Math Problem Statement

Fi(Xi-μ)2

Solution

It looks like you’re asking for the calculation of fi(Xiμ)2\sum f_i(X_i - \mu)^2, which is part of the variance formula. Let's break this down step by step using the data provided.

Step 1: Recap of Midpoints and Mean

We've already calculated the midpoints (XiX_i) for each score range and the mean (μ\mu) of the distribution:

  • μ70.12\mu \approx 70.12

Step 2: Calculate (Xiμ)2(X_i - \mu)^2 for Each Midpoint

We need to subtract the mean from each midpoint, square the result, and then multiply by the frequency (fif_i):

Score RangeFrequency (fif_i)Midpoint (XiX_i)(Xiμ)2(X_i - \mu)^2fi(Xiμ)2f_i(X_i - \mu)^2
30-39834.5(34.570.12)21275.57(34.5 - 70.12)^2 \approx 1275.578×1275.57=10204.568 \times 1275.57 = 10204.56
40-491044.5(44.570.12)2641.64(44.5 - 70.12)^2 \approx 641.6410×641.64=6416.410 \times 641.64 = 6416.4
50-591254.5(54.570.12)2243.40(54.5 - 70.12)^2 \approx 243.4012×243.4=2920.812 \times 243.4 = 2920.8
60-691564.5(64.570.12)231.65(64.5 - 70.12)^2 \approx 31.6515×31.65=474.7515 \times 31.65 = 474.75
70-792074.5(74.570.12)219.08(74.5 - 70.12)^2 \approx 19.0820×19.08=381.620 \times 19.08 = 381.6
80-892584.5(84.570.12)2205.90(84.5 - 70.12)^2 \approx 205.9025×205.9=5147.525 \times 205.9 = 5147.5
90-991594.5(94.570.12)2597.57(94.5 - 70.12)^2 \approx 597.5715×597.57=8963.5515 \times 597.57 = 8963.55

Step 3: Sum the Values

Now sum up all the values in the fi(Xiμ)2f_i(X_i - \mu)^2 column:

fi(Xiμ)2=10204.56+6416.4+2920.8+474.75+381.6+5147.5+8963.55=34509.16\sum f_i(X_i - \mu)^2 = 10204.56 + 6416.4 + 2920.8 + 474.75 + 381.6 + 5147.5 + 8963.55 = 34509.16

Step 4: Calculate the Variance (σ2\sigma^2)

Finally, divide by the total frequency (which is 105) to get the variance:

σ2=34509.16105328.66\sigma^2 = \frac{34509.16}{105} \approx 328.66

Step 5: Calculate the Standard Deviation

The standard deviation (σ\sigma) is the square root of the variance:

σ=328.6618.12\sigma = \sqrt{328.66} \approx 18.12

Step 6: Calculate the Coefficient of Variation (CV)

The coefficient of variation is:

CV=σμ×100%=18.1270.12×100%25.85%\text{CV} = \frac{\sigma}{\mu} \times 100\% = \frac{18.12}{70.12} \times 100\% \approx 25.85\%


Summary for Question 2

  • Variance: 328.66
  • Standard Deviation: 18.12
  • Coefficient of Variation: 25.85%

Do you need any further details or explanations? Here are some related questions:

  1. How do you interpret the coefficient of variation in a dataset?
  2. What are the advantages of using standard deviation over variance?
  3. How does the range compare to the standard deviation as a measure of spread?
  4. Can the coefficient of variation be used to compare variability between different datasets?
  5. What is the effect of extreme values on the range and standard deviation?

Tip: Always remember that variance is sensitive to outliers since it squares the deviations from the mean.

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Math Problem Analysis

Mathematical Concepts

Statistics
Variance
Standard Deviation
Coefficient of Variation

Formulas

Variance formula
Standard deviation formula
Coefficient of variation formula

Theorems

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Suitable Grade Level

Undergraduate