Probability and Distributions Exam Notes

A function $f(x)$ is a probability density function for a continuous random variable $X$ if the total area under its curve is equal to 1.

In a joint probability distribution, the mean, median, and mode are not necessarily equal.

Problem: In a group of 5 boys and 3 girls, two are selected at random. Let $x$ be the number of boys and $y$ be the number of girls. Find the probability of $x = 2$ .
Solution:
- Total number of ways to select 2 people from 8: ${8 <br>choose 2} = \frac{8!}{2!6!} = 28$
- Number of ways to select 2 boys from 5: ${5 <br>choose 2} = \frac{5!}{2!3!} = 10$
- Probability of selecting 2 boys: $\frac{10}{28} = \frac{5}{14}$

In the empirical rule of normal distribution, approximately 68% of values are within 1 standard deviation from the mean.

A probability distribution is a representation of different values of a random variable with their corresponding probabilities.

The standard normal distribution, also called the z-distribution, is a special normal distribution where the mean is 0 and the standard deviation is 1.

Problem: A continuous random variable $X$ can assume values between $x = 2$ and $x = 4$ and has a density function. Find P(2.4 < x < 3.5).
Note: The density function $f(x)$ is not provided in the question, so a numerical answer cannot be computed without this information. If $f(x)$ was provided, the probability P(2.4 < x < 3.5) would be found by calculating the definite integral of $f(x)$ from 2.4 to 3.5: $\int_{2.4}^{3.5} f(x) dx$

The number of students in a classroom is an example of a discrete random variable, not a continuous random variable.

Continuous probability distributions deal with intervals rather than point values.

Problem: If the value of random variable is 2, mean is 5 and the standard deviation is 4, then find the probability density function of the gaussian distribution.
Solution: The probability density function (PDF) of a Gaussian (normal) distribution is given by: $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} (\frac{x - \mu}{\sigma})^2}$ where:
- $x$ is the value of the random variable (2 in this case)
- $\mu$ is the mean (5 in this case)
- $\sigma$ is the standard deviation (4 in this case)
  Plugging in the values: $f(2) = \frac{1}{4 \sqrt{2\pi}} e^{-\frac{1}{2} (\frac{2 - 5}{4})^2}$
  $f(2) = \frac{1}{4 \sqrt{2\pi}} e^{-\frac{1}{2} (\frac{-3}{4})^2}$
  $f(2) = \frac{1}{4 \sqrt{2\pi}} e^{-\frac{1}{2} (0.5625)}$
  $f(2) = \frac{1}{4 \sqrt{2\pi}} e^{-0.28125}$
  $f(2) \approx \frac{1}{4 \times 2.5066} \times 0.754 \approx 0.0997$

Problem: A continuous random variable $X$ can assume values between $x = 2$ and $x = 4$ and has a density function given by: $f(x) = \frac{x+1}{8}$ . Find P(2 < x < 4).
Solution: To find P(2 < x < 4), integrate the density function $f(x)$ from 2 to 4:
P(2 < x < 4) = \int{2}^{4} \frac{x+1}{8} dx P(2 < x < 4) = \frac{1}{8} \int{2}^{4} (x+1) dx
P(2 < x < 4) = \frac{1}{8} [\frac{x^2}{2} + x]_{2}^{4}
P(2 < x < 4) = \frac{1}{8} [(\frac{4^2}{2} + 4) - (\frac{2^2}{2} + 2)]
P(2 < x < 4) = \frac{1}{8} [(8 + 4) - (2 + 2)]
P(2 < x < 4) = \frac{1}{8} [12 - 4]
P(2 < x < 4) = \frac{1}{8} [8]
P(2 < x < 4) = 1.0

The probability distribution is used to define the random variable's probability coming within a distinct range of values.