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AP Statistics 4.9 Combining Random Variables Study Notes

AP Statistics 4.9 Combining Random Variables Study Notes- New syllabus

AP Statistics 4.9 Combining Random Variables Study Notes -As per latest AP Statistics Syllabus.

LEARNING OBJECTIVE

  • Probability distributions may be used to model variation in populations.

Key Concepts:

  • Combining Random Variables
  • Linear Transformations of Random Variables

AP Statistics -Concise Summary Notes- All Topics

Combining Random Variables

Combining Random Variables

When working with two or more random variables, we often need to determine the combined mean and standard deviation. These rules apply when the random variables are independent.

Mean of Combined Random Variables

The mean of the sum or difference is the sum or difference of the means.

  • Formula: \( \mu_{X+Y} = \mu_X + \mu_Y \).
  • Formula: \( \mu_{X-Y} = \mu_X – \mu_Y \).
  • Interpretation: Expected value behaves linearly.

Variance and Standard Deviation of Combined Random Variables

If \( X \) and \( Y \) are independent, variances add whether combining with a plus or minus.

  • Formula: \( \sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y \).
  • Formula: \( \sigma^2_{X-Y} = \sigma^2_X + \sigma^2_Y \).

Standard deviation: \( \sigma_{X \pm Y} = \sqrt{\sigma^2_X + \sigma^2_Y} \).

Important:

You cannot simply add or subtract standard deviations , you must work with variances first.

Example 

Suppose the time it takes Student A to finish a test is modeled by random variable \( X \) with mean \( \mu_X = 40 \) minutes. Student B’s test time is modeled by random variable \( Y \) with mean \( \mu_Y = 50 \) minutes.

Find the mean of the combined time \( X + Y \).

▶️ Answer / Explanation

Step 1: Use rule for combined means: \( \mu_{X+Y} = \mu_X + \mu_Y \).

Step 2: \( \mu_{X+Y} = 40 + 50 = 90 \).

Answer: On average, the total time for both students is 90 minutes.

Example 

Suppose Student A’s test time has standard deviation \( \sigma_X = 5 \) minutes, and Student B’s test time has standard deviation \( \sigma_Y = 8 \) minutes. Assume \( X \) and \( Y \) are independent.

Find the standard deviation of the total test time \( X + Y \).

▶️ Answer / Explanation

Step 1: Variances add: \( \sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y \).

Step 2: \( \sigma^2_{X+Y} = 5^2 + 8^2 = 25 + 64 = 89 \).

Step 3: \( \sigma_{X+Y} = \sqrt{89} \approx 9.43 \).

Answer: The standard deviation of the total time is about 9.43 minutes.

Combining Random Variables

Linear Transformations of Random Variables

When we apply a linear transformation to a random variable, such as multiplying by a constant or adding a constant, the mean and standard deviation are affected in predictable ways.

Effects on the Mean

If \( Y = a + bX \), then \( \mu_Y = a + b\mu_X \).

  • Adding a constant \( a \) shifts the mean but does not change variability.
  • Multiplying by a constant \( b \) stretches or shrinks the mean by factor \( b \).

Effects on the Variance and Standard Deviation

If \( Y = a + bX \), then \( \sigma^2_Y = b^2\sigma^2_X \).

  • Variance is affected only by multiplication, not addition.
  • Standard deviation: \( \sigma_Y = |b|\sigma_X \).
  • Adding a constant does not change spread; multiplying scales spread by \( |b| \).

Example

A random variable \( X \) has mean \( \mu_X = 50 \) and standard deviation \( \sigma_X = 10 \). Define \( Y = X + 5 \).

Find the new mean and standard deviation of \( Y \).

▶️ Answer / Explanation

Step 1: Mean: \( \mu_Y = \mu_X + 5 = 50 + 5 = 55 \).

Step 2: Standard deviation: \( \sigma_Y = \sigma_X = 10 \) (unchanged).

Answer: The mean increases to 55, but the standard deviation remains 10.

Example

A random variable \( X \) has mean \( \mu_X = 50 \) and standard deviation \( \sigma_X = 10 \). Define \( Y = 2X \).

Find the new mean and standard deviation of \( Y \).

▶️ Answer / Explanation

Step 1: Mean: \( \mu_Y = 2 \cdot \mu_X = 2 \cdot 50 = 100 \).

Step 2: Standard deviation: \( \sigma_Y = |2| \cdot 10 = 20 \).

Answer: The mean doubles to 100, and the standard deviation also doubles to 20.

Example

A random variable \( X \) has mean \( \mu_X = 30 \) and standard deviation \( \sigma_X = 4 \). Define \( Y = 3X + 10 \).

Find the new mean and standard deviation of \( Y \).

▶️ Answer / Explanation

Step 1: Mean: \( \mu_Y = 3 \cdot 30 + 10 = 100 \).

Step 2: Standard deviation: \( \sigma_Y = |3| \cdot 4 = 12 \).

Answer: The mean becomes 100, and the standard deviation becomes 12.

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