\( Mean , \overline{X} = \dfrac{\displaystyle\sum_{i=1}^{a} X_i}{n}\)
So, x̅ = \( \dfrac{ (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)}{8}\)
= \( \dfrac{72}{8}\)
= 9
Let us make the table of the given data and append other columns after calculations.
We know that Variance,
\(σ^2= \dfrac{1}{N} \displaystyle\sum_{i=1}^{a} f_i (X_i-\overline {X} )^2\)
σ2= \( \dfrac{1}{8}\times 74\)
= 9.2
Mean = 9 and Variance = 9.25
Answered by Pragya Singh | 1 year 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 |