2 Marks Questions

Question 12 Marks

A student noted the number of cars passing through a spot on a road for $100$ periods each of $3$ minutes and summarized it in the table given below. Find the mode of the data:

Number of cars	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$7$	$14$	$13$	$12$	$20$	$11$	$15$	$8$

Answer

Here, the maximum class frequency is $20$, and the class corresponding to this frequency is $40-50$.
So, the modal class is $40-50$.
Therefore $h = 10, l = 40, f_1= 20, f_0= 12 , f_2= 11$
Mode $= l + \left[ {\frac{{{f_1}\; - {f_0}}}{{2{f_1} - \;{f_0} - {f_2}}}} \right] \times$ $h = 40$ + $\left[ {\frac{{20 - 12}}{{2(20) - \;12 - 11}}} \right] \times$ $10$
$= 40 + \frac{{80}}{{17}}= 40 + 4.7 = 44.7$
Hence the mode of the data is $44.7 $ cars.

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Question 22 Marks

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored	Number of batsmen
$3000-4000$	$4$
$4000-5000$	$18$
$5000-6000$	$9$
$6000-7000$	$7$
$7000-8000$	$6$
$8000-9000$	$3$
$9000-10000$	$1$
$10000-11000$	$1$

Find the mode of the data.

Answer

Since the maximum number of batsman have their runs scored in the interval $4000-5000,$ the modal class is $4000-5000.$
Therefore, $\mathrm{I}=4000, \mathrm{~h}=1000, \mathrm{f}_1=18, \mathrm{f}_0=4, \mathrm{f}_2=$
Mode $= l + \left[ {\frac{{f1\; - f0}}{{2f1 - \;f0 - f2}}} \right] \times $ $h = 4000 + \left[ {\frac{{18 - 4}}{{2(18) - \;4 - 9}}} \right] \times 1000$
$= 4000 + \frac{{14000}}{{23}} = 4000 + 608.7 = 4608.7$
Hence, the mode of the data is $4608.7$

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Question 32 Marks

The following data gives the information on the observed lifetimes (in hours) of $225$ electrical components:

Lifetimes (in hours)	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
Frequency	$10$	$35$	$52$	$61$	$38$	$29$

Determine the modal lifetimes of the components.

Answer

Here, the maximum class frequency is $61$, and the class corresponding to this frequency is $60-80$. So, the modal class is $60-80$.
Therefore$h=20, I=60, f_1=61, f_0=52, f_2=38$
${Mode = \;l\; + \left[ {\frac{{{f_1}\; - {f_0}}}{{2{f_1} - \;{f_0} - {f_2}}}} \right] \times h = \;60 + \left[ {\frac{{61 - 52}}{{2(61) - \;52 - 38}}} \right] \times 20 = }$ $\begin{array}{*{20}{l}} {\;60 + \left[ {\frac{9}{{122 - 90}}} \right] \times 20 = 60 + \;\frac{{180}}{{32}} = 60\; + \;5.625 = 65.625} \end{array}$
Therefore, the modal lifetime of the components is $65.625$ hours.

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Question 42 Marks

The marks distribution of $30$ students in a mathematics examination. Find the mode of this data. Also, compare and interpret the mode and the mean.

Class interval	Number of students $(f_i)$	Classmark $( x_i)$	$f_ix_i$
$10 - 25$	$2$	$17.5$	$35.0$
$25 - 40$	$3$	$32.5$	$97.5$
$40 - 55$	$7$	$47.5$	$332.5$
$55 - 70$	$6$	$62.5$	$375.0$
$70 - 85$	$6$	$77.5$	$465.0$
$85 - 100$	$6$	$92.5$	$555.0$
Total	$\sum f_{i}$ $= 30$		$\sum f_{i}x_i$ $= 1860.0$

Answer

Since the maximum number of students $($i.e., $7)$ have got marks in the interval $40 - 55$, the modal class is $40 - 55$
Therefore, the lower limit $(l)$ of the modal class $= 40$
the class size $(h) = 15$
the frequency $\left(f_1\right)$ of modal class $= 7$
the frequency $(f_0)$ of the class preceding the modal class $= 3$
the frequency $\left(f_2\right)$ of the class succeeding in the modal class $= 6$
Now, using the formula:
Mode $=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h$
we get Mode = $40+\left(\frac{7-3}{14-6-3}\right) \times 15=52$
So, the mode mark is $52$
Now $\bar{x}=\frac{\sum f_{i} x_{i}}{\sum f_{i}}$
$\bar{x}=\frac{1860.0}{30}=62$
So, the maximum number of students obtained $52$ marks, while on average a student obtained $62$ marks.

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Question 52 Marks

A survey conducted on $20$ households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

Family size	$1-3$	$3-5$	$5-7$	$7-9$	$9-11$
Number of families	$7$	$8$	$2$	$2$	$1$

Find the mode of this data.

Answer

The frequency distribution table is given as:

Family size$(x_i)$	$1-3$	$3-5$	$5-7$	$7-9$	$9-11$
No. of families$(f_i)$	$f_0=7$	$f_1=8$	$f_2=2$	$2$	$1$

From the given frequency table, the maximum class frequency is $8$, and the class corresponding to this frequency is $3 – 5$. So, the modal class is $3 - 5.$
Now modal class $= 3 - 5$, lower limit $(l)$ of modal class $= 3,$ class size $(h) = 2$
frequency $(f_1)$ of the modal class $= 8,$
frequency $(f_0)$ of class preceding the modal class $= 7$
frequency $(f_2)$ of class succeeding the modal class $= 2$
Now, let us substitute these values in the formula :
$\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h$
$=3+\left(\frac{8-7}{2 \times 8-7-2}\right) \times 2=3+\frac{2}{7}=3.286$
Therefore, the mode of the data above is $3.286$

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Question 62 Marks

The wickets taken by a bowler in $10$ cricket matches are as follows :
$2, 6, 4, 5, 0, 2,1, 3, 2, 3$
Find the mode of the above data.

