4 Marks Questions

Question 14 Marks

The following distribution gives the state-wise teachers-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures:

Number of students per teacher	Number of states/U.T.
$15 - 20$	$3$
$20 - 25$	$8$
$25 - 30$	$9$
$30 - 35$	$10$
$35 - 40$	$3$
$40 - 45$	$0$
$45 - 50$	$0$
$50 - 55$	$2$

Answer

$WE$ may observe from the given data that maximum class frequency is $10$ belonging to class interval $30 - 35.$
So, modal class $= 30 - 35$
Class size $(h) = 5$
Lower limit $(l)$ of modal class $= 30$
Frequency $(f)$ of modal class $= 10$
Frequency $(f_1)$ of class preceding modal class $= 9$
Frequency $(f_2)$ of class succeeding modal class $= 3$
Mode$ = l + \frac { f - f _ { 1 } } { 2 f - f _ { 1 } - f _ { 2 } } \times h$
$= 30 + \frac { 10-9 } { 2 \times 10 - 9 - 3 } \times h$
$= 30 + \frac { 1 } { 20 - 12 } \times 5$
$= 30 + \frac {5} {8}$
$= 30.625$
Mode $= 30.6$
It represents that most of states/U.T have a teacher-student ratio as $30.6$
Now we may find class marks by using the relation
Class mark $= \frac { \text { upper class limit } + \text {lower class limit} } { 2 }$
Now taking $32.5$ as assumed mean $(a)$ we may calculate $d_i, u_i,$ and $f_iu_i$ as following

Number of students per teacher	Number of states/U.T $(f_i)$	$x_i$	$d_i= x_i- 32.5$	$U_i$	$f_iu_i$
$15 – 20$	$3$	$17.5$	$-15$	$-3$	$-9$
$20 – 25$	$8$	$22.5$	$-10$	$-2$	$-16$
$25 – 30$	$9$	$27.5$	$-5$	$-1$	$-9$
$30 – 35$	$10$	$32.5$	$0$	$0$	$0$
$35 – 40$	$3$	$37.5$	$5$	$1$	$3$
$40 – 45$	$0$	$42.5$	$10$	$2$	$0$
$45 – 50$	$0$	$47.5$	$15$	$3$	$0$
$50 – 55$	$2$	$52.5$	$20$	$4$	$8$
Total	$35$				$-23$

Now, Mean $\overline { x } = a + \frac { \Sigma f _ { i } u _ { i } } { \Sigma f _ { i } } \times h$
$= 32.5 + \frac { - 23 } { 35 } \times$ 5
$= 32.5 - \frac {23} {7} $
$= 32.5 - 3.28$
$= 29.22$
So mean of data is $29.2.$
It represents that on an average teacher-student ratio was $29.2$

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Question 24 Marks

The following data gives the distribution of total monthly household expenditure of $200$ families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in ₹)	Frequency
$1000-1500$	$24$
$1500-2000$	$40$
$2000-2500$	$33$
$2500-3000$	$28$
$3000-3500$	$30$
$3500-4000$	$22$
$4000-4500$	$16$
$4500-5000$	$7$

Answer

We may observe from the given data that maximum class frequency is $40$ belonging to $1500 - 2000$ interval.
Class size $(h) = 500$
Mode $= l + \frac { f - f _ { 1 } } { 2 f - f _ { 1 } - f _ { 2 } } \times h$
Lower limit $(l) $of modal class $= 1500$
Frequency $(f)$ of modal class $= 40$
Frequency $(f_1)$ of class preceding modal class $= 24$
Frequency $(f_2)$ of class succeeding modal class $= 33$
mode = 1500 + $\frac { 40 - 24 } { 2 \times 40 - 24 - 33 } \times 500$
$= 1500 + \frac { 16 } { 80 - 57 } \times 500$
$= 1500 + 347.826$
$= 1847.826 ≈ 1847.83$

Expenditure (in ₹.)	Number of families $f_i$	$x_i$	$d_i= x_i- 2750$	$u_i$	$u_if_i$
$1000-1500$	$24$	$1250$	$-1500$	$-3$	$-72$
$1500-2000$	$40$	$1750$	$-1000$	$-2$	$-80$
$2000-2500$	$33$	$2250$	$-500$	$-1$	$-33$
$2500-3000$	$28$	$2750=a$	$0$	$0$	$0$
$3000-3500$	$30$	$3250$	$500$	$1$	$30$
$3500-4000$	$22$	$3750$	$1000$	$2$	$44$
$4000-4500$	$16$	$4250$	$1500$	$3$	$48$
$4500-5000$	$7$	$4750$	$2000$	$4$	$28$
	$\Sigma f _ { i }= 200$				$\Sigma f _ { i } d _ { i } = - 35$

