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प्रश्न
Marks obtained (in mathematics) by 9 students are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56
- Find the arithmetic mean.
- If marks of each student be increased by 4; what will be the new value of arithmetic mean?
उत्तर
a. Here, n = 9,
`barx = (x_1 + x_2 + ...... + x_n)/n`
∴ `barx = (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/9`
= `531/9`
= 59
b. If marks of each student be incresed by 4 then new arithmetic mean will be = 59 + 4 = 63
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संबंधित प्रश्न
Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.
| Marks | 5 | 6 | 7 | 8 | 9 |
| Number of Students | 6 | a | 16 | 13 | b |
If the mean of the distribution is 7.2, find a and b.
The following distribution represents the height of 160 students of a school.
| Height (in cm) | No. of Students |
| 140 – 145 | 12 |
| 145 – 150 | 20 |
| 150 – 155 | 30 |
| 155 – 160 | 38 |
| 160 – 165 | 24 |
| 165 – 170 | 16 |
| 170 – 175 | 12 |
| 175 – 180 | 8 |
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
- The median height.
- The interquartile range.
- The number of students whose height is above 172 cm.
The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of 'f'.
| Height (cm) | 50 | 55 | 58 | 60 | 65 | 70 | 71 |
| No. of plants | 2 | 4 | 10 | f | 5 | 4 | 3 |
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| Marks | No. of boys |
| 30 – 40 | 10 |
| 40 – 50 | 12 |
| 50 – 60 | 14 |
| 60 – 70 | 12 |
| 70 – 80 | 9 |
| 80 – 90 | 7 |
| 90 – 100 | 6 |
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Step-deviation method
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The median of the following observations arranged in ascending order is 64. Find the value of x:
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