Math Problem Statement

  1. The following data give the number of power outages suffered by a town for each of the last 12 months.

4 5 7 3 2 0 2 3 2 1 2 4

a. Calculate the Mean

b. Find the Range

c. Calculate the Variance

d. Calculate the Standard Deviation

Solution

Let's go through each calculation step-by-step:

a. Calculate the Mean

The mean (average) is calculated by summing all the values and dividing by the number of values.

Mean=Sum of all valuesNumber of values\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}

Mean=4+5+7+3+2+0+2+3+2+1+2+412\text{Mean} = \frac{4 + 5 + 7 + 3 + 2 + 0 + 2 + 3 + 2 + 1 + 2 + 4}{12}

Mean=3512=2.92 (rounded to 2 decimal places)\text{Mean} = \frac{35}{12} = 2.92 \ (\text{rounded to 2 decimal places})

b. Find the Range

The range is the difference between the highest and lowest values.

Range=Max valueMin value\text{Range} = \text{Max value} - \text{Min value}

Range=70=7\text{Range} = 7 - 0 = 7

c. Calculate the Variance

Variance is the average of the squared differences from the mean.

  1. Subtract the mean from each value, and square the result.
  2. Find the average of those squared differences.

Steps:

  1. Subtract the mean from each value: 42.92=1.08,52.92=2.08,72.92=4.08,32.92=0.08,22.92=0.92,02.92=2.924 - 2.92 = 1.08, \quad 5 - 2.92 = 2.08, \quad 7 - 2.92 = 4.08, \quad 3 - 2.92 = 0.08, \quad 2 - 2.92 = -0.92, \quad 0 - 2.92 = -2.92 Continue for all values.

  2. Square the results: 1.082,2.082,4.082,(0.92)2,1.08^2, \quad 2.08^2, \quad 4.08^2, \quad (-0.92)^2, \quad \dots

  3. Find the average of the squared differences. We will compute this in the next step.

d. Calculate the Standard Deviation

Standard deviation is the square root of the variance.

Let me calculate the variance and standard deviation for you.Here are the results:

a. Mean

Mean=2.92 (rounded to 2 decimal places)\text{Mean} = 2.92 \ (\text{rounded to 2 decimal places})

b. Range

Range=7\text{Range} = 7

c. Variance

Variance=3.24 (rounded to 2 decimal places)\text{Variance} = 3.24 \ (\text{rounded to 2 decimal places})

d. Standard Deviation

Standard Deviation=1.80 (rounded to 2 decimal places)\text{Standard Deviation} = 1.80 \ (\text{rounded to 2 decimal places})

Would you like further details or have any questions?

Here are five related questions:

  1. How does variance change if the dataset had fewer extreme values?
  2. Can you calculate the median for this dataset?
  3. What is the interpretation of standard deviation in this context?
  4. How does the variance formula differ for samples vs. populations?
  5. What would be the effect of adding a new data point to the dataset?

Tip: Always check whether your dataset is a sample or population before calculating variance or standard deviation, as the formulas differ slightly!

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Statistics
Measures of Central Tendency
Measures of Spread

Formulas

Mean = (Sum of values) / (Number of values)
Range = Max value - Min value
Variance = (Σ (x - mean)^2) / N
Standard Deviation = √Variance

Theorems

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Suitable Grade Level

Grades 8-10