Mean $\overline { x } = a + \frac { \Sigma f _ { i } d _ { i } } { \Sigma f _ { i } } \times$ h
$\overline { x } = 2750 +\frac { - 35 } { 200 } \times 500$
$\overline { x }= 2750 - 87.5$
$\overline { x } = 2662.5$

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Question 34 Marks

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	$5-15$	$15-25$	$25-35$	$35-45$	$45-55$	$55-65$
Number of patients	$6$	$11$	$21$	$23$	$14$	$5$

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer

Mode:
Here, the maximum frequency is $23$ and the class corresponding to this frequency is $35 - 45.$
So, the modal class is $35 - 45.$
Now, size $(h) = 10$
lower limit it $(l)$ of modal class $= 35$
frequency $(f_1)$ of the modal class $= 23$
frequency $(f_0)$ of class previous the modal class $= 21$
frequency $(f_2)$ of class succeeding the modal class $= 14$
$\therefore$ Mode $= l + \frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h = 35 + \frac{{23 - 21}}{{2 \times 23 - 21 - 14}} \times 10$
$= 35 + \frac{2}{{11}} \times 10 = 35 + \frac{{20}}{{11}}$
$= 35 + 1.8 ($approx.$)$
$= 36.8$ years $($approx.$)$
Mean:-
Take $a = 40, h = 10.$

Age (in years)	Number of patients $(f_i)$	Class marks $(x_i)$	$d_i= x_i- 40$	${u_i} = \frac{{{x_i} - 40}}{{10}}$	$f_iu_i$
$5-15$ $15-25$ $25-35$ $35-45$ $45-55$ $55-65$	$6$ $11$ $21$ $23$ $14$ $5$	$10$ $20$ $30$ $40$ $50$ $60$	$–30$ $–20$ $–10$ $0$ $10$ $20$	$–3$ $–2$ $–1$ $0$ $1$ $2$	$–18$ $–22$ $–21$ $0$ $14$ $10$
Total	$\sum {{f_i}} = 80$				$\sum {{f_i}{u_i}} = - 37$

Using the step deviation method,
$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times$ h = 40 + $ \left( {\frac{{ - 37}}{{80}}} \right) \times 10$
$= 40 - \frac{{37}}{8} = 40 - 4.63$
$= 35.37$ years
Interpretation:- Maximum number of patients admitted in the hospital are of the age $36.8$ years $($approx.$),$ while on an average the age of a patient admitted to the hospital is $35.37$ years.

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Question 44 Marks

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	$50-52$	$53-55$	$56-58$	$59-61$	$62-64$
Number of boxes	$15$	$110$	$135$	$115$	$25$

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose$?$

Answer

Since value of number of mangoes and number of boxes are large numerically. So we use step-deviation method

True Class Interval	No. of boxes$(f_i)$	Class mark$(x_i)$	$u _ { i } = \frac { x _ { i } - a } { h }$	$f_iu_i$
$49.5-52.5$	$15$	$51$	$-2$	$-30$
$52.5-55.5$	$110$	$54$	$-1$	$-110$
$55.5-58.5$	$135$	$57$	$0$	$0$
$58.5-61.5$	$115$	$60$	$1$	$115$
$61.5-64.5$	$25$	$63$	$2$	$50$
	$\sum f _ { i } = 400$			$\sum f_iu_i= 25$

Let assumed mean $(a) = 57,$
$h = 3 ,$
$\therefore \overline { u } = \frac { \sum f _ { i } u _ { i } } { \sum f _ { i } } = \frac { 25 } { 400 } = 0.0625 ($approx$.)$
Using formula, Mean $( \overline { x } ) = a + h \overline { u }$
$= 57 + 3 (0.0625$
$= 57 + 0.1875$
$= 57.1875$
$= 57.19($ approx$)$
Therefore, the mean number of mangoes is $57.19$

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Question 54 Marks

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is $₹\ 18.$ Find the missing frequency $f.$

Daily pocket allowance (in ₹)	$11-13$	$13-15$	$15-17$	$17-19$	$19-21$	$21-23$	$23-25$
Number of children	$7$	$6$	$9$	$13$	$f$	$5$	$4$

Answer

Daily pocket allowance (in ₹)	Number of Children $(f_i)$	Class mark $(x_i)$	$f_ix_i$
$11-13$ $13-15$ $15-17$ $17-19$ $19-21$ $21-23$ $23-25$	$7$ $6$ $9$ $13$ $f$ $5$ $4$	$12$ $14$ $16$ $18$ $20$ $22$ $24$	$84$ $84$ $144$ $234$ $20f$ $110$ $96$
Total	$\sum f_i = f + 44$		$\sum f_iu_i - 20f +752$

Using the direct method,
${\bar x\; = \;\frac{{\mathop {\mathop {\sum {{f_i}} {x_i}}\limits_{} }\limits^{} }}{{\sum\limits_{}^{} {{f_i}} }}}$ ${18 = \frac{{20f + \;752}}{{f + 44}}}$
$⇒ 20f + 752 = 18(f +4))$
$⇒ 20f + 752 = 18f + 792$
$⇒20f - 18f = 792 - 752$
$⇒ 2f = 40$
$⇒ f = \frac{40}2= 20$
Hence, the missing frequency is $20.$

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Question 64 Marks

$100$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	$1-4$	$4-7$	$7-10$	$10-13$	$13-16$	$16-19$
Number of surnames	$6$	$30$	$40$	$16$	$4$	$4$

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames$?$ Also, find the modal size of the surnames.

Answer

First, we will convert the graph given into tabular form as shown below:

Class interval	Frequency $(f_i)$	Mid value $(x_i)$	$f_ix_i$	Cumulative Frequency
$1 – 4$	$6$	$2.5$	$15$	$6$
$4 – 7$	$30$	$5.5$	$165$	$36$
$7 – 10$	$40$	$8.5$	$340$	$76$
$10 – 13$	$16$	$11.5$	$184$	$92$
$13 – 16$	$4$	$14.5$	$58$	$96$
$16 – 19$	$4$	$17.5$	$70$	$100$
	$N = \sum f_i= 100$		$\Sigma f_i x_i = 832$

$N = 100$
Mean $= \frac { \Sigma f_i x_i } { N } = \frac { 832 } { 100 } = 8.32$
$ \frac{N}{2} = \frac{{100}}{2} = 50$
The cumulative frequency just greater than $\frac {N} {2}$ is $76,$ then the median class is $7 - 10$ such that
$l = 7, h = 10 - 7 = 3, f = 40, F = 36$
Median $= l + \frac { \frac { N } { 2 } - F } { f } \times h$
$= 7 + \frac { 50 - 36 } { 40 } \times 3$
$= 7 + \frac {42} {40} = 7 + 1.05 = 8.05$
Mode $=\ 3$ Median $-\ 2$ Mean
$= 3 \times 8.05 - 2 \times 8.32 = 7.51$

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Question 74 Marks

The following table gives the distribution of the life time of 400 neon lamps:

Lite time (in hours)	Number of lamps
1500-2000	14
2000-2500	56
2500-3000	60
3000-3500	86
3500-4000	74
4000-4500	62
4500-5000	48

Find the median life time of a lamp.

Answer

Life time	Number of lamps $(f_i)$	Cumulative frequency
1500-2000	14	14
2000-2500	56	14 + 56 = 70
2500-300	60	70 + 60 = 130
3000-3500	86	130 + 86 = 216
3500-4000	74	216 + 74 = 290
4000-4500	62	290 + 62 = 352
4500-5000	48	352 + 48 = 400
	400

N = 400
Now we may observe that cumulative frequency just greater than $ \frac n2$ (ie., $\frac { 400 } { 2 }$ = 200) is 216
Median class = 3000 - 3500
Median = l + $\left( \frac { \frac { n } { 2 } - c f } { f } \right) \times$ h
Here,
l = Lower limit of median class
F = Cumulative frequency of class prior to median class.
f = Frequency of median class.
h = Class size.
Lower limit (l) of median class = 3000
Frequency (f) of median class 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
Median = 3000 + $\left( \frac { 200 - 130 } { 86 } \right) \times$ 500
= 3000 + $ \frac { 70 \times 500 } { 86 }$
= 3406.98

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Question 84 Marks

The lengths of $40$ leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)	Number of leaves
$118-126$	$3$
$127-135$	$5$
$136-144$	$9$
$145-153$	$12$
$154-162$	$5$
$163-171$	$4$
$172-180$	$2$

Find the median length of the leaves.
$($Hints: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to $117.5 - 126.5, 126.5 -135.5, …, 171.5 -180.5).$

