1 Answer
Solution :-
\(\)_1638888042.png)
Mean,
\(\overline{x}= A +\dfrac{\displaystyle\sum_{i=1}{f_i}u_i }{N}× h\)
Where, A = 92.5, h = 5
So, x̅ = 92.5 + ((\( \dfrac{6}{60}\)) × 5)
= 92.5 + \( \dfrac{1}{2}\)
= 92.5 + 0.5
= 93
Then, Variance,
\( σ^2= \dfrac{h^2}{N^2}[N\displaystyle\sum_{i=1}{f_i}u_i^2-(\displaystyle\sum_{i=1}{f_i}u_i)^2] \)
σ2 = \( ( \dfrac{5^2}{60^2})\) [60(254) – 62]
= (\( \dfrac{1}{144}\)) [15240 – 36]
= \( \dfrac{15204}{144}\)
= \( \dfrac{1267}{22}\)
= 105.583
Hence, standard deviation = σ = \( \sqrt{105.583}\)
= 10.275
Mean = 93, variance = 105.583 and Standard Deviation = 10.275
Answered by Abhisek | 1 year agoRelated Questions
Find 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 |