#### 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 =
| 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**