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Ex-7.2 Statistics, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download
1. The number of telephone calls received at an exchange per interval for 250 successive one- minute intervals are given in the following frequency table:
| No. of calls(x): | 0 1 2 3 4 5 6 |
| No. of intervals (f): | 15 24 29 46 54 43 39 |
Compute the mean number of calls per interval.
Sol: Let be assumed mean (A) = 3
| No. of calls xi | No. of intervals fi | u1 = xi−A = xi = 3 | fiui |
| 0 | 15 | -3 | -45 |
| 1 | 24 | -2 | -47 |
| 2 | 29 | -1 | -39 |
| 3 | 46 | 0 | 0 |
| 4 | 54 | 54 | |
| 5 | 43 | 2 | 43(2) = 86 |
| 6 | 39 | 3 | 47 |
| N = 250 | Sum = 135 |
Mean number of cells = 3+frac135250 = frac885250 = 3.54
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.
| No of heads per toss (x): | 0 1 2 3 4 5 |
| No of tosses (f): | 38 144 342 287 164 25 |
Sol:Let the assumed mean (A) = 2
No. of heads per toss xi | No of intervals fi | ui = Ai−x = Ai−2 | fiui |
| 0 | 38 | -2 | -7 |
| 1 | 144 | -1 | -144 |
| 2 | 342 | 0 | 0 |
| 3 | 287 | 1 | 287 |
| 4 | 164 | 2 | 328 |
| 5 | 25 | 3 | 75 |
| N = 1000 | Sum = 470 |
Mean number of per toss = 2 +470/1000 = 2 + 0.47 = 2.47
3. The following table gives the number of branches and number of plants in the garden of a school.
| No of branches (x): | 2 3 4 5 6 |
| No of plants (f): | 49 43 57 38 13 |
Calculate the average number of branches per plant.
Sol: Let the assumed mean (A) = 4
| No of branches xi | No of plants fi | ui = xi−A = xi−4 | fiui |
| 2 | 49 | -2 | -98 |
| 3 | 43 | -1 | -43 |
| 4 | 57 | 0 | 0 |
| 5 | 38 | 1 | 38 |
| 6 | 13 | 2 | 26 |
| N = 200 | Sum = -77 |
Average number of branches per plant = 4 + (-77/200) = 4 -77/200 = (800 -77)/200 = 3.615
4. The following table gives the number of children of 150 families in a village
| No of children (x): | 0 1 2 3 4 5 |
| No of families (f): | 10 21 55 42 15 7 |
Find the average number of children per family.
Sol: Let the assumed mean (A) = 2
| No of children xi | No of families fi | ui = xi−A = xi−2 | fiui |
| 0 | 10 | -2 | -20 |
| 1 | 21 | -1 | -21 |
| 3 | 42 | 1 | 42 |
| 4 | 15 | 2 | 30 |
| 5 | 7 | 5 | 35 |
| N = 20 | Sum = 52 |
Average number of children for family = 2 + 52/150 = (300 +52)/150 = 352/150 = 2.35 (approx)
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.
Sol: Let the assumed mean (A) = 25
| Marks xi | Frequency fi | ui = xi−A = xi−2 | fiui |
| 15 | 5 | -10 | -50 |
| 20 | 8 | -5 | -40 |
| 22 | 8 | -3 | -24 |
| 24 | 20 | -1 | -20 |
| 25 | 23 | 0 | 0 |
| 30 | 18 | 5 | 90 |
| 33 | 13 | 8 | 104 |
| 38 | 3 | 12 | 36 |
| 45 | 3 | 20 | 60 |
| N = 122 | Sum = 110 |
Average number of marks = 25 + 110/102
= (2550 + 110)/102
= 2660/102
= 26.08 (Approx)
6.The number of students absent in a class was recorded every day for 120 days and the information is given in the following
| No of students absent (x): | 0 1 2 3 4 5 6 7 |
| No of days (f): | 1 4 10 50 34 15 4 2 |
Find the mean number of students absent per day.
