Continuous Probability Distributions: A Note

Notation: $X \sim N(\mu, \sigma^2)$

Probability Density Function (PDF): $$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x - \mu}{\sigma})^2}, \quad -\infty < x < \infty$$ The PDF defines the shape of the curve. Unlike discrete distributions, you don’t use the PDF to directly calculate the probability of a specific value (since the probability of any single point in a continuous distribution is 0). Instead, you find probabilities for ranges of values by calculating the area under the curve (using integration or, more commonly, tables or software).

Mean (Expected Value): $$E(X) = \mu$$

Variance: $$Var(X) = \sigma^2$$

Example: Heights of adult women are normally distributed with a mean of 65 inches and a standard deviation of 3.5 inches. You can’t use the PDF to find P(height = 68 inches), but you can find P(67 < height < 69 inches) using standardization and a Z-table (or software).

Notation: $Z \sim N(0, 1)$

Probability Density Function (PDF): $$f(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z^2}, \quad -\infty < z < \infty$$ This is a simplified version of the normal PDF, with μ = 0 and σ = 1.

Mean (Expected Value): $$E(Z) = 0$$

Variance: $$Var(Z) = 1$$

Standardization: To convert any normal random variable X (with mean μ and standard deviation σ) to a standard normal variable Z, use the formula: $$Z = \frac{X - \mu}{\sigma}$$

Example: Using the latte example (mean = 2 minutes, standard deviation = 0.5 minutes), a latte that takes 2.75 minutes has a Z-score of: $Z = \frac{2.75 - 2}{0.5} = 1.5$. You can then use a Z-table to find the probability of a Z-score being less than 1.5, which corresponds to the probability of a latte taking less than 2.75 minutes to make.

Notation: $X \sim Bin(n, p)$ and $X$ can be approximated as $Y \sim N(np, np(1-p))$

Condition: $np > 5$ and $n(1-p) > 5$

Continuity Correction $P(X=x) \approx P(x-0.5 < Y < x+0.5)$

Z-score Calculation (with Continuity Correction):

To find probabilities, we use the Z-score formula, incorporating the continuity correction. For example, to find $P(X \le x)$:

$$Z = \frac{(x + 0.5) - np}{\sqrt{np(1-p)}}$$

You would then find $P(Z \le z)$ using a standard normal table.

Example: 300 lattes (n=300), probability of soy milk request = 0.2 (p=0.2). Find P(more than 70 soy lattes). This is P(X > 70).

1. Check conditions: np = 300 * 0.2 = 60 > 5. n(1-p) = 300 * 0.8 = 240 > 5. The normal approximation is appropriate.
2. Apply continuity correction: P(X > 70) becomes P(X > 70.5).
3. Calculate Z-score: $Z = \frac{70.5 - (300 \times 0.2)}{\sqrt{300 \times 0.2 \times 0.8}} = \frac{10.5}{\sqrt{48}} \approx 1.52$
4. Find probability: Use a Z-table (or software) to find P(Z > 1.52) = 1 - P(Z ≤ 1.52) ≈ 1 - 0.9357 = 0.0643. So, the approximate probability of more than 70 soy lattes is 6.43%.

1. Check condition: λ = 60 > 5. The normal approximation is appropriate.
2. Apply continuity correction: $P(X < 50)$ becomes $P(X < 49.5)$
3. Calculate the Z-score: $Z = \frac{(49.5 - 60)}{\sqrt{60}} \approx -1.36$
4. Find probability: Use a Z-table to find P(Z < -1.36) ≈ 0.0869.

Notation: $X \sim Exp(\lambda)$

Probability Density Function (PDF): $$f(x) = \lambda e^{-\lambda x}, \quad x \ge 0, \lambda > 0$$

Cumulative Distribution Function (CDF): $$F(x) = P(X \le x) = 1 - e^{-\lambda x}$$ This is often more useful than the PDF for direct probability calculations.

Mean (Expected Value): $$E(X) = \frac{1}{\lambda}$$

Variance: $$Var(X) = \frac{1}{\lambda^2}$$

Example: Time between customer arrivals is exponential with λ = 1 arrival per minute. What’s the probability that more than 2 minutes pass between arrivals? This is P(X > 2).

• We use the complement of the CDF: $P(X > 2) = 1 - P(X \le 2) = 1 - (1 - e^{-1 \times 2}) = e^{-2} \approx 0.135$