Answer

At first, we arrange the given data in ascending order :- $0, 1, 2, 2, 2, 3, 3, 4, 5, 6$
Now find the most occurring number i.e. $2$
Hence, mode $= 2$

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Question 72 Marks

The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Number of Wickets	$20-60$	$60-100$	$100-150$	$150-250$	$250-350$	$350-450$
Number of Bowlers	$7$	$5$	$16$	$12$	$2$	$3$

Answer

Here, the class size varies, and hence $x_i$, so $f_ix_i$ will be large. Let us apply the step deviation method with $a = 200$ and $h = 20$. Then, we obtain the data as in Table

Number of tickets taken	Number of bowlers $ (f_i)$	$x_i$	$d_i= x_i– 200$	$u_{i}=\frac{d_{i}}{20}$	$u_i f_i$
$20-60$	$7$	$40$	$-160$	$-8$	$-56$
$60-100$	$5$	$80$	$-120$	$-6$	$-30$
$100-150$	$16$	$125$	$-75$	$-3.75$	$-60$
$150-250$	$12$	$200$	$0$	$0$	$0$
$250-350$	$2$	$300$	$100$	$5$	$10$
$350-450$	$3$	$400$	$200$	$10$	$30$
Total	$45$				$-106$

Therefore, $\bar{x}=200+20\left(\frac{-106}{45}\right)=200-47.11=152.89$
This tells us that, on average, the number of wickets taken by these 45 bowlers in one-day cricket is $152.89$.

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Question 82 Marks

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories $(U.T.)$ of India. Find the mean percentage of female teachers by all the three methods discussed in this section.

Percentage of female teachers	$15 - 25$	$25 - 35$	$35 - 45$	$45 - 55$	$55 - 65$	$65 - 75$	$75 - 85$
Number of states/ $U.T.$	$6$	$11$	$7$	$4$	$4$	$2$	$1$

Answer

Let, $a = 50$

C.I.	Number of states/ U.T. $(f_i)$	$x_i$	$d_i= x_i- 50$	$f_id_i$
$15 - 25$	$6$	$20$	$-30$	$-180$
$25 - 35$	$11$	$30$	$-20$	$-220$
$35 - 45$	$7$	$40$	$-10$	$-70$
$45 - 55$	$4$	$50$	$0$	$0$
$55 - 65$	$4$	$60$	$10$	$40$
$65 - 75$	$2$	$70$	$20$	$40$
$75 - 85$	$1$	$80$	$30$	$30$

From table, $\Sigma f_id_i= -360 , \Sigma f_i=36$
we know that, mean=$\overline { x } = a + \frac { \Sigma f _ { i } d _ { i } } { \Sigma f _ { i } } $
$= 50 + \frac { - 360 } { 35 } $
$= 39.71$

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Question 92 Marks

The marks obtained by $30$ students of Class $X$ of a certain school in a mathematics paper consisting of $100$ marks are presented in the table given below. Find the mean of the marks obtained by the students in mathematics paper.

Marks obtained $(x_i)$	$10$	$20$	$36$	$40$	$50$	$56$	$60$	$70$	$72$	$80$	$88$	$92$	$95$
Number of Students $(f_i)$	$1$	$1$	$3$	$4$	$3$	$2$	$4$	$4$	$1$	$1$	$2$	$3$	$1$

Answer

Recall that to find the mean marks, we require the product of each $x_i$ with the corresponding frequency $f_i$. So, let us put them in a column as shown in Table given below:

Marks obtained $(x_i)$	Number of students $(f_i)$	$f_ix_i$
$10$	$1$	$10$
$20$	$1$	$20$
$36$	$3$	$108$
$40$	$4$	$160$
$50$	$3$	$150$
$56$	$2$	$112$
$60$	$4$	$240$
$70$	$4$	$280$
$72$	$1$	$72$
$80$	$1$	$80$
$88$	$2$	$176$
$92$	$3$	$276$
$95$	$1$	$95$
Total	$Σf_i= 30$	$Σf_ix_i= 1779$

Now, $\bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{1779}{30}=59.3$
Therefore, the mean marks obtained by students in mathematics paper is $59.3$.

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Question 102 Marks

Find median if n = 53 l = 60 f = 7 cf = 22 and h = 10.

Answer

Median $=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h$
$\frac{n}{2}$
$\frac{53}{2}=26.5$
Median $=60+\left(\frac{26.5-22}{7}\right) \times 10$
Simplify the expression:
Median=$60+\left(\frac{4.5}{7}\right) \times 10$"
Median $\approx 60+0.6429 \times 10$
Median $\approx 60+6.429$
Median $\approx 66.43$
The median is approximately 66.43.

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Question 112 Marks

If the mean of observations x, x + 3 x + 6 x + 9 and x + 12 is 10 then find x.

Answer

Mean $=\frac{\text { Sum of Observations }}{\text { Number of Observations }}$
Sum $=(x)+(x+3)+(x+6)+(x+9)+(x+12)$
Sum $=5 x+30$
The number of observations is 5, and the mean is 10.
$10=\frac{5 x+30}{5}$
Multiply both sides by 5:
$50=5 x+30$
Subtract 30 from both sides:
$20=5 x$
Divide by 5:
$x=\frac{20}{5}$
$x=4$
The value of $x$ is 4.

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