A standard normal distribution is a special type of normal distribution that has:
- Mean (μ) = 0
- Standard deviation (σ) = 1
Example 1: Find the probability density function of the standard normal distribution of the following data. x = 2, μ = 3 and σ = 4.
Given,
- Variable (x) = 2
- Mean = 3
- Standard Deviation = 4
Using formula of probability density of standard normal distribution
F(Z) = -∞∫Z
e^{(-Z^2/2)} / √(2π) and Z = [(x-μ)/σ]Here Z = -0.25
Simplifying using Standard Normal Distribution Table, we get
F(-0.25) = 1 - F(0.25) = 1 - 0.5987 = 0.4013
Example 2: If the value of the random variable is 1, the mean is 0 and the standard deviation is 1, then find the probability density function of the Gaussian distribution.
Given,
- Variable (x) = 1
- Mean = 0
- Standard Deviation = 1
Using formula of probability density of standard normal distribution
F(Z) = -∞∫Z
e^{(-Z^2/2)} / √(2π) and Z = [(x-μ)/σ]Here Z = 1
Simplifying using Standard Normal Distribution Table, we get
F(1) = 0.8413
Example 3: Find the z-score for a value of 40 if the mean is 50 and the standard deviation is 5.
z=\frac{40-50}{5}=\frac{-10}{5}=-2 The z-score is −2.
Practice Problem
1. The heights of students in a class are normally distributed with a mean of 170 cm and a standard deviation of 5 cm. What is the z-score of a student whose height is 180 cm?
2. A z-score of −1.5 is obtained for a test score, Is the score above or below the mean?
3. Using a standard normal table, find the probability that a standard normal random variable is less than: z = 1.25
4. Find the probability that a standard normal variable lies between:z = −1 and z = 1
5. A normal distribution has a mean of 50 and a standard deviation of 8. According to the 68–95–99.7 rule: What percentage of observations lie between 42 and 58?
