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AP Statistics 4.8 Mean and Standard Deviation of Random Variables Study Notes

AP Statistics 4.8 Mean and Standard Deviation of Random Variables Study Notes- New syllabus

AP Statistics 4.8 Mean and Standard Deviation of Random Variables Study Notes -As per latest AP Statistics Syllabus.

LEARNING OBJECTIVE

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

Key Concepts:

  • Mean and Standard Deviation of Random Variables
  • Interpret Parameters of a Discrete Random Variable

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Mean and Standard Deviation of Random Variables

Mean and Standard Deviation of Random Variables

For a discrete random variable, the mean and standard deviation describe the center and spread of its probability distribution.

Mean (Expected Value)

The mean represents the long-run average value if the random process were repeated many times.

  • Formula: \( \mu_X = E[X] = \sum x \cdot P(X = x) \).
  • Interpretation: Balance point of the probability distribution.

Variance and Standard Deviation

Variance measures how far the values of \( X \) are spread from the mean.

Formula: \( \sigma_X^2 = \sum (x – \mu_X)^2 \cdot P(X = x) \).

Standard deviation is the square root of variance:

Formula: \( \sigma_X = \sqrt{\sigma_X^2} \).

Interpretation: Typical distance of a random variable from its mean.

Example 

A random variable \( X \) represents the number of defective bulbs in a box of 3. The probability distribution is:

  • \( P(X = 0) = 0.6 \)
  • \( P(X = 1) = 0.3 \)
  • \( P(X = 2) = 0.1 \)

Find the mean (expected value) of \( X \).

▶️ Answer / Explanation

Step 1: Use formula \( \mu_X = \sum x \cdot P(X = x) \).

Step 2: \( \mu_X = (0)(0.6) + (1)(0.3) + (2)(0.1) \).

Step 3: \( \mu_X = 0 + 0.3 + 0.2 = 0.5 \).

Answer: The mean number of defective bulbs is \( \mu_X = 0.5 \).

Example

Using the same probability distribution:

  • \( P(X = 0) = 0.6 \)
  • \( P(X = 1) = 0.3 \)
  • \( P(X = 2) = 0.1 \)

Find the standard deviation of \( X \).

▶️ Answer / Explanation

Step 1: Recall formula \( \sigma_X^2 = \sum (x – \mu_X)^2 \cdot P(X = x) \).

Step 2: We already found \( \mu_X = 0.5 \).

Step 3: Compute:

  • \( (0 – 0.5)^2 (0.6) = 0.25 \times 0.6 = 0.15 \)
  • \( (1 – 0.5)^2 (0.3) = 0.25 \times 0.3 = 0.075 \)
  • \( (2 – 0.5)^2 (0.1) = 2.25 \times 0.1 = 0.225 \)

Step 4: \( \sigma_X^2 = 0.15 + 0.075 + 0.225 = 0.45 \).

Step 5: \( \sigma_X = \sqrt{0.45} \approx 0.67 \).

Answer: The standard deviation is approximately \( \sigma_X = 0.67 \).

Interpret Parameters of a Discrete Random Variable

Interpret Parameters of a Discrete Random Variable

Mean ( \( \mu_X \) ):

Represents the expected long-term average of the outcomes. It tells us the “center” of the distribution.

Standard Deviation ( \( \sigma_X \) ):

Measures variability of outcomes. A small standard deviation means outcomes cluster close to the mean; a large one means more spread out.

Individual Probabilities:

Each \( P(X = x) \) shows the likelihood of a specific outcome, helping us interpret what is typical versus unusual.

Unusual Outcomes:

If an outcome lies more than 2 standard deviations away from the mean, it is often considered unusual.

Contextual Meaning: Parameters must be interpreted in the real-world context of the problem (e.g., “On average, 0.5 defective bulbs per box, with a typical variation of 0.67 bulbs”).

Example :

A company produces light bulbs. Let \( X \) be the number of defective bulbs in a randomly selected box of 10 bulbs. The probability distribution of \( X \) has:

  • Mean \( \mu_X = 0.5 \)
  • Standard deviation \( \sigma_X = 0.67 \)

Interpret the parameters of \( X \) in context.

▶️ Answer / Explanation
  • Mean (\( \mu_X = 0.5 \)): On average, there are 0.5 defective bulbs per box of 10. Over many boxes, the long-run average number of defective bulbs will approach 0.5.
  • Standard deviation (\( \sigma_X = 0.67 \)): The number of defective bulbs typically varies by about 0.67 from the mean. Most boxes will have between 0 and 1 defective bulbs.
  • Unusual outcomes: Values more than 2 standard deviations away (i.e., \( 0.5 \pm 2(0.67) \) → below 0 or above 1.84) are rare. Having 3 or more defective bulbs would be unusual.

Conclusion: The production process is consistent, with few defective bulbs and little variation across boxes.

Example :

Let \( X \) be the number of goals a soccer player scores in a game. The probability distribution of \( X \) is given by:

Goals (\(x\))0123
\( P(X = x) \)0.500.300.150.05

Using this distribution, \( \mu_X = 0.75 \) and \( \sigma_X \approx 0.90 \).

Interpret the mean and standard deviation in context.

▶️ Answer / Explanation
  • Mean (\( \mu_X = 0.75 \)): On average, the player scores about 0.75 goals per game — roughly 3 goals every 4 games.
  • Standard deviation (\( \sigma_X = 0.90 \)): The number of goals varies typically by about 0.9 from the average. In most games, the player’s goal count will be between 0 and 2.
  • Unusual outcomes: Scores more than 2 SDs above the mean → \( 0.75 + 2(0.90) = 2.55 \). Scoring 3 goals or more is unusual.

Conclusion: The player typically scores between 0–2 goals, with an average close to 1 goal per game. A hat trick (3 goals) is rare and exceptional.

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