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Example 10 - Calculate mean, variance, standard deviation - Standard deviation and variance - Continuous frequency (grouped data)

Example 10 - Chapter 15 Class 11 Statistics - Part 2
Example 10 - Chapter 15 Class 11 Statistics - Part 3


Transcript

Example 10 - Chapter 15 Class 11 Statistics - NCERT Calculate the mean, variance and standard deviation for the following distribution : Finding Variance and Standard Deviation Class Frequency (fi) Mid – point (x_i) fixi 30 – 40 3 35 35 × 3 = 105 40 – 50 7 45 45 × 7 = 315 50 – 60 12 55 55 × 12 = 660 60 – 70 15 65 65 × 15 = 975 70 – 80 8 75 75 × 8 = 600 80 – 90 3 85 85 × 3 = 255 90 – 100 2 95 95 × 2 = 190 ∑▒fi = 50 ∑▒fixi = 3100 ∑▒𝑓𝑖𝑥𝑖 = 3100 ∑▒𝑓𝑖 = 50 Mean (𝒙 ̅) = (∑▒𝑓𝑖𝑥𝑖)/(∑▒𝑓𝑖) = 3100/50 = 62 Finding Variance and Standard Deviation Class Frequency (fi) Mid – point (x_i) (xi -x ̅ )^2 fi(xi -x ̅ )^2 30 – 40 3 35 〖"(35 - 62)" 〗^2 = 729 3 × 729 =2187 40 – 50 7 45 〖"(45 - 62)" 〗^2 = 289 7 × 289 = 2023 50 – 60 12 55 〖"(55 - 62)" 〗^2 = 49 12 × 49 = 588 60 – 70 15 65 〖"(65 - 62)" 〗^2 = 9 15 × 9 = 135 70 – 80 8 75 〖"(75 - 62)" 〗^2 = 169 8 × 169 = 1352 80 – 90 3 85 〖"(85 - 62)" 〗^2 = 529 3 × 529 = 1589 90 – 100 2 95 〖"(95 - 62)" 〗^2 = 1089 2 × 1089 = 2187 ∑▒fi = 50 Sum = 10050 ∑▒〖𝑓𝑖(𝑥𝑖 −𝑥 ̅ )^2 〗 = 10050 ∑▒𝑓𝑖 = 50 Variance ("σ" ^2) = 1/𝑁 ∑▒〖𝑓𝑖(𝑥𝑖 −𝑥 ̅ )^2 〗 = 1/50 ×10050 = 201 Standard deviation (σ) = √("201" ) (σ) = 14.17

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.