We know that Mean =

\( \dfrac{Sum\; of\; all \;observations}{Number\; of \;observations}\)

Mean, x̅ = \( \dfrac{ ((n(n + 1))2)}{n}\)

= \( \dfrac{ (n + 1)}{2}\)

and also Variance,

By substitute that value of x̅ we get,

We know that,

\( (a-b)^2=a^2-2ab+b^2\)

(a + b)(a – b) = a^{2} – b^{2}

σ^{2} = \(
\dfrac{(n^2 – 1)}{12}\)

Mean = \( \dfrac{ (n + 1)}{2}\) and Variance

\( \dfrac{(n^2 – 1)}{12}\)

Answered by Pragya Singh | 11 months agoFind the mean deviation from the mean and from a median of the following distribution:

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

No. of students | 5 | 8 | 15 | 16 | 6 |

The age distribution of 100 life-insurance policy holders is as follows

Age (on nearest birthday) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |

No. of persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |

Compute mean deviation from mean of the following distribution:

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |

No. of students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |

Find the mean deviation from the mean for the following data:

Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |

Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |

Find the mean deviation from the mean for the following data:

Classes | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |

Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |