Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.2

Question1: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:

Number of calls (x)	0	1	2	3	4	5	6
Number of intervals (f)	15	24	29	46	54	43	39

Compute the mean number of calls per interval.

Solution: 

Let the assumed mean(A) be =3 (Generally we choose the middle element to be the assumed mean, but it's not mandatory),

hence, the table is,
 

Number of calls (x_i)	Number of intervals (f_i)	u_{i}=x_{i}-A=x_{i}-3	f_{i}*u_{i}
0	15	-3	-45
1	24	-2	-48
2	29	-1	-29
3	46	0	0
4	54	1	54
5	43	2	86
6	39	3	117
	\displaystyle\sum_{}^{} f_{i}=250		\displaystyle\sum_{}^{} f_{i}*u_{i}=135

hence, mean of the calls = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                        = 3+\frac{135}{250}

                                        = 3.54

Therefore, mean number of calls per interval is 3.54

Question 2: Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

Number of heads per toss (x)	0	1	2	3	4	5
Number of tosses (f)	38	144	342	287	164	25

Solution:

Let the assumed mean (A) be = 2

hence, the table is,

Number of heads per toss (x_i)	Number of tosses (f_i)	u_{i}=x_{i}-A=x_{i}-2	f_{i}*u_{i}
0	38	-2	-76
1	144	-1	-144
2	342	0	0
3	287	1	287
4	164	2	328
5	25	3	75
	\displaystyle\sum_{}^{} f_{i}=1000		\displaystyle\sum_{}^{} f_{i}*u_{i}=470

Mean number of head per toss = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                   = 2+\frac{470}{1000}

                                                   = 2.47

Therefore, mean number of head per toss is 2.47

Question 3: The following table gives the number of branches and number of plants in the garden of a school.

Number of branches (x)	2	3	4	5	6
Number of plants (f)	49	43	57	38	13

Calculate the average number of branches per plant.

Solution:

Let the assumed mean(A) be = 4

hence, the table is,

Number of branches (x_i)	Number of plants (f_i)	u_{i}=x_{i}-A=x_{i}-4	f_{i}*u_{i}
2	49	-2	-98
3	43	-1	-43
4	57	0	0
5	38	1	38
6	13	2	26
	\displaystyle\sum_{}^{} f_{i}=200		\displaystyle\sum_{}^{} f_{i}*u_{i}=-77

Average Number of branches per plant = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                                = 2+(\frac{-77}{200})

                                                                = 3.615

Therefore, mean number of branches per plant is 3.615

Question 4: The following table gives the number of children of 150 families in a village

Number of children (x)	0	1	2	3	4	5
Number of families (f)	10	21	55	42	15	7

Find the average number of children per family.

Solution:

Let the assumed mean(A) be = 2

Hence, the table is,

Number of children (x_i)	Number of families (f_i)	u_{i}=x_{i}-A=x_{i}-2	f_{i}*u_{i}
0	10	-2	-20
1	21	-1	-21
2	55	0	0
3	42	1	42
4	15	2	30
5	7	3	21
	\displaystyle\sum_{}^{} f_{i}=150		\displaystyle\sum_{}^{} f_{i}*u_{i}=52

Average number of children per family = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                               = 2+(\frac{52}{150})

                                                               = 2.35 (approximately)

Therefore, the average number of children per family is 2.35 (approximately)

Question 5: The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:

Marks (x)	15	20	22	24	25	30	33	38	45
Frequency (f)	5	8	11	20	23	18	13	3	1

Find the average number of marks.

Solution:

Let the assume mean (A) be = 25

hence, the table is,

Marks (x_i)	Frequency (f_i)	u_{i}=x_{i}-A=x_{i}-25	f_{i}*u_{i}
15	5	-10	-50
20	8	-5	-40
22	11	-3	-33
24	20	-1	-20
25	23	0	0
30	18	5	90
33	13	8	104
38	3	13	39
45	1	20	20
	\displaystyle\sum_{}^{} f_{i}=102		\displaystyle\sum_{}^{} f_{i}*u_{i}=110

Average number of marks =  A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                           = 25+(\frac{102}{110})

                                           = 26.08 (approximately)

Therefore, average number of marks is 26.08 (approximately)

Question 6: The number of students absent in a class was recorded every day for 120 days and the information is given in the following

Number of students absent (x)	0	1	2	3	4	5	6	7
Number of Days (f)	1	4	10	50	34	15	4	2

Find the mean number of students absent per day.

Solution:

Let mean assumed mean (A) be = 3

Number of students absent (x_i)	Number of Days (f)	u_{i}=x_{i}-A=x_{i}-3	f_{i}*u_{i}
0	1	-3	-3
1	4	-2	-8
2	10	-1	-10
3	50	0	0
4	34	1	34
5	15	2	30
6	4	3	12
7	2	4	8
	\displaystyle\sum_{}^{} f_{i}=120		\displaystyle\sum_{}^{} f_{i}*u_{i}=63

Mean number of students absent per day = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                                    = 3+(\frac{63}{120})

                                                                    = 3.525

Therefore, the mean number of students absent per day is 3.525

Question 7: In the first proof of reading of a book containing 300 pages the following distribution of misprints was obtained:

Number of misprints per page (x)	0	1	2	3	4	5
Number of page (f)	154	95	36	9	5	1

Find the average number of misprints per page.

Solution:

Let the assumed mean (A) be = 2

Number of misprints per page (x_i)	Number of page (f_i)	u_{i}=x_{i}-A=x_{i}-2	f_{i}*u_{i}
0	154	-2	-308
1	95	-1	-95
2	36	0	0
3	9	1	9
4	5	2	10
5	1	3	3
	\displaystyle\sum_{}^{} f_{i}=300		\displaystyle\sum_{}^{} f_{i}*u_{i}=-381

Average number of misprints per day = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                             = 2+(\frac{-381}{300})

                                                             = 0.73

Therefore, the average number of misprints per day is 0.73

Question 8: Find the mean from the following frequency distribution of marks at a test in statistics:

Number of accidents (x)	0	1	2	3	4
Number of workers (f)	70	52	34	3	1

Find the average number of misprints per page.

Solution:

Let the assumed mean (A) = 2

Number of accidents (x_i)	Number of workers (f_i)	u_{i}=x_{i}-A=x_{i}-2	f_{i}*u_{i}
0	70	-2	-140
1	52	-1	-52
2	34	0	0
3	3	1	3
4	1	2	2
	\displaystyle\sum_{}^{} f_{i}=100		\displaystyle\sum_{}^{} f_{i}*u_{i}=-187

Average no of accidents per day workers = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                                  = 2+(\frac{-187}{100})

                                                                  = 0.83

Therefore, average no of accidents per day workers 0.83

Question 9: Find the mean from the following frequency distribution of marks at a test in statistics:

Marks (x)	5	10	15	20	25	30	35	40	45	50
Number of students (f)	15	50	80	76	72	45	39	9	8	6

Solution:

Let the assumed mean (A) be = 25

Marks (x_i)	Number of students (f_i)	u_{i}=x_{i}-A=x_{i}-25	f_{i}*u_{i}
5	15	-20	-300
10	50	-15	-750
15	80	-10	-800
20	76	-5	-380
25	72	0	0
30	45	5	225
35	39	10	390
40	9	15	135
45	8	20	160
50	6	25	150
	\displaystyle\sum_{}^{} f_{i}=400		\displaystyle\sum_{}^{} f_{i}*u_{i}=-1170

Mean = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

          = 25+(\frac{-1170}{400})

          = 22.075

Therefore, the mean is 22.075