Math Problem Statement

What is the standard deviation of the grade distribution provided in the table?

Solution

To find the standard deviation of the distribution, let's go through the following steps using the table provided.

Given Data

  • Grades XX are assigned values as follows: A=4A = 4, B=3B = 3, C=2C = 2, D=1D = 1, F=0F = 0.
  • Probabilities for each grade are:
    • P(X=4)=0.4P(X=4) = 0.4
    • P(X=3)=0.32P(X=3) = 0.32
    • P(X=2)=0.17P(X=2) = 0.17
    • P(X=1)=0.08P(X=1) = 0.08
    • P(X=0)=0.03P(X=0) = 0.03

Steps to Calculate the Standard Deviation

  1. Calculate the Mean (μ\mu) of the Distribution:

    μ=(X×P(X))=(4×0.4)+(3×0.32)+(2×0.17)+(1×0.08)+(0×0.03)\mu = \sum (X \times P(X)) = (4 \times 0.4) + (3 \times 0.32) + (2 \times 0.17) + (1 \times 0.08) + (0 \times 0.03)

    μ=1.6+0.96+0.34+0.08+0=2.98\mu = 1.6 + 0.96 + 0.34 + 0.08 + 0 = 2.98

  2. Calculate the Variance (σ2\sigma^2) of the Distribution:

    The formula for variance is:

    σ2=((Xμ)2×P(X))\sigma^2 = \sum \left((X - \mu)^2 \times P(X)\right)

    Breaking it down:

    • For X=4X = 4: (42.98)2×0.4=(1.02)2×0.4=1.0404×0.4=0.41616(4 - 2.98)^2 \times 0.4 = (1.02)^2 \times 0.4 = 1.0404 \times 0.4 = 0.41616
    • For X=3X = 3: (32.98)2×0.32=(0.02)2×0.32=0.0004×0.32=0.000128(3 - 2.98)^2 \times 0.32 = (0.02)^2 \times 0.32 = 0.0004 \times 0.32 = 0.000128
    • For X=2X = 2: (22.98)2×0.17=(0.98)2×0.17=0.9604×0.17=0.163268(2 - 2.98)^2 \times 0.17 = (-0.98)^2 \times 0.17 = 0.9604 \times 0.17 = 0.163268
    • For X=1X = 1: (12.98)2×0.08=(1.98)2×0.08=3.9204×0.08=0.313632(1 - 2.98)^2 \times 0.08 = (-1.98)^2 \times 0.08 = 3.9204 \times 0.08 = 0.313632
    • For X=0X = 0: (02.98)2×0.03=(2.98)2×0.03=8.8804×0.03=0.266412(0 - 2.98)^2 \times 0.03 = (-2.98)^2 \times 0.03 = 8.8804 \times 0.03 = 0.266412

    Summing these values:

    σ2=0.41616+0.000128+0.163268+0.313632+0.266412=1.1596\sigma^2 = 0.41616 + 0.000128 + 0.163268 + 0.313632 + 0.266412 = 1.1596

  3. Calculate the Standard Deviation (σ\sigma) by taking the square root of the variance:

    σ=1.15961.08\sigma = \sqrt{1.1596} \approx 1.08

Answer

The standard deviation of the distribution is approximately 1.08, matching option (B).


Would you like further clarification on any of these steps?

Here are some related questions to deepen your understanding:

  1. How would the standard deviation change if one of the probabilities was different?
  2. What would happen to the mean if the grading scale changed, such as using A = 5 instead of A = 4?
  3. Can you calculate the variance without using the standard deviation formula?
  4. What is the purpose of standard deviation in interpreting data?
  5. How does this discrete probability distribution differ from a continuous one?

Tip: Remember, the mean provides a central value, while the standard deviation shows how spread out the values are around the mean.

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Math Problem Analysis

Mathematical Concepts

Probability distribution
Expected value (mean)
Variance
Standard deviation

Formulas

Mean (μ) of a discrete distribution: μ = Σ(X * P(X))
Variance (σ²) of a discrete distribution: σ² = Σ((X - μ)² * P(X))
Standard deviation (σ): σ = √σ²

Theorems

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Suitable Grade Level

College Introductory Statistics