron Rodgers completes a pass 65.1% of the time. Find the probability that (a) the first pass he completes is the second pass, (b) the first pass he completes is the first or second pass, and (c) he does not complete his first two passes. (Source: National Football League)

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Textbook Question

Constructing and Graphing Binomial Distributions In Exercises 27–30, (a) construct a binomial distribution, (b) graph the binomial distribution using a histogram and describe its shape, and (c) identify any values of the random variable x that you would consider unusual. Explain your reasoning.


Workplace Cleanliness Fifty-seven percent of employees judge their peers by the cleanliness of their workspaces. You randomly select 10 employees and ask them whether they judge their peers by the cleanliness of their workspaces. The random variable represents the number who judge their peers by the cleanliness of their workspaces. (Source: Adecco)

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