Math Problem Statement

14 12 18 11 20 17 13 15 13 15 Calculate the range, population variance, and population standard deviation for the following data set. If necessary, round to one more decimal place than the largest number of decimal places given in the data.

Solution

Let's solve for the range, population variance, and population standard deviation for the given data set:

Step 1: Organize the Data

The data set is:
14,12,18,11,20,17,13,15,13,1514, 12, 18, 11, 20, 17, 13, 15, 13, 15

Step 2: Calculate the Range

The range is the difference between the maximum and minimum values in the data set.

  • Maximum value: 20
  • Minimum value: 11

Range=2011=9\text{Range} = 20 - 11 = 9

Step 3: Calculate the Population Variance

The formula for the population variance is:

σ2=(xiμ)2N\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}

Where:

  • xix_i is each data point
  • μ\mu is the mean of the data set
  • NN is the number of data points

Step 3.1: Find the Mean (μ\mu)

μ=14+12+18+11+20+17+13+15+13+1510=14810=14.8\mu = \frac{14 + 12 + 18 + 11 + 20 + 17 + 13 + 15 + 13 + 15}{10} = \frac{148}{10} = 14.8

Step 3.2: Find the Squared Differences from the Mean and Sum Them

(1414.8)2=(0.8)2=0.64(14 - 14.8)^2 = (-0.8)^2 = 0.64 (1214.8)2=(2.8)2=7.84(12 - 14.8)^2 = (-2.8)^2 = 7.84 (1814.8)2=(3.2)2=10.24(18 - 14.8)^2 = (3.2)^2 = 10.24 (1114.8)2=(3.8)2=14.44(11 - 14.8)^2 = (-3.8)^2 = 14.44 (2014.8)2=(5.2)2=27.04(20 - 14.8)^2 = (5.2)^2 = 27.04 (1714.8)2=(2.2)2=4.84(17 - 14.8)^2 = (2.2)^2 = 4.84 (1314.8)2=(1.8)2=3.24(13 - 14.8)^2 = (-1.8)^2 = 3.24 (1514.8)2=(0.2)2=0.04(15 - 14.8)^2 = (0.2)^2 = 0.04 (1314.8)2=(1.8)2=3.24(13 - 14.8)^2 = (-1.8)^2 = 3.24 (1514.8)2=(0.2)2=0.04(15 - 14.8)^2 = (0.2)^2 = 0.04

Sum of squared differences:

0.64+7.84+10.24+14.44+27.04+4.84+3.24+0.04+3.24+0.04=71.60.64 + 7.84 + 10.24 + 14.44 + 27.04 + 4.84 + 3.24 + 0.04 + 3.24 + 0.04 = 71.6

Step 3.3: Calculate the Population Variance

σ2=71.610=7.16\sigma^2 = \frac{71.6}{10} = 7.16

Step 4: Calculate the Population Standard Deviation

The population standard deviation is the square root of the variance.

σ=7.162.68\sigma = \sqrt{7.16} \approx 2.68

Summary:

  • Range: 9
  • Population Variance: 7.16
  • Population Standard Deviation: 2.68

Would you like more details or have any questions?

Here are 5 related questions:

  1. How would the calculations change if the data were considered a sample instead of a population?
  2. What does the standard deviation tell us about the data set's spread?
  3. How can we use variance and standard deviation to identify outliers?
  4. What would be the effect on the variance if we added a constant to each data point?
  5. How is the range affected by outliers in the data set?

Tip: To reduce rounding errors, carry more decimal places during intermediate steps and round only at the final answer.

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Math Problem Analysis

Mathematical Concepts

Descriptive Statistics
Range
Variance
Standard Deviation

Formulas

Range = Max value - Min value
Population variance (σ^2) = Σ(x_i - μ)^2 / N
Population standard deviation (σ) = √Variance

Theorems

None required

Suitable Grade Level

Grades 9-12