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

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.

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.

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) = ?

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

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!

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.