Shortcut Trick
Find the Mean (μ) of the marks: (65 + 47 + 59 + 75 + 82 + 91 + 55 + 78) ÷ 8 = 552 ÷ 8 = 69.
Calculate the sum of squared deviations from the mean: 16 + 484 + 100 + 36 + 169 + 484 + 196 + 81 = 1566.
Find the Variance (σ2): 1566 ÷ 8 = 195.75.
Calculate the Standard Deviation (σ): √195.75 ≈ 13.99.
∴ The correct answer is 13.99.
Alternate Method
Given:
Marks obtained by students: 65, 47, 59, 75, 82, 91, 55, 78
Total number of students (N) = 8
Formula Used:
Mean (μ) = Σx ÷ N
Variance (σ2) = Σ(x − μ)2 ÷ N
Standard Deviation (σ) = √Variance
Calculations:
⇒ Mean (μ) = (65 + 47 + 59 + 75 + 82 + 91 + 55 + 78) ÷ 8
⇒ Mean (μ) = 552 ÷ 8 = 69
⇒ Now, calculate the squared deviations (x − μ)2 for each student:
⇒ (65 − 69)2 = (−4)2 = 16
⇒ (47 − 69)2 = (−22)2 = 484
⇒ (59 − 69)2 = (−10)2 = 100
⇒ (75 − 69)2 = 62 = 36
⇒ (82 − 69)2 = 132 = 169
⇒ (91 − 69)2 = 222 = 484
⇒ (55 − 69)2 = (−14)2 = 196
⇒ (78 − 69)2 = 92 = 81
⇒ Sum of squared deviations Σ(x − μ)2 = 16 + 484 + 100 + 36 + 169 + 484 + 196 + 81 = 1566
⇒ Variance (σ2) = 1566 ÷ 8 = 195.75
⇒ Standard Deviation (σ) = √195.75 ≈ 13.99
Option Analysis:
⇒ Option (1) 15.05 is incorrect because 15.052 ≈ 226.5, which is much higher than the calculated variance of 195.75.
⇒ Option (2) 13.99 is correct because 13.992 ≈ 195.72, which matches our calculated variance of 195.75.
⇒ Option (3) 14.25 is incorrect because 14.252 ≈ 203.06, which does not match the variance.
⇒ Option (4) 12.53 is incorrect because 12.532 ≈ 157.00, which is much lower than the calculated variance.
∴ The correct answer is 13.99.
Additional Information
Standard Deviation (σ)
It is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.
Variance (σ2)
Variance is the expectation of the squared deviation of a random variable from its mean. It is calculated as the average of the squared differences from the Mean.
Population vs Sample Standard Deviation
For population data, the denominator is N. For sample data, Bessel's correction is applied, and the denominator becomes (N − 1) to correct bias in the estimation of the population variance.