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 = 0
  • σ = 1
  • σ^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)

• 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)

  • Find 1.2 in the row and 0.03 in the column.
  • The entry equals 0.8907, so P(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 ≤ 1.23 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) = 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(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)
  • = 0.8944 – 0.3085 = 0.5859
• P(-1<Z< 1) = ?

Exercise 3: P( Z > a)

• P(Z > a) = 1 - P(Z < a)
• = 1 – (table value for a)
• OR
• = P(Z < -a)
• = (table value for –a)
• Example: Find P(Z > 0.75)
  • P(Z > 0.75) = 1 – 0.7734 = 0.2266
  • P(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 = 1.28

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 - µ
  2. Compute the numbers of standard deviations away from the mean: (X - µ) / σ
• If X follows a normal distribution with mean µ and variance σ^2, i.e., X ∼ N(µ, σ^2)
• Then the transformed variable Z = ats X−µ / σ will be distributed normally with mean 0 and variance 1, i.e., 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.