Math Problem Statement

Compute the (sample) variance and standard deviation of the data sample. (Round your answers to two decimal places.) 3.8, −3.5, 3.9, −0.2, −0.2 variance Incorrect: Your answer is incorrect. standard deviation

Solution

To compute the sample variance and standard deviation, let's go through the steps systematically:

Step 1: Calculate the Mean

First, we find the mean xˉ\bar{x} of the data sample.

Given dataTo compute the sample variance and standard deviation, let's go through the steps systematically:

Step 1: Calculate the Mean

First, we find the mean xˉ\bar{x} of the data sample.

Given 3.8,3.5,3.9,0.2,0.23.8, -3.5, 3.9, -0.2, -0.2

xˉ=3.8+(3.5)+3.9+(0.2)+(0.2)5=4.05=0.8\bar{x} = \frac{3.8 + (-3.5) + 3.9 + (-0.2) + (-0.2)}{5} = \frac{4.0}{5} = 0.8

Step 2: Calculate the Squared Deviations

Now, we calculate each value's deviation from the mean, square those deviations, and sum them up.

  1. (3.80.8)2=(3.0)2=9.0(3.8 - 0.8)^2 = (3.0)^2 = 9.0
  2. (3.50.8)2=(4.3)2=18.49(-3.5 - 0.8)^2 = (-4.3)^2 = 18.49
  3. (3.90.8)2=(3.1)2=9.61(3.9 - 0.8)^2 = (3.1)^2 = 9.61
  4. (0.20.8)2=(1.0)2=1.0(-0.2 - 0.8)^2 = (-1.0)^2 = 1.0
  5. (0.20.8)2=(1.0)2=1.0(-0.2 - 0.8)^2 = (-1.0)^2 = 1.0

Sum of squared deviations: 9.0+18.49+9.61+1.0+1.0=39.19.0 + 18.49 + 9.61 + 1.0 + 1.0 = 39.1

Step 3: Calculate the Sample Variance

To find the sample variance s2s^2, divide the sum of squared deviations by n1n - 1, where nn is the sample size (in this case, 5):

s2=39.151=39.14=9.775s^2 = \frac{39.1}{5 - 1} = \frac{39.1}{4} = 9.775

Rounding to two decimal places, the variance is: s2=9.78s^2 = 9.78

Step 4: Calculate the Sample Standard Deviation

The sample standard deviation ss is the square root of the variance:

s=9.7753.13s = \sqrt{9.775} \approx 3.13

Rounding to two decimal places, the standard deviation is: s3.13s \approx 3.13

Final Answers:

  • Variance: 9.789.78
  • Standard Deviation: 3.133.13

Would you like more details or have any questions?

Here are five related questions:

  1. How is sample variance different from population variance?
  2. Why do we divide by n1n - 1 in sample variance calculations?
  3. What does a high standard deviation indicate about data dispersion?
  4. How does rounding affect the accuracy of variance and standard deviation?
  5. Can we calculate the variance and standard deviation without knowing the mean?

Tip: For a larger dataset, consider using statistical software or a calculator to avoid manual calculation errors.

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Math Problem Analysis

Mathematical Concepts

Statistics
Variance
Standard Deviation
Data Analysis

Formulas

Mean: \( \bar{x} = \frac{\sum x}{n} \)
Sample Variance: \( s^2 = \frac{\sum (x - \bar{x})^2}{n - 1} \)
Sample Standard Deviation: \( s = \sqrt{s^2} \)

Theorems

The formula for sample variance and standard deviation in statistics

Suitable Grade Level

Grades 10-12