Class 10 RD Sharma Solutions Chapter 7 Statistics Exercise 7.2 (original) (raw)
Last Updated : 21 Dec, 2020
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