Continuous Probability Distributions Notes

Continuous Probability Distributions

  • A continuous random variable can assume any value within an interval.

  • It is not possible to talk about the probability of the random variable assuming a particular value. Instead, we discuss the probability of the random variable assuming a value within a given interval.

Continuous Distributions

  • The probability of a random variable assuming a value within a given interval from x<em>1x<em>1 to x</em>2x</em>2 is defined as the area under the graph of the probability density function between x<em>1x<em>1 and x</em>2x</em>2.

  • The area under the graph of f(x)f(x) and probability are identical.

The Uniform Distribution

  • Two numbers, min (a) and max (b), specify the distribution.

  • All outcomes between a and b are equally likely.

  • The probability density function is given by:
    f(x)=1b−af(x) = \frac{1}{b - a}

  • The area under the curve is calculated as:
    area=width×height=(b−a)×1b−a=1area = width \times height = (b - a) \times \frac{1}{b - a} = 1

Expected Value and Variance of Uniform Distribution

  • Expected Value of x:
    E(x)=a+b2E(x) = \frac{a + b}{2}

  • Variance of x:
    Var(x)=(b−a)212Var(x) = \frac{(b - a)^2}{12}

Example: Flight Time

  • Consider the flight time from Chicago to New York, which can be any value in the interval from 120 minutes to 140 minutes.

Uniform Probability Density Function

  • xx = Flight time of an airplane traveling from Chicago to New York

  • Expected Value of x: 130

  • Variance of x: 33.33

  • Standard Deviation: 33.33=5.77\sqrt{33.33} = 5.77

Probability Calculation

  • Probability of a flight time between 120 and 130 minutes:
    P(120 < x < 130) = \frac{1}{20}(10) = 0.5

Mini Workshop 1

  • The amount of gas sold daily at a service station is uniformly distributed between 2000 and 5000 gallons.

  • What is the probability that tomorrow between 2500 and 3000 gallons are sold?

  • Algebraically: what is P(2500≤X≤3000)P(2500 \le X \le 3000) ?

Mini Workshop 2: Uniform Distribution Example

  • A friend is always late, with the lateness (X) being between 0 and 30 minutes, with all 1-minute intervals equally likely. This follows a uniform probability distribution.

  • What is the probability your friend is more than 12 minutes late?

The Normal Distribution

  • Symmetric and bell-shaped.

  • Mean and median are both in the middle.

  • Probabilities are intervals under the curve, with the total area being 100%.

  • Follows the Empirical Rule.

  • The normal probability distribution is the most important distribution for describing a continuous random variable.

Characteristics of Normal Distribution

  • The normal probability distribution is defined by its mean (μ\mu) and its standard deviation (σ\sigma).

  • The highest point on the normal curve is at the mean, which is also the median and mode.

Characteristics: Mean

  • The mean can be any numerical value: negative, zero, or positive.

Characteristics: Standard Deviation

  • The standard deviation determines the width of the curve; larger values result in wider, flatter curves.

Characteristics: Probabilities and Area

  • Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (0.5 to the left of the mean and 0.5 to the right).

Empirical Rule

  • For normally distributed data:

    • 68.3% of the data falls within ±1 standard deviation from the mean.

    • 95.5% of the data falls within ±2 standard deviations from the mean.

    • 99.7% of the data falls within ±3 standard deviations from the mean.

Example: Grear Tire Company Problem

  • Grear Tire Company developed a new steel-belted radial tire. The mean tire mileage is estimated to be 36,500 miles with a standard deviation of 5000 miles.

  • The manager wants to know the probability that the tire mileage (x) will exceed 40,000 miles.

Steps to Solve the Grear Tire Company Problem

  1. Convert x to a standard normal distribution (z).

  2. Find the area under the standard normal curve to the left of z = 0.7.

Solving for the Probability

  • Compute the area under the standard normal curve to the right of z = 0.7.

Determining Guaranteed Mileage

  • What should be the guaranteed mileage if Grear wants no more than 10% of tires to be eligible for the discount guarantee?

  • (Hint: Use the standard normal table in an inverse fashion to find the corresponding z value.)

Solving for Guaranteed Mileage

  • Find the z value that cuts off an area of 0.1 in the left tail of the standard normal distribution.

Step 2: Convert Z to the corresponding Value of X

  • Thus, a guarantee of 30,100 miles will meet the requirement that approximately 10% of the tires will be eligible for the guarantee.

Mini Workshop 4

  • A company has determined that the distribution of customer demand is normal with a mean of 750 and a standard deviation of 100 units per month.

    • Determine the probability that next month’s demand will be less than 900 units. Draw a picture to visualize. p(x< 900) = ? p(Z< ? ) = ?

    • Determine the probability that next month’s demand will be more than 700 units. Draw a picture to visualize.

    • Determine the probability that next month’s demand will be between 700 and 900 units. Draw a picture to visualize.

    • What level of demand would we need to prepare for to ensure that we would have the capacity to meet it 90% of the months? Only 10% of the time would demand be more than this amount (and we would not have enough capacity). Draw a picture to visualize.

Using Excel to Compute Standard Normal Probabilities

  • Excel has two functions for computing probabilities and z values for a standard normal probability distribution.

  • “S” in the function names reminds us that these functions relate to the standard normal probability distribution.

Excel Formulas for Probabilities

  • P(z≤1)P(z \le 1) : =NORM.S.DIST(1,TRUE)

  • P(−0.5≤z≤1.25)P(-0.5 \le z \le 1.25) : =NORM.S.DIST(1.25, TRUE) - NORM.S.DIST(-0.5, TRUE)

  • P(−1≤z≤1)P(-1 \le z \le 1) : =NORM.S.DIST(1, TRUE) - NORM.S.DIST(-1, TRUE)

  • P(z≥1.58)P(z \ge 1.58) : =1-NORM.S.DIST(1.58,TRUE)

Excel Values for Probabilities

  • P(z≤1)P(z \le 1) : 0.8413

  • P(−0.5≤z≤1.25)P(-0.5 \le z \le 1.25) : 0.5858

  • P(−1≤z≤1)P(-1 \le z \le 1) : 0.6827

  • P(z≥1.58)P(z \ge 1.58) : 0.0571

Finding z-values given Probabilities - Excel Formulas

  • z value with 0.10 in the upper tail: =NORM.S.INV(0.9)

  • z value with 0.025 in the upper tail: =NORM.S.INV(0.975)

  • z value with 0.025 in the lower tail: =NORM.S.INV(0.025)

Finding z-values given Probabilities - Excel Values

  • z value with 0.10 in the upper tail: 1.28

  • z value with 0.025 in the upper tail: 1.96

  • z value with 0.025 in the lower tail: -1.96