Using the frequency distribution provided below, calculate the standard deviation using the formula, where x denotes the class midpoint, f is the class frequency, and n represents the total number of sample values. <math alttext="s equals the square root of the fraction with numerator n times open bracket Sigma f times x squared close bracket minus open bracket Sigma times open paren f times x close paren close bracket squared and denominator n times open paren n minus 1 close paren" data-value="s=\sqrt{\frac{n\left\lbrack\Sigma f\cdot x^2\right\rbrack-\left\lbrack\Sigma\left(f\cdot x\right)\right\rbrack^2}{n\left(n-1\right)}}"> s = n [ Σ f ⋅ x 2 ] − [ Σ ( f ⋅ x ) ] 2 n ( n − 1 ) s=\sqrt{\frac{n\left\lbrack\Sigma f\cdot x^2\right\rbrack-\left\lbrack\Sigma\left(f\cdot x\right)\right\rbrack^2}{n\left(n-1\right)}} Then, compare your calculated standard deviation with the value of 14.90 obtained using the sum of squares of deviations formula.

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1. Intro to Stats and Collecting Data(0)

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Levels of Measurement
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Intro to Collecting Data
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Sampling Methods
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2. Describing Data with Tables and Graphs(0)

Visualizing Qualitative vs. Quantitative Data
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Frequency Distributions
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Histograms
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Bar Graphs and Pareto Charts
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Pie Charts
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Frequency Polygons
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Time-Series Graph
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3. Describing Data Numerically(0)

Mean
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Median
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Mode
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Standard Deviation
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Interpreting Standard Deviation
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Percentiles & Quartiles
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3. Describing Data Numerically

Standard Deviation

3. Describing Data Numerically

Standard Deviation: Videos & Practice Problems

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Problem 23Multiple Choice

Using the frequency distribution provided below, calculate the standard deviation using the formula, where x denotes the class midpoint, f is the class frequency, and n represents the total number of sample values.
$s = \sqrt{\frac{n [Σ f \cdot x^{2}] - {[Σ (f \cdot x)]}^{2}}{n (n - 1)}}$

Then, compare your calculated standard deviation with the value of 14.90 obtained using the sum of squares of deviations formula.

$s equals 14.90$

The standard deviation from the frequency distribution is identical to the standard deviation from the original data.

$s equals 13.95$

The standard deviation from the frequency distribution is slightly lower than the standard deviation from the original data due to class grouping.

$s equals 19.50$

The standard deviation from the frequency distribution is significantly higher than the standard deviation from the original data.

$s equals 6.75$

The standard deviation from the frequency distribution is significantly lower than the standard deviation from the original data

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