Math Problem Statement

The length (in pages) of math research projects is given below. Using this information, calculate the range, variance, and standard deviation.

29, 17, 15, 34, 24, 299, 23, 17, 20

range =  variance =  standard deviation = 

Solution

To calculate the range, variance, and standard deviation of the given data set:

Data set:
29, 17, 15, 34, 24, 299, 23, 17, 20


1. Range

The range is the difference between the maximum and minimum values.

Range=Maximum valueMinimum value\text{Range} = \text{Maximum value} - \text{Minimum value} Maximum value = 299
Minimum value = 15

Range=29915=284\text{Range} = 299 - 15 = 284


2. Variance

To calculate variance, we first need to find the mean (average) of the dataTo calculate the range, variance, and standard deviation of the given data set:

Data set:
29, 17, 15, 34, 24, 299, 23, 17, 20


1. Range

The range is the difference between the maximum and minimum values.

Range=Maximum valueMinimum value\text{Range} = \text{Maximum value} - \text{Minimum value} Maximum value = 299
Minimum value = 15

Range=29915=284\text{Range} = 299 - 15 = 284


2. Variance

To calculate variance, we first need to find the mean (average) of the

Mean=Data valuesN\text{Mean} = \frac{\sum \text{Data values}}{N} Mean=29+17+15+34+24+299+23+17+209=478953.11\text{Mean} = \frac{29 + 17 + 15 + 34 + 24 + 299 + 23 + 17 + 20}{9} = \frac{478}{9} \approx 53.11

Now, to calculate the variance, we use the formula:

Variance(σ2)=(ValueMean)2N\text{Variance} (\sigma^2) = \frac{\sum (\text{Value} - \text{Mean})^2}{N}

Let’s compute the squared differences from the mean for each data point:

(2953.11)2=(24.11)2=581.30(29 - 53.11)^2 = (-24.11)^2 = 581.30 (1753.11)2=(36.11)2=1304.00(17 - 53.11)^2 = (-36.11)^2 = 1304.00 (1553.11)2=(38.11)2=1451.98(15 - 53.11)^2 = (-38.11)^2 = 1451.98 (3453.11)2=(19.11)2=365.19(34 - 53.11)^2 = (-19.11)^2 = 365.19 (2453.11)2=(29.11)2=847.47(24 - 53.11)^2 = (-29.11)^2 = 847.47 (29953.11)2=(245.89)2=60464.32(299 - 53.11)^2 = (245.89)^2 = 60464.32 (2353.11)2=(30.11)2=906.61(23 - 53.11)^2 = (-30.11)^2 = 906.61 (1753.11)2=(36.11)2=1304.00(17 - 53.11)^2 = (-36.11)^2 = 1304.00 (2053.11)2=(33.11)2=1096.26(20 - 53.11)^2 = (-33.11)^2 = 1096.26

Now, sum up all the squared differences:

=581.30+1304.00+1451.98+365.19+847.47+60464.32+906.61+1304.00+1096.26=68720.13\sum = 581.30 + 1304.00 + 1451.98 + 365.19 + 847.47 + 60464.32 + 906.61 + 1304.00 + 1096.26 = 68720.13

Finally, divide by the number of data points (9):

Variance=68720.1397635.57\text{Variance} = \frac{68720.13}{9} \approx 7635.57


3. Standard Deviation

The standard deviation is the square root of the variance:

Standard deviation(σ)=7635.5787.39\text{Standard deviation} (\sigma) = \sqrt{7635.57} \approx 87.39


Summary:

Range=284\text{Range} = 284 Variance7635.57\text{Variance} \approx 7635.57 Standard deviation87.39\text{Standard deviation} \approx 87.39

Would you like further clarification on any of these steps?


Here are some related questions you might want to explore:

  1. How does removing the outlier (299) affect the variance and standard deviation?
  2. What is the difference between population variance and sample variance?
  3. How would the range change if you added more data points?
  4. Can you explain why the standard deviation is more robust than the range?
  5. What is the significance of a high standard deviation in this data set?

Tip: Variance gives a rough idea of how data spreads around the mean, but standard deviation provides a more intuitive sense of this spread because it has the same units as the data.

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Math Problem Analysis

Mathematical Concepts

Descriptive Statistics
Measures of Dispersion

Formulas

Range = Maximum value - Minimum value
Variance (σ²) = Σ(Value - Mean)² / N
Standard Deviation (σ) = √Variance

Theorems

No specific theorems involved, general statistical formulas

Suitable Grade Level

Grades 9-12