How will you construct a 15° angle?

First, we make and then its bisector.
The steps of constructions are:
1. Draw a ray OA.
2. Taking O as the center and any radius of your own choice, draw an arc cutting OA at B.
3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
4. Draw the ray OD passing through the C.
Thus,
Now, we draw bisector of
5. Taking C and...

**Q. 6. ** Give one example of a situation in which

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

The mean is not an appropriate measure of central tendency in case the observations are not close to each other. An example of such a case is prices of the toys in a toy shop.

Give one example of a situation in which (i) the mean is an appropriate measure of central tendency.

**Q. 6 ** Give one example of a situation in which

(i) the mean is an appropriate measure of central tendency.

The mean is an appropriate measure of central tendency in case the observations are close to each other. An example of such a case is height of the students in a class.

**Q. 5** Find the mean salary of 60 workers of a factory from the following table:

Salary ( in Rs)(xi)
Number of workers(fi)
fixi
3000
16
48000
4000
12
48000
5000
10
50000
6000
8
48000
7000
6
42000
8000
4
32000
9000
3
27000
10000
1
10000
Total
The mean of the above data is given by
The mean salary of the workers working in the factory is Rs 5083.33

**Q. 4. ** Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

In the given data 14 is occuring the maximum number of times (4)
Mode of the given data is therefore 14.

**Q. 3. ** The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.

29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

The given data is already in ascending order
Number of observations, n = 10 (even)
x + 1 = 63
x = 62

**Q. 2. ** In a mathematics test given to 15 students, the following marks (out of 100) are recorded:

41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60

Find the mean, median and mode of this data.

Number of observations, n = 15
Mean is 54.8
To find the median we have to arrange the given data in ascending order as follows:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
n = 15 (odd)
In the given data 52 occurs the maximum number of times ()
Therefore, Mode = 52

**Q. 1. ** The following number of goals were scored by a team in a series of 10 matches:

2, 3, 4, 5, 0, 1, 3, 3, 4, 3

Find the mean, median and mode of these scores.

Number of observations, n = 10
Mean is 2.8
To find the median we have to arrange the given data in ascending order as follows:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
n = 10 (even)
In the given data 3 occurs the maximum number of times (4)
Therefore, Mode = 3

**Q. 9. ** 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(ii) Write the class interval in which the maximum number of surnames lie.

The class interval in which the maximum number of surnames lie is 6 - 8
The weighted frequency of this class interval (taking 2 as the minimum class size) is 44.

**Q. 9. ** 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.

Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width
Minimum class size = 6 - 4 = 2
The modified table showing the weighted frequency as per the size of the class intervals is as follows.
The histogram representing the information given in the above table is as follows.

**Q. 8. ** A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width
Minimum class size = 2 - 1 = 1
The modified table showing the weighted frequency as per the size of the class intervals is as follows.
The histogram representing the information given in the above table is as follows.

**Q. 6. ** The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections

To make the frequency polygon we first modify the table as follows
To make the frequency polygon we mark the marks on the x-axis and the number of students on the y-axis.
The representation of the given information in the form of frequency polygon is as follows.
From the frequency polygon we can see that the performance of section A is better.

**Q. 5. ** The following table gives the life times of 400 neon lamps:

(ii) How many lamps have a life time of more than 700 hours?

Lamps having life time in the range 700 - 800 = 74
Lamps having life time in the range 800 - 900 = 62
Lamps having life time in the range 900 - 1000 = 48
Lamps having a life time of more than 700 hours = 74 + 62 + 48 = 184.

**Q. 5. ** The following table gives the life times of 400 neon lamps:

(i) Represent the given information with the help of a histogram.

The representation of the given information in the form of a histogram is as follows.

**Q. 4. ** The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

` (iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

No it is certainly not correct to conclude that the maximum number of leaves are 153 mm long because the given data does not tell us about the exact length of the leaves. It only tells us about the range in which their lengths lie. We can only conclude that the maximum number of leaves (12) have their lengths in the region 145 - 153.

**Q. 4. ** The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

(ii) Is there any other suitable graphical representation for the same data?

A frequency polygon could be another suitable graphical representation for the same data.

**Q. 4. **The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]

As we can see from the given table that the data is discontinous and the difference between the upper limit of a class and the lower limit of the next class is 1 and therefore we change both of them by a value 1/2.
e.g 127 - 135 would become 126.5 - 235.5
The modified table therefore is
The representation of the above data through a histogram is as follows

**Q. 3. ** Given below are the seats won by different political parties in the polling outcome of state assembly elections:

(ii) Which political party won the maximum number of seats?

Party A has won the maximum number of seats. Party A has won 75 seats.

**Q. 3. ** Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) Draw a bar graph to represent the polling results.

The representation of the given data in the form of a bar graph is as follows.

**Q. 2. ** The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

(ii) In the classroom discuss what conclusions can be arrived at from the graph

From the graph, we can see that the number of girls per thousand boys is the least in urban society and the highest in the Scheduled Tribes.
910 in case of urban society and 970 in that of Scheduled Tribes.

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