Econ 120A - Continuous Random Variables and Standard Normal Distribution

Standard Normal Distribution

  • Random variable Z follows a standard normal distribution with a mean of 0 and a standard deviation of 1.
    • µ=0µ = 0
    • σ=1σ = 1
    • σ2=1σ^2 = 1

Types of Probability Exercises

  • Three types of probabilities need to be computed:
    1. Probability that the standard normal random variable z will be less than or equal to a given value.
      • Directly given by the table!
    2. Probability that z will be between two values.
    3. Probability that z will be greater than or equal to a given value.
  • Finding Z value.

Exercise 1: P(Z<a)P(Z < a)

  • The table provides cumulative probabilities for the standard normal distribution.

  • Entries in the table give the area under the curve to the left of the z value.

  • Example: Find P(Z≤1.23)P(Z ≤ 1.23)

    • Find 1.2 in the row and 0.03 in the column.
    • The entry equals 0.8907, so P(Z≤1.23)=0.8907P(Z ≤ 1.23) = 0.8907
    • The area under the curve to the left of 1.23 is 89.07% of the total area under the curve.
    • The probability of obtaining a value of Z≤1.23Z ≤ 1.23 is 89.07%.

  • The table gives the probability of Z being less than or equal to a specific value, a.

  • The table has integers, and the first decimal (tenth place) is listed in the rows, and the second decimal (hundredths place) value is listed in columns.

  • Entries in the body of the table are the probability of Z being less than the value located in the crossing of the corresponding row and column.

  • P(Z≤a)P(Z ≤ a) = probability of Z being less than or equal to a.

  • This is the area under the curve (the density function of random variable Z that follows a N(0,1)) to the left of z.

Exercise 2: P(a<Z<b)P (a < Z < b)

  • P(a<Z<b)=P(Z<b)–P(Z<a)P(a< Z < b) = P(Z < b) – P (Z < a)
  • == (table value for b) – (table value for a)
  • Example:
    • P(−0.5<Z<1.25)=P(Z<1.25)–P(Z<−0.5)P(-0.5< Z < 1.25) = P(Z < 1.25) – P (Z < -0.5)
    • =0.8944–0.3085=0.5859= 0.8944 – 0.3085 = 0.5859
  • P(−1<Z<1)=?P(-1<Z< 1) = ?

Exercise 3: P(Z>a)P( Z > a)

  • P(Z>a)=1−P(Z<a)P(Z > a) = 1 - P(Z < a)
  • =1= 1 – (table value for a)
  • OR
  • =P(Z<−a)= P(Z < -a)
  • == (table value for –a)
  • Example: Find P(Z>0.75)P(Z > 0.75)
    • P(Z>0.75)=1–0.7734=0.2266P(Z > 0.75) = 1 – 0.7734 = 0.2266
    • P(Z>0.75)=P(Z<−0.75)=0.2266P(Z > 0.75) = P(Z < -0.75) = 0.2266

Exercise 4: Find z value

  • Find a z value such that the probability of obtaining a larger z value is 0.1
  • z=1.28z = 1.28

Transforming Normal to Standard Normal

  • Any normally distributed random variable X can be transformed into a standard normal.
    1. Take deviations from the mean: X−µX - µ
    2. Compute the numbers of standard deviations away from the mean: (X−µ)/σ(X - µ) / σ
  • If X follows a normal distribution with mean µ and variance σ2σ^2, i.e., X∼N(µ,σ2)X ∼ N(µ, σ^2)
  • Then the transformed variable Z=atsX−µ/σZ = ats X−µ / σ will be distributed normally with mean 0 and variance 1, i.e., Z∼N(0,1)Z ∼ N(0,1)
  • Now we can use the same old standard normal distribution tables for probabilities!

Homework Questions

  • Question 1: Exam grades are normally distributed with a mean of 73 and a standard deviation of 11. The top 12.3% receive grades of A. What is the minimum score needed to receive a grade of A?
  • Question 2: A vaccine must be kept cold. If the temperature rises above 40 degrees Fahrenheit, the vials can no longer be used. The freezer fluctuates following a normal distribution with mean µ and standard deviation 2. What is the highest µ that the pharmacist can set to guarantee that the vials are damaged with probability no higher than 0.02?
  • Question 3: Monthly fuel consumption is normally distributed with a mean of 20,000 liters and a standard deviation of 2,500 liters. Which fuel consumption level is surpassed in 95% of months? Round your answer to the nearest integer.