Mean and Variance of Normal Distribution with Given Density Function

Mean of Normal Distribution
The mean of a normal distribution with density function f(x) = k.e=(-(x-μ)^2)/(2σ^2) is given by the parameter μ. Therefore, we need to manipulate the given density function to obtain it in this form.

Given that f(x) = k.e = (x^2-6x+9)/8, we can simplify it as follows:

f(x) = (1/8)(x^2-6x+9)
= (1/8)(x-3)^2

Comparing with the general form of the density function, we see that μ = 3. Therefore, the mean of the distribution is μ = 3.

Variance of Normal Distribution
The variance of a normal distribution with density function f(x) = k.e=(-(x-μ)^2)/(2σ^2) is given by the parameter σ^2. Therefore, we need to manipulate the given density function to obtain it in this form.

Given that f(x) = k.e = (x^2-6x+9)/8, we can simplify it as follows:

f(x) = (1/8)(x-3)^2

Expanding the square term, we get:

f(x) = (1/8)(x^2 - 6x + 9)
= (1/8)x^2 - (3/4)x + (9/8)

Comparing with the general form of the density function, we see that σ^2 = 4/8 = 1/2. Therefore, the variance of the distribution is σ^2 = 1/2.

Conclusion
Thus, the mean and variance of the normal distribution with density function f(x) = (x^2-6x+9)/8 are μ = 3 and σ^2 = 1/2, respectively.

For a random variable x, with mean “μ” and standard deviation “σ”, the normal distribution formula is given by:

f(x) = (1/√(2πσ2)) (e[-(x-μ)^2]/2σ^2)

x²-6x+9 = (x-3)²

μ = 3

σ = √8÷2 = √4 = 2

Mean = 3 Variance = 2