Calculating Mean and Standard Deviation in Normal Distribution

Calculating Mean and Standard Deviation

The marks obtained in a statistics examination are normally distributed. If 15% of students scored ≥ 60 marks and 40% scored < 30 marks, find the mean (μ) and standard deviation (σ).

Given

• P(X ≥ 60) = 0.15
• P(X < 30) = 0.40

We need to find the values of μ and σ.

Step 1: Convert to Z-values

(i) P(X ≥ 60) = 0.15

Since P(X ≥ 60) = 0.15, then P(X < 60) = 0.85. From the standard normal table, P(Z < 1.04) = 0.85. Therefore:

(60 − μ) / σ = 1.04
μ = 60 − 1.04σ (Equation 1)

(ii) P(X < 30) = 0.40

From the standard normal table, P(Z < -0.25) = 0.40. Therefore:

(30 − μ) / σ = -0.25
μ = 30 + 0.25σ (Equation 2)

Step 3: Solve the Equations

Equating (1) and (2):

60 − 1.04σ = 30 + 0.25σ
30 = 1.29σ
σ ≈ 23.26

Substitute σ into (2):

μ = 30 + 0.25(23.26)
μ ≈ 35.82

Final Answer

• Mean (μ): ≈ 35.8 marks
• Standard Deviation (σ): ≈ 23.3 marks

Proof: Mean = Median = Mode

For a normal distribution, the three measures of central tendency are equal: Mean = Median = Mode = μ.

Proof

The probability density function (PDF) of the normal distribution is:

f(x) = (1 / σ√2π) * e^(-(x-μ)² / 2σ²)

• μ = mean
• σ = standard deviation

1. Mean

The mean is defined as E(X) = ∫ x f(x) dx. By substituting x = μ + (x - μ), we find that E(X) = μ.

2. Median

The median M satisfies P(X ≤ M) = 0.5. Because the normal distribution is perfectly symmetric about μ, P(X ≤ μ) = 0.5, therefore Median = μ.

3. Mode

The mode is the value of x where f(x) is at its maximum. By differentiating f(x) and setting the derivative to zero, we find the maximum occurs at x = μ.

Conclusion

Since Mean = μ, Median = μ, and Mode = μ, we conclude that Mean = Median = Mode = μ.