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MATHS 2 Marks Questions

Statistics · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 6 questions.

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Question 12 Marks
Let us now consider the following frequency distribution table which gives the weights of 38 students of a class:

Weights (in kg) Number of students
31-35 9
36-40 5
41-45 14
46-50 3
51-55 1
56-60 2
61-65 2
66-70 1
71-75 1
Total 38

Now, if two new students of weights 35.5 kg and 40.5 kg are admitted in this class, then in which interval will we include them. Create a frequency distribution table for this class interval.

Answer
For this, we find the difference between the upper limit of a class and the lower limit of its succeeding class. For example, consider the classes 31 - 35 and 36 - 40. The lower limit of 36 - 40 = 36 The upper limit of 31 - 35 = 35 The difference = 36 – 35 = 1 So, half the difference = $\begin{equation} \frac{1}{2}=0.5 \end{equation}$
So the new class interval formed from 31 - 35 is (31 – 0.5) - (35 + 0.5), i.e., 30.5 - 35.5. Similarly, the new class formed from the class 36 - 40 is (36 – 0.5) - (40 + 0.5), i.e., 35.5 - 40.5. Continuing in the same manner, the continuous classes formed are: 30.5-35.5, 35.5-40.5, 40.5-45.5, 45.5-50.5, 50.5-55.5, 55.5-60.5, 60.5 - 65.5, 65.5 - 70.5, 70.5 - 75.5.
Now, with these assumptions, the new frequency distribution table will be as shown below:
Weights (in kg) Number of students
30.5-35.5 9
35.5-40.5 6
40.5-45.5 15
45.5-50.5 3
50.5-55.5 1
55.5-60.5 2
60.5-65.5 2
65.5-70.5 1
70.5-78.5 1
Total 40
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Question 22 Marks
100 plants each were planted in 100 schools during Van Mahotsava. After one month, the number of plants that survived were recorded as : 95, 67, 28, 32, 65, 65, 69, 33, 98, 96, 76, 42, 32, 38, 42, 40, 40, 69, 95, 92, 75, 83, 76, 83, 85, 62, 37, 65, 63, 42, 89, 65, 73, 81, 49, 52, 64, 76, 83, 92, 93, 68, 52, 79, 81, 83, 59, 82, 75, 82, 86, 90, 44, 62, 31, 36, 38, 42, 39, 83, 87, 56, 58, 23, 35, 76, 83, 85, 30, 68, 69, 83, 86, 43, 45, 39, 83, 75, 66, 83, 92, 75, 89, 66, 91, 27, 88, 89, 93, 42, 53, 69, 90, 55, 66, 49, 52, 83, 34, 36
Create a frequency distribution table with tally number.
Answer
The frequency table is given as
Number of plants survived Tally Marks Number of schools (frequency)
20-29 ||| 3
30-39 |||| |||| |||| 14
40-49 |||| |||| ||| 12
50-59 |||| ||| 8
60-69 |||| |||| |||| ||| 18
70-79 |||| |||| 10
80-89 |||| |||| |||| |||| ||| 23
90-99 |||| |||| || 12
Total   100
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Question 32 Marks
Consider the marks obtained (out of 100 marks) by 30 students of Class IX of a school and create a frequency distribution table: 10, 20, 36, 92, 95, 40, 50, 56, 60, 70, 92, 88, 80, 70, 72, 70, 36, 40, 36, 40, 92, 40, 50, 50, 56, 60, 70, 60, 60, 88
Answer
Marks Number of students (i.e., the frequency)
10 1
20 1
36 3
40 4
50 3
56 2
60 4
70 4
72 1
80 1
88 2
92 3
95 1
Total 30

the above table is the required frequency distribution table.

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Question 42 Marks
The points scored by a Kabaddi team in a series of matches are as follows:
17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28
Find the median of the points scored by the team.
Answer
First of all we Arranging the points scored by the team in ascending order, we get 2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48.
There are 16 terms. So the median is given by the average of the $\frac{16}{2}$ th and $\left(\frac{16}{2}+1\right)th$, i.e., the 8th and 9th terms.
So, the median is the mean of the values of the 8th and 9th terms.
i.e, the median = $\begin{equation} \frac{10+14}{2}=12 \end{equation}$
So, the medial point scored by the Kabaddi team is 12.
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Question 52 Marks
The heights (in cm) of 9 students of a class are as follows:
155, 160, 145, 149, 150, 147, 152, 144, 148
Find the median of this data.
Answer
In order to calculate median .First of all we arrange the data in ascending order, as follows:
144, 145, 147, 148, 149, 150, 152, 155, 160
Since the number of students is 9, an odd number, we find out the median by finding the height of the $\begin{equation} \left(\frac{n+1}{2}\right) \mathrm{th}=\left(\frac{9+1}{2}\right) \mathrm{th} \end{equation}$ = the 5th student, which is 149 cm.
So, the median, i.e., the medial height is 149 cm.
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Question 62 Marks
5 people were asked about the time in a week they spend in doing social work in their community. They said 10, 7, 13, 20 and 15 hours, respectively. Find the mean (or average) time in a week devoted by them for social work.
Answer
The mean of the above data is goven by.$\bar{x}=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}$
$\begin{equation} =\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5} \end{equation}$
$\begin{equation} =\frac{10+7+13+20+15}{5}=\frac{65}{5}=13 \end{equation}$
So, the mean time spent by these 5 people in doing social work is 13 hours in a week.
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