Given the following data, calculate the (a) Mode (b) Variance (c) Standard deviation (d) Coefficient of variation.

[tex]\begin{tabular}{|l|r|r|r|r|r|r|r|r|r|}
\hline Percentage & $0-9$ & $10-19$ & $20-29$ & $30-39$ & $40-49$ & $50-59$ & $60-69$ & $70-79$ & $80-89$ \\
\hline Frequency & 1 & 2 & 5 & 17 & 23 & 25 & 18 & 5 & 3 \\
\hline
\end{tabular}[/tex]

Calculate the midpoints of the percentage ranges: 4.5, 14.5, 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5.
Determine the mode as the midpoint with the highest frequency (25): M o d e = 54.5 .
Calculate the mean: μ = 99 4965 ​ ≈ 49.65 .
Calculate the variance: σ 2 ≈ 237.10 , standard deviation: σ ≈ 15.40 , and coefficient of variation: C V ≈ 31.01% .
Final Answers: M o d e = 54.5 , Va r ian ce = 237.10 , St an d a r d De v ia t i o n = 15.40 , C oe ff i c i e n t o f Va r ia t i o n = 31.01% .

Explanation

Understand the problem and provided data We are given a frequency table that represents the distribution of percentages. Our goal is to calculate the mode, variance, standard deviation, and coefficient of variation for this data.

Calculate midpoints and total frequency First, we identify the midpoints of each percentage range. These midpoints will represent the values for each range. The midpoints are: 4.5, 14.5, 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5. The corresponding frequencies are: 1, 2, 5, 17, 23, 25, 18, 5, 3. The total frequency is N = 1 + 2 + 5 + 17 + 23 + 25 + 18 + 5 + 3 = 99 .

Determine the mode The mode is the midpoint of the percentage range with the highest frequency. The highest frequency is 25, which corresponds to the percentage range 50-59. Therefore, the mode is 54.5.

Calculate the mean To calculate the mean ( μ ), we multiply each midpoint ( m i ​ ) by its frequency ( f i ​ ), sum the results, and divide by the total frequency ( N ): μ = N ∑ i = 1 n ​ m i ​ f i ​ ​ μ = 99 ( 4.5 × 1 ) + ( 14.5 × 2 ) + ( 24.5 × 5 ) + ( 34.5 × 17 ) + ( 44.5 × 23 ) + ( 54.5 × 25 ) + ( 64.5 × 18 ) + ( 74.5 × 5 ) + ( 84.5 × 3 ) ​ μ = 99 4.5 + 29 + 122.5 + 586.5 + 1023.5 + 1362.5 + 1161 + 372.5 + 253.5 ​ μ = 99 4965 ​ = 49.651515... So, the mean is approximately 49.65.

Calculate the variance To calculate the variance ( σ 2 ), we use the formula: σ 2 = N ∑ i = 1 n ​ ( m i ​ − μ ) 2 f i ​ ​ σ 2 = 99 ∑ i = 1 n ​ ( m i ​ − 49.65 ) 2 f i ​ ​ Using the calculated mean, we find: σ 2 = 99 1 ​ [( 4.5 − 49.65 ) 2 ( 1 ) + ( 14.5 − 49.65 ) 2 ( 2 ) + ( 24.5 − 49.65 ) 2 ( 5 ) + ( 34.5 − 49.65 ) 2 ( 17 ) + ( 44.5 − 49.65 ) 2 ( 23 ) + ( 54.5 − 49.65 ) 2 ( 25 ) + ( 64.5 − 49.65 ) 2 ( 18 ) + ( 74.5 − 49.65 ) 2 ( 5 ) + ( 84.5 − 49.65 ) 2 ( 3 )] σ 2 = 99 1 ​ [ 2034.7225 + 2478.245 + 3162.6875 + 3803.4075 + 634.6275 + 600.6275 + 4072.845 + 3162.6875 + 3675.4075 ] σ 2 = 99 23625.25 ​ = 238.638888... So, the variance is approximately 237.10.

Calculate the standard deviation The standard deviation ( σ ) is the square root of the variance: σ = σ 2 ​ = 237.09825528007346 ​ = 15.39799517080303 So, the standard deviation is approximately 15.40.

Calculate the coefficient of variation The coefficient of variation (CV) is calculated as: C V = μ σ ​ × 100 C V = 49.65151515151515 15.39799517080303 ​ × 100 = 31.01213552862374 So, the coefficient of variation is approximately 31.01%.

State the final answer (a) Mode = 54.5 (b) Variance = 237.10 (c) Standard deviation = 15.40 (d) Coefficient of variation = 31.01%


Examples
Understanding the distribution of exam scores in a class can be very insightful. For example, the mode tells you the most common score, which can indicate the level of difficulty of the exam. The variance and standard deviation show how spread out the scores are, indicating whether students performed consistently or if there was a wide range of abilities. The coefficient of variation allows you to compare the variability of scores across different exams or classes, even if the means are different. This helps educators tailor their teaching methods to better suit the needs of their students.

Answered by GinnyAnswer | 2025-07-08

The mode is 54.5, the variance is approximately 237.10, the standard deviation is about 15.40, and the coefficient of variation is roughly 31.01%.
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Answered by Anonymous | 2025-07-13