Sol: Let mean assumed mean (A) = 3
| No of students absent xi | No of days fi | ui = xi−A = xi−3 | fiui |
| 3 | 1 | -3 | -3 |
| 1 | 4 | -2 | -8 |
| 2 | 10 | -1 | -10 |
| 3 | 50 | 0 | 0 |
| 4 | 34 | 1 | 24 |
| 5 | 15 | 2 | 30 |
| 6 | 4 | 3 | 12 |
| 7 | 2 | 4 | 8 |
| N = 120 | Sum = 63 |
Mean number of students absent per day = 3 + 63/120
= (360 + 63)/120
= 423/120
= 3.53
7. In the first proof of reading of a book containing 300 pages the following distribution of misprints was obtained:
| No of misprints per page (x): | 0 1 2 3 4 5 |
| No of pages (f): | 154 96 36 9 5 1 |
Find the average number of misprints per page.
Sol: Let the assumed mean (A) = 2
| No of misprints per page xi | No of days fi | ui = xi−A = xi−3 | fiui |
| 0 | 154 | -2 | -308 |
| 1 | 95 | -1 | -95 |
| 2 | 36 | 0 | 0 |
| 3 | 9 | 1 | 9 |
| 4 | 5 | 2 | 1 |
| 5 | 1 | 3 | 3 |
| N = 300 | Sum = 381 |
Average number of misprints per day = 2 + (-381/300)
= 2 – 381/300
= (600-381)/300
= 219/300
= 0.73
8. The following distribution gives the number of accidents met by 160 workers in a factory during a month.
| No of accidents (x): | 0 1 2 3 4 |
| No of workers (f): | 70 52 34 3 1 |
Find the average number of accidents per worker.
Sol: Let the assumed mean (A) = 2
| No of accidents | No of workers fi | ui = xi−A = xi−3 | fiui |
| 0 | 70 | -2 | -140 |
| 1 | 52 | -1 | -52 |
| 2 | 34 | 0 | 0 |
| 3 | 3 | 1 | 3 |
| 4 | 1 | 2 | 2 |
| N = 100 | Sum = -187 |
Average no of accidents per day workers
= > x + (-187/160)
= 133/160
= 0.83
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 |
| No of students (f): | 15 50 80 76 72 45 39 9 8 6 |
Sol:Let the assumed mean (A) = 25
| Marks xi | No of students fi | ui = xi−A = xi−3 | fiui< |
| 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 |
| N = 400 | Sum = -1170 |
Mean = 25 + (-1170)/400 = 22.075
The document Ex-7.2 Statistics, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 is a part of the Class 10 Course Extra Documents, Videos & Tests for Class 10.
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FAQs on Ex-7.2 Statistics, Class 10, Maths RD Sharma Solutions - Extra Documents, Videos & Tests for Class 10
| 1. What are the different measures of central tendency in statistics? |  |
Ans. In statistics, there are three main measures of central tendency: mean, median, and mode. The mean is calculated by adding up all the values and dividing by the total number of values. The median is the middle value when the data is arranged in ascending or descending order. The mode is the value that appears most frequently in the data set.
| 2. How is the mean calculated in statistics? | |
Ans. The mean is calculated by adding up all the values in the data set and dividing the sum by the total number of values. For example, if we have the data set {2, 4, 6, 8, 10}, the mean would be (2 + 4 + 6 + 8 + 10) / 5 = 6.
| 3. What is the median in statistics? | |
Ans. The median is a measure of central tendency in statistics. It is the middle value when the data is arranged in ascending or descending order. If the data set has an odd number of values, the median is the middle value. If the data set has an even number of values, the median is the average of the two middle values.
| 4. How is the mode calculated in statistics? | |
Ans. The mode is the value that appears most frequently in a data set. To calculate the mode, you need to determine which value occurs most frequently. If there is more than one value that appears with the same highest frequency, the data set is said to have multiple modes.
| 5. What is the importance of statistics in everyday life? | |
Ans. Statistics plays a crucial role in everyday life. It helps in making informed decisions by providing a way to analyze and interpret data. Statistics is used in various fields such as economics, finance, healthcare, sports, and social sciences. It helps in understanding trends, making predictions, and evaluating the reliability of information.
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