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Maths 2 Marks Questions

Statistics · Jharkhand Board (JAC) STD 9 (Jharkhand - 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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