AP Statistics 5.2 The Normal Distribution, Revisited Study Notes
AP Statistics 5.2 The Normal Distribution, Revisited Study Notes- New syllabus
AP Statistics 5.2 The Normal Distribution, Revisited Study Notes -As per latest AP Statistics Syllabus.
LEARNING OBJECTIVE
- The normal distribution may be used to model variation
Key Concepts:
- Normal Distributions: Interval Probabilities
- Normal Distributions: Finding Intervals from Areas
- Approximating with a Normal Distribution
Normal Distributions: Interval Probabilities
Normal Distributions: Interval Probabilities
In statistics, many real-world variables (like heights, weights, or test scores) follow a normal distribution. We can use the normal model to calculate the probability that a value lies within a given interval.
Key Concepts
Standardization (Z-scores):
Convert a raw score \( x \) to a standard normal score using \( z = \dfrac{x – \mu}{\sigma} \).
Standard Normal Distribution:
A normal distribution with mean 0 and standard deviation 1.
Probabilities are found using z-tables or technology.
Interval Probabilities:
To find \( P(a \leq X \leq b) \), compute the corresponding z-scores and subtract:
\(P(a \leq X \leq b) = P(Z \leq z_b) – P(Z \leq z_a). \)
Key Takeaways
- Always convert raw values into z-scores before finding probabilities.
- Use technology (normalcdf on calculator, statistical software, or tables) for more accurate results.
- Interval probabilities are found by subtraction: upper probability minus lower probability.
Example
The heights of adult men in a city are normally distributed with mean \( \mu = 70 \) inches and standard deviation \( \sigma = 3 \) inches. What is the probability that a randomly selected man has a height between 68 and 74 inches?
Find \( P(68 \leq X \leq 74) \).
▶️ Answer / Explanation
Step 1: Standardize both values.
- \( z_1 = \dfrac{68 – 70}{3} = \dfrac{-2}{3} \approx -0.67 \)
- \( z_2 = \dfrac{74 – 70}{3} = \dfrac{4}{3} \approx 1.33 \)
Step 2: Find probabilities from the standard normal distribution.
- \( P(Z \leq -0.67) \approx 0.2514 \)
- \( P(Z \leq 1.33) \approx 0.9082 \)
Step 3: Subtract to find the interval probability.
\( P(68 \leq X \leq 74) = 0.9082 – 0.2514 = 0.6568 \).
Answer: The probability is about 65.7% that a randomly selected man has a height between 68 and 74 inches.
Normal Distributions: Finding Intervals from Areas
Normal Distribution: Intervals from Areas
Intervals in a normal distribution are found by connecting probabilities (areas under the curve) to boundary values of \( x \). These boundaries depend on whether we are interested in the left tail, right tail, middle region, or extreme values.
Key Cases
Left-Tail Interval:
\( P(X < x_a) = \dfrac{p}{100} \)
This means that the lowest \( p\% \) of values lie to the left of \( x_a \).
Example: The 10th percentile is the value below which 10% of observations fall.
Middle Interval:
\( P(x_a < X < x_b) = \dfrac{p}{100} \)
This means that \( p\% \) of values lie between two cutoffs \( x_a \) and \( x_b \).
Example: The middle 80% of values is bounded by the 10th and 90th percentiles.
Right-Tail Interval:
\( P(X > x_b) = \dfrac{p}{100} \)
This means that the highest \( p\% \) of values lie to the right of \( x_b \).
Example: The top 5% cutoff (like elite test scores).
Extreme Tails (Two-Sided):
To capture the most extreme \( p\% \) of values, split into two equal tails:
\( P(X < x_a) = \dfrac{1}{2}\dfrac{p}{100}, \quad P(X > x_b) = \dfrac{1}{2}\dfrac{p}{100} \)
This means half of the \( p\% \) most extreme values lie below \( x_a \), and half lie above \( x_b \).
Example: The most extreme 2% of observations means 1% in each tail.
How to Solve:
- Identify whether the problem is left-tail, right-tail, central, or extreme.
- Translate the probability into the corresponding z-score(s) using tables or technology.
- Convert back to raw scores with: \( x = \mu + z \cdot \sigma \).
Example
The scores on a standardized test are normally distributed with mean \( \mu = 500 \) and standard deviation \( \sigma = 100 \). Find the test score that marks the 90th percentile.
What value of \( x \) corresponds to the top 10% cutoff?
▶️ Answer / Explanation
Step 1: The 90th percentile means \( P(X \leq x) = 0.90 \).
Step 2: From z-tables or technology, \( z_{0.90} \approx 1.28 \).
Step 3: Convert back to raw score: \( x = \mu + z \cdot \sigma = 500 + (1.28)(100) = 628 \).
Answer: The 90th percentile score is approximately 628.
Example
The weights of bags of chips are normally distributed with mean \( \mu = 14 \) oz and standard deviation \( \sigma = 0.3 \) oz. Find the interval that contains the middle 95% of all bag weights.
What interval \( (a, b) \) contains the middle 95% of weights?
▶️ Answer / Explanation
Step 1: Middle 95% leaves 2.5% in each tail.
Step 2: Critical z-scores: \( z = \pm 1.96 \).
Step 3: Convert back to raw values:
- Lower bound: \( a = 14 + (-1.96)(0.3) = 14 – 0.588 \approx 13.41 \)
- Upper bound: \( b = 14 + (1.96)(0.3) = 14 + 0.588 \approx 14.59 \)
Answer: The middle 95% of bag weights lie between 13.41 oz and 14.59 oz.
Approximating with a Normal Distribution
Approximating with a Normal Distribution
The normal distribution can be used to approximate the probabilities of other distributions when certain conditions are met. This is because normal distributions are symmetric and bell-shaped, making them good models for many real-world situations.
Key Points:
Shape Requirement: The unknown distribution should be approximately symmetric and unimodal (bell-shaped). If the distribution is heavily skewed, the normal approximation may not be appropriate.
Sample Size: With larger samples, many distributions become approximately normal by the Central Limit Theorem (CLT). The CLT states that the sampling distribution of the sample mean approaches normality as the sample size increases.
Rule of Thumb for Binomial Approximation: For a binomial distribution, the normal approximation is reasonable if both conditions hold: \( n \cdot p \geq 10 \quad \text{and} \quad n \cdot (1-p) \geq 10 \) This ensures the distribution is not too skewed.
Continuity Correction: Since the normal distribution is continuous and binomial is discrete, a continuity correction (adding or subtracting 0.5) is applied when approximating binomial probabilities.
Limitations: Normal approximations are not appropriate when distributions are strongly skewed, have small sample sizes, or when extreme tail probabilities are of interest.
Example
A factory produces items where the probability of a defect is \( p = 0.2 \). If we take a sample of \( n = 100 \) items, can we approximate the binomial distribution with a normal distribution?
Verify the conditions and find the mean and standard deviation for the normal approximation.
▶️ Answer / Explanation
Step 1: Check conditions for normal approximation.
- \( n \cdot p = 100 \cdot 0.2 = 20 \geq 10 \)
- \( n \cdot (1-p) = 100 \cdot 0.8 = 80 \geq 10 \)
Since both conditions are satisfied, the normal approximation is appropriate.
Step 2: Find the parameters of the normal distribution.
- Mean: \( \mu = n \cdot p = 100 \cdot 0.2 = 20 \)
- Standard deviation: \( \sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{100 \cdot 0.2 \cdot 0.8} = \sqrt{16} = 4 \)
Conclusion: The binomial distribution \( B(100, 0.2) \) can be approximated by a normal distribution with \( \mu = 20 \) and \( \sigma = 4 \).