9th Standard Syllabus & Materials
9th Standard
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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
In a distribution, the mean and mode are 66 and 60 respectively. Calculate the median.
2.
Find the mode for the following data
| Marks | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 |
|---|---|---|---|---|---|
| No. of students | 7 | 10 | 16 | 32 | 24 |
3.
The Median of the following data is 24. Find the value of x.
| Class Interval (CI) | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
|---|---|---|---|---|---|
| Frequency (f) | 6 | 24 | x | 16 | 9 |
4.
The following table gives the weekly expenditure of 200 families. Find the median of the weekly expenditure.
| Weekly expenditure(Rs) | 0-1000 | 1000-2000 | 2000-3000 | 3000-4000 | 4000-500 |
|---|---|---|---|---|---|
| Number of families | 28 | 46 | 54 | 42 | 30 |
5.
Calculate the median for the following data:
| Height (cm) | 160 | 150 | 152 | 161 | 156 | 154 | 155 |
|---|---|---|---|---|---|---|---|
| No. of Students | 12 | 8 | 4 | 4 | 3 | 3 | 7 |
1.
Given, Mean = 66 and Mode = 60.
Using, Mode ≈ 3Median – 2Mean
60 ≈ 3Median – 2(66)
3 Median ≈ 60 +132
Therefore, Median ≈ \({192\over 3}тЙИ64\)
2.
| Marks | f |
|---|---|
| 0.5-5.5 | 7 |
| 5.5-10.5 | 10 |
| 10.5-15.5 | 16 |
| 15.5-20.5 | 32 |
| 20.5-25.5 | 24 |
Modal class is 16 -20 since it has the maximum frequency.
l = 15.5, f = 32, f1 = 16, f2 = 24, c = 20.5–15.5 = 5
Mode \(=l+{\left(f-f_1\over 2f-f_1-f_2\right)}\times c\)
\(=15.5+\left(32-16\over64-16-24\right)\times 5\)
\(=15.5+\left(16\over24\right)\times = 15.5 + 3.33 =18.83\)
3.
| Class Interval (CI) | Frequency (f) | Cumulative frequency (cf) |
|---|---|---|
| 0-10 | 6 | 6 |
| 10-20 | 24 | 30 |
| 20-30 | x | 30 + x |
| 30-40 | 16 | 46 + x |
| 40-50 | 9 | 55 + x |
| N = 55 + x |
Since the median is 24 and median class is 20 – 30
l = 20 N = 55 + x, m = 30, c = 10, f = x
Median = \(l+{\left({N\over2}-m\right)\over f}\times c\)
\(24=20+{\left({55+x\over2}-3\right)\over x}\times 10\)
\(4={5x-25\over x}\) (after simplification)
4x = 5x – 25
5x – 4x = 25
x = 25
4.
| Weekly Expenditure | Number of families (f) | Cumulative frequency (cf) |
|---|---|---|
| 0-1000 | 28 | 28 |
| 1000-2000 | 46 | 74 |
| 2000-3000 | 54 | 128 |
| 3000-4000 | 42 | 170 |
| 4000-5000 | 30 | 200 |
| N = 200 |
Median class = \(\left(N\over 2\right)^{th}\) value=\(\left(200\over 2\right)^{th}\) value
= 100th value
Median class = 2000 – 3000
\({N\over 2}=100\ l=2000\)
m = 74, c = 1000, f = 54
Median = \(l+{\left({n\over2}-m\right)\over f}\times c\)
\(=2000+\left(100-74\over54\right)\times1000\)
\(=2000+({26\over54})\times1000=2000+481.5\)
= 2481.5
5.
Let us arrange the marks in ascending order and prepare the following data:
| Height (cm) | Number of students (f) | Cumulative frequency (cf) |
|---|---|---|
| 150 | 8 | 8 |
| 152 | 4 | 12 |
| 154 | 3 | 15 |
| 155 | 7 | 22 |
| 156 | 3 | 25 |
| 160 | 12 | 37 |
| 160 | 12 | 37 |
Here N = 41
Median = size of \(\left(N+1\over 2\right)^{th}\) value = size of \(\left(41+1\over2\right)^{th}\) value = size of 21st value.
If the 41 students were arranged in order (of height), the 21st student would be the middle most one, since there are 20 students on either side of him/her. We therefore need to find the height against the 21st student. 15 students (see cumulative frequency) have height less than or equal to 154 cm. 22 students have height less than or equal to 155 cm. This means that the 21st student has a height 155 cm.
Therefore, Median = 155 cm
9th Standard Syllabus & Materials
9th Standard
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NEW9th Standard
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NEW9th Standard
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Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards
