Statistics – Exercise 6.1
Solve the following using step deviation of the following data.
- Calculate the standard deviation of the following data.
x | 3 | 8 | 13 | 18 | 23 |
f | 7 | 10 | 15 | 10 | 8 |
- The number of books bought by 200 students in a book exhibition is given below.
No. of books | 0 | 1 | 2 | 3 | 4 |
No. of students | 35 | 64 | 68 | 18 | 15 |
Find the variance and standard variation.
- The marks scored by 60 students in a science test are gven below.
Marks(x) | 10 | 20 | 30 | 40 | 50 | 60 |
No. of students | 8 | 12 | 20 | 10 | 7 | 3 |
Calculate the variance and standard deviation
- The daily wages of Rs. 40 workers of a factory are given in the following table.
Wages in Rs. | 30 – 34 | 34 – 38 | 38 – 42 | 42 – 46 | 46 – 50 | 50 – 54 |
No. of workers | 4 | 7 | 9 | 11 | 6 | 3 |
Calculate (i) Mean (ii) variance and (iii) standard deviation of wages and interpret the findings
- Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the items and the sum of squares of all the items.
- In study of diabetic patients in a village, the following observations were noted.
Age in years | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
No. of patients | 2 | 5 | 12 | 19 | 9 | 3 |
Calculate the mean and standard deviation. Also interpret the results.
Statistics – Exercise 6.1 – Solution
Solve the following using step deviation of the following data.
- Calculate the standard deviation of the following data.
x | 3 | 8 | 13 | 18 | 23 |
f | 7 | 10 | 15 | 10 | 8 |
Solution:
Let the assumed mean, A = 13
The common factor of the scores, C = 5
x | f | Step deviation d = ^{x-A}/_{C} |
fd | d^{2} | fd^{2} |
3 | 7 | -2 | -14 | 4 | 28 |
8 | 10 | -1 | -10 | 1 | 10 |
13 | 15 | 0 | 0 | 0 | 0 |
18 | 10 | 1 | 10 | 1 | 10 |
23 | 8 | 2 | 16 | 4 | 32 |
n = 50 | ⅀fd = 2 | ⅀fd^{2} = 80 |
Therefore standard deviation is 6.32
- The number of books bought by 200 students in a book exhibition is given below.
No. of books | 0 | 1 | 2 | 3 | 4 |
No. of students | 35 | 64 | 68 | 18 | 15 |
Find the variance and standard variation.
Solution:
Let the assumed mean, A = 2
The common factor of the scores, C = 1
No. of books | No. of students | Standard deviation, d= ^{x-A}/_{C} | fd | d^{2} | fd^{2} |
0 | 35 | -2 | -70 | 4 | 140 |
1 | 64 | -1 | -64 | 1 | 64 |
2 | 68 | 0 | 0 | 0 | 0 |
3 | 18 | 1 | 18 | 1 | 18 |
4 | 15 | 2 | 30 | 4 | 60 |
⅀f = 200 | ⅀fd = -86 | ⅀d^{2} = 10 | ⅀fd^{2 }= 282 |
Therefore, standard deviation is 1.107 and variance is 1.23
- The marks scored by 60 students in a science test are gven below.
Marks(x) | 10 | 20 | 30 | 40 | 50 | 60 |
No. of students | 8 | 12 | 20 | 10 | 7 | 3 |
Calculate the variance and standard deviation.
Solution:
Let assumed mean , A = 40
C = 10
Marks(x) | No. of students | d | fd | d^{2} | fd^{2} |
10 | 8 | -3 | -24 | 9 | 72 |
20 | 12 | -2 | -24 | 4 | 48 |
30 | 20 | -1 | -20 | 1 | 20 |
40 | 10 | 0 | 0 | 0 | 0 |
50 | 7 | 1 | 7 | 1 | 7 |
60 | 3 | 2 | 6 | 4 | 12 |
n = 60 | ⅀d = -3 | ⅀fd = -55 | ⅀fd^{2} = 159 |
Therefore, standard deviation is 13.4 and variance is 179.23
- The daily wages of Rs. 40 workers of a factory are given in the following table.
Wages in Rs. | 30 – 34 | 34 – 38 | 38 – 42 | 42 – 46 | 46 – 50 | 50 – 54 |
No. of workers | 4 | 7 | 9 | 11 | 6 | 3 |
Calculate (i) Mean (ii) variance and (iii) standard deviation of wages and interpret the findings
Solution:
Assumed mean, A = 40
C = 34 – 30 = 4
Wages in Rs. | No. of workers(f) | x | fx | d = ^{x-A}/_{C} | fd | d^{2} | fd^{2} |
30 – 34 | 4 | 32 | 128 | -2 | -8 | 4 | 16 |
34 – 38 | 7 | 36 | 252 | -1 | -7 | 1 | 7 |
38 – 42 | 9 | 40 | 360 | 0 | 0 | 0 | 0 |
42 – 46 | 11 | 44 | 484 | 1 | 11 | 1 | 11 |
46 – 50 | 6 | 48 | 288 | 2 | 12 | 4 | 24 |
50 – 54 | 3 | 52 | 156 | 3 | 9 | 9 | 27 |
n = 40 | ⅀fx = 1668 | ⅀fd = 17 | ⅀fd^{2} = 85 |
This means each score deviates from the mean value 41.7 by 5.58
5. Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the items and the sum of squares of all the items.
Solution:
The number of items = n = 100
Mean of 100 items = 48
Standard deviation, = 10
Sum of scores ,
Standard deviation of 100 items, = 10
Therefore, sum of all the items is 4800 and the sum of squares of all the items is 2,40,400
- In study of diabetic patients in a village, the following observations were noted.
Age in years | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
No. of patients | 2 | 5 | 12 | 19 | 9 | 3 |
Calculate the mean and standard deviation. Also interpret the results.
Solution:
Assumed mean A = 35
C = 10
Age in years | No. of patients | x | fx | d =
^{x-A}/_{C} ^{ } |
d^{2} | fd | fd^{2} |
10 – 20 | 2 | 15 | 30 | -2 | 4 | -4 | 8 |
20 – 30 | 5 | 25 | 50 | -1 | 1 | -5 | 5 |
30 – 40 | 12 | 35 | 420 | 0 | 0 | 0 | 0 |
40 – 50 | 19 | 45 | 855 | 1 | 1 | 19 | 19 |
50 – 60 | 9 | 55 | 495 | 2 | 4 | 18 | 36 |
60 – 70 | 3 | 65 | 195 | 3 | 9 | 9 | 27 |
n = 50 | ⅀fx = 2045 | ⅀fd = 37 | ⅀fd^{2} = 95 |
This means each score deviates from the mean value 41.7 by 11.62
Next exercise – Statistics – Exercise 6.2 – Class X