Stats/Module3-1

Uniform & Normal Distributions

Learning Goals

By the end of this lesson, you should be able to:

1. Explain what a density curve is and its three key properties.
2. Compute probabilities and quantiles for Uniform and Normal distributions.
3. Standardize to z-scores and move between areas and values.
4. Apply the Empirical Rule (68–95–99.7) appropriately.
5. Understand how shifting and scaling data change measures of position and spread.

Density Curves

A density curve models the distribution of a continuous random variable. A graphical representation of the probability density function (PDF) for continuous data.

Histogram and Density Curve

• Density curves can be represented alongside histograms to visualize the distribution of data.
• Example data representation demonstrates the density of values across a defined interval.

Properties of Density Curves

A density curve has the following key properties:

1. Nonnegativity: For all values of $x$, $f(x)
   eq 0$.
2. Total Area = 1: The integral of the density function across all possible values equals 1:
   $ext{Total Area} = ext{Area under the curve} = \int_{- ext{∞}}^{ ext{∞}} f(x) \, dx = 1$
3. Probability = Area: For any interval $A$, the probability that the random variable $X$ falls within this interval is given by: $P(X ext{ in } A) = \\frac{(X ext{ in } A) ext{ area under the curve}}{ ext{ area under the curve }} = \int_A f(x) \, dx$
   • Additionally, for any constant $c$, $P(X = c) = 0$.

Uniform Distribution

A random variable $X$ is said to be Uniform on $[a, b]$, denoted as $X extasciitilde ext{Unif}(a, b)$, if it is equally likely to fall anywhere in that interval.

Probabilities of Uniform Distribution

Let $X$ be a continuous uniform random variable such that $X extasciitilde ext{Unif}(a, b)$.

• The probability that $X$ lies between $c$ and $d$ (with $a
  eq c, d
  eq b$) is given by:
  $P(c ext{ ≤ } X ext{ ≤ } d) = \frac{d - c}{b - a}$

Example: Waiting Time in a Stadium

• At a stadium's ticket booth, the waiting time $X$ (in minutes) is uniformly distributed between $0.7$ and $5.2$. To find the probability that a customer waits less than $2.4$ minutes, compute:
  $P(X < 2.4) = \frac{2.4 - 0.7}{5.2 - 0.7} = \frac{1.7}{4.5} ext{ which is approximately } 0.3778, ext{ or } 37.78\%.$

Further Examples

1. In an amusement park, fast-pass wait time $X extasciitilde ext{Unif}(3, 5)$:
   • Probability that a guest waits between $1.2$ and $2.7$ minutes:
     $P(1.2 < X < 2.7) = 0$ (since $1.2$ and $2.7$ are outside the support).
   • Probability that a guest waits less than $1.2$ minutes:
     $P(X < 1.2) = 0$ (outside support).
   • Probability that a guest waits more than $2.7$ minutes:
     $P(X > 2.7) = 1$ (all wait times are greater than $2.7$).

Mean and Variance of the Uniform Distribution

• The mean ($ ext{μ}$) and variance ($σ²$) of a uniform distribution are calculated as follows:
  • Mean: $μ = \frac{a + b}{2}$
  • Variance: $σ² = \frac{(b - a)²}{12}$

Normal Distribution

The Normal distribution is characterized by its symmetric shape around the mean. In a normal distribution:

1. The mean, median, and mode are all equal and denote the center of the distribution.
2. The spread of the distribution is controlled by the standard deviation ($σ$). A larger $σ$ means a flatter and more spread-out distribution.

Standard Normal Distribution

In the standard normal distribution, parameters are set as:

• Mean ($μ$) = 0
• Standard Deviation ($σ$) = 1

Comparison of Scores: Ann vs. Tom

• The mean and standard deviations for total scores on the SAT and ACT are as follows:
  • SAT: Mean ($μ$) = 1500, Standard deviation ($σ$) = 300
  • ACT: Mean ($μ$) = 21, Standard deviation ($σ$) = 5

Standardization of Scores

To compare Ann and Tom's scores, we standardize both distributions:

• Standardization formula: If $X extasciitilde N(μ, σ)$, then
  $Z = \frac{X - μ}{σ} ext{ where } Z extasciitilde N(0, 1).$

• Ann scored 1800 on the SAT:
  $Z = \frac{1800 - 1500}{300} = 1.0$

• Tom scored 24 on the ACT:
  $Z = \frac{24 - 21}{5} = 0.6$

Thus, interpretation: Ann is 1 standard deviation above the SAT mean, while Tom is only 0.6 standard deviations above the ACT mean, indicating that Ann performed better relative to her group.

Calculating Probabilities under the Normal Distribution

To find probabilities under the normal distribution, we compute areas under the curve (PDF). Reports give:

• For P(65 < x < 80): Area calculated from Normal curve.
• For P(X < 75): Left-tail probability area.

Example: Left-Tail Probability

• Given that exam scores are normally distributed:
  • Mean = 70, Standard Deviation = 10.
  • Calculating the probability a student scores below 85:
    $Z = \frac{85 - 70}{10} = 1.5$. Therefore:
    $P(X < 85) = P(Z < 1.5) ≈ 0.933.$

Example: Right-Tail Probability

• Weights of apples are typically distributed with:
  • Mean = 150, Standard Deviation = 20.
• Find probability that a randomly selected apple weighs more than 180 grams:
  $Z = \frac{180 - 150}{20} = 1.5$ and
  $P(X > 180) = P(Z > 1.5) = 1 - P(Z < 1.5) ≈ 0.0668, ext{ or } 6.7\%.$

Example: Probability of an Interval

• Weekly weights of apples: Mean = 70, Standard Deviation = 10.
• Find the probability apple weighs between 65 grams and 80 grams:
  • Standardize both bounds:
    $z<em>1 = \frac{65 - 70}{10} = -0.5, z</em>2 = \frac{80 - 70}{10} = 1.0.$
  • $P(65 < X < 80) = P(-0.5 < Z < 1.0) ≈ P(Z < 1) - P(Z < -0.5) ≈ 0.8413 - 0.3085 = 0.5328 ext{ or } 53\%.$

Finding the 90th Percentile

• Exam scores are distributed with:
  • Average = 70 points, Standard deviation = 10 points. To find score that marks the 90th percentile:
  1. Use the standard normal table to find $z_{0.90} ≈ 1.2816$.
  2. Convert back:
     $x = μ + zσ = 70 + (1.2816)(10) ≈ 82.82 (≈ 83).$

Empirical Rule (68–95–99.7)

Approximately: