Math Problem Statement

Complete the probability distribution and calculate the expected value, variance, and standard deviation.

Solution

To complete the provided probability distribution and solve the given questions, we will follow these steps:

1. Complete the Probability Distribution:

First, we need to find the missing values in the table, specifically for (Xμ)2P(X)(X - \mu)^2 P(X). Given values are:

  • μ=E(X)=3.8200\mu = E(X) = 3.8200

The formula for each term in (Xμ)2P(X)(X - \mu)^2 P(X) is: (Xμ)2P(X)(X - \mu)^2 P(X) Where XX is the value from the first column and P(X)P(X) is the probability from the second column.

Let's compute these values for each row:

For X=2X = 2:

(Xμ)2=(23.82)2=(1.82)2=3.3124(X - \mu)^2 = (2 - 3.82)^2 = (-1.82)^2 = 3.3124 (Xμ)2P(X)=3.3124×0.25=0.8281(X - \mu)^2 P(X) = 3.3124 \times 0.25 = 0.8281

For X=3X = 3:

(Xμ)2=(33.82)2=(0.82)2=0.6724(X - \mu)^2 = (3 - 3.82)^2 = (-0.82)^2 = 0.6724 (Xμ)2P(X)=0.6724×0.14=0.0941(X - \mu)^2 P(X) = 0.6724 \times 0.14 = 0.0941

For X=4X = 4:

(Xμ)2=(43.82)2=0.0324(X - \mu)^2 = (4 - 3.82)^2 = 0.0324 (Xμ)2P(X)=0.0324×0.2=0.0065(X - \mu)^2 P(X) = 0.0324 \times 0.2 = 0.0065

For X=5X = 5:

(Xμ)2=(53.82)2=1.3924(X - \mu)^2 = (5 - 3.82)^2 = 1.3924 (Xμ)2P(X)=1.3924×0.38=0.5291(X - \mu)^2 P(X) = 1.3924 \times 0.38 = 0.5291

For X=6X = 6:

(Xμ)2=(63.82)2=4.7524(X - \mu)^2 = (6 - 3.82)^2 = 4.7524 (Xμ)2P(X)=4.7524×0.01=0.0475(X - \mu)^2 P(X) = 4.7524 \times 0.01 = 0.0475

For X=7X = 7:

(Xμ)2=(73.82)2=10.1824(X - \mu)^2 = (7 - 3.82)^2 = 10.1824 (Xμ)2P(X)=10.1824×0.02=0.2036(X - \mu)^2 P(X) = 10.1824 \times 0.02 = 0.2036

Adding these values will give us (Xμ)2P(X)\sum (X - \mu)^2 P(X).

2. Find the Variance:

The variance is given by: σ2=(Xμ)2P(X)\sigma^2 = \sum (X - \mu)^2 P(X) Adding up the calculated values: 0.8281+0.0941+0.0065+0.5291+0.0475+0.2036=1.70890.8281 + 0.0941 + 0.0065 + 0.5291 + 0.0475 + 0.2036 = 1.7089

3. Find the Standard Deviation:

The standard deviation is the square root of the variance: σ=1.70891.3073\sigma = \sqrt{1.7089} \approx 1.3073

Summary:

  • (Xμ)2P(X)=1.7089\sum (X - \mu)^2 P(X) = 1.7089
  • Variance (σ2\sigma^2) = 1.7089
  • Standard Deviation (σ\sigma) ≈ 1.3073

Would you like further details or have additional questions?

Related Questions:

  1. How do you find the mean of a probability distribution?
  2. What is the significance of variance in probability theory?
  3. How would the calculations change if the probability distribution was not valid (i.e., didn't sum to 1)?
  4. Can variance be a negative value? Why or why not?
  5. How can the variance and standard deviation help understand the spread of data?

Tip:

Always check that the sum of probabilities in a probability distribution equals 1. This ensures that the distribution is valid.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Variance
Standard Deviation

Formulas

Expected value: E(X) = Σ[X * P(X)]
Variance: σ^2 = Σ[(X - μ)^2 * P(X)]
Standard deviation: σ = sqrt(σ^2)

Theorems

Law of Total Probability

Suitable Grade Level

Grades 11-12 / College Level