Answer

We shall first convert the given data to continuous classes. Then, the data become

Length (in mm)	Number of leaves	Cumulative frequency
$117.5-126.5$	$3$	$3$
$126.5-135.5$	$5$	$8$
$135.5-144.5$	$9$	$17$
$144.5-153.5$	$12$	$29$
$153.5-162.5$	$5$	$34$
$162.5-171.5$	$4$	$38$
$171.5-180.5$	$2$	$40$

Now, $n = 40$
So, $\frac{n}{2} = \frac{{40}}{2} = 20$
This observation lies in the class $144.5 - 153.5.$
So, $144.5 - 153.5$ is the median class.
Therefore,
$l = 144.5$
$h = 9$
$cf = 17$
$f = 12$
$\therefore $ Median $= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h = 144.5 + \left( {\frac{{20 - 17}}{{12}}} \right) \times 9$
$= 144.5 + 2.25 = 146.75\ mm$
Hence, the median length of the leaves is $146.75\ mm$

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Question 94 Marks

A life insurance agent found the following data for distribution of ages of $100$ policy holders. Calculate the median age, if policies are only given to persons having age $18$ years onwards but less than $60 $ year.

Age (in years)	Number of policyholders
Below $20$	$2$
Below $25$	$6$
Below $30$	$24$
Below $35$	$45$
Below $40$	$78$
Below $45$	$89$
Below $50$	$92$
Below $55$	$98$
Below $60$	$100$

Answer

To calculate the median age, we need to find the class intervals and their corresponding frequencies.
It is shown below:

Class interval	Frequency	Cumulative Frequency
Below $20$	$2$	$2$
$20-25$	$4$	$6$
$25-30$	$18$	$24$
$30-35$	$21$	$45$
$35-40$	$33$	$78$
$40-45$	$11$	$89$
$45-50$	$3$	$92$
$50-55$	$6$	$98$
$55-60$	$2$	$100$

Now, $n = 100$
So, $\frac{n}{2} = \frac{{100}}{2}= 50$
This observation lies in class $35 - 40.$
So, $35 - 40$ is the median class.
Therefore,
$l = 35$
$h = 5$
$cf = 45$
$f = 33$
$\therefore $ Median $= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h = 35 + \left( {\frac{{50 - 45}}{{33}}} \right) \times 5$
$= 35 + \frac{{25}}{{33}} = 35 + 0.76 = 35.76$ years
Hence, the median age is $35.76$ years.

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Question 104 Marks

If the median of the distribution given below is $28.5,$ then find the values of $x$ and $y.$

Class Interval	frequency
$0-10$	$5$
$10-20$	$x$
$20-30$	$20$
$30-40$	$15$
$40-50$	$y$
$50-60$	$5$
Total	$60$

Answer

Monthly Consumption	Number of consumers $\left( f _ { i } \right)$	Cumulative Frequency
$0-10$	$5$	$5$
$10-20$	$x$	$5 + x$
$20-30$	$20$	$25 + x$
$30-40$	$15$	$40 + x$
$40-50$	$y$	$40 + x + y$
$50-60$	$5$	$45 + x + y$
Total	$\sum f _ { i } = n = 60$

Here, $\sum f _ { i } = n = 60$, then $\frac { n } { 2 } = \frac { 60 } { 2 } = 30$, also, median of the distribution is $28.5,$ which lies in interval $20 – 30.$
$\therefore$ Median class $= 20 – 30$
So, $l = 20, n = 60, f = 20, cf = 5 + x$ and $h = 10$
$\because 45 + x + y = 60$
$\Rightarrow x + y = 15 ………...........(i)$
Now, Median $= l + \left[ \frac { \frac { n } { 2 } - c f } { f } \right] \times h$
$\Rightarrow { 28.5 = 20 + \left[ \frac { 30 - ( 5 + x ) } { 20 } \right] \times 10 }$
$\Rightarrow 28.5 = 20 + \frac { 30 - 5 - x } { 2 }$
$\Rightarrow { 28.5 } = \frac { 40 + 25 - x } { 2 }$
$\Rightarrow 57.0 = 65 - x$
$\Rightarrow x = 65 - 57 = 8$
$\Rightarrow x = 8$
Putting the value of $x$ in eq.$ (i)$, we get,
$8 + y = 15$
$\Rightarrow y = 7$
Hence the value of $x$ and $y$ are $8$ and $7$ respectively.

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Question 114 Marks

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	Number of Consumers
$65-85$	$4$
$85-105$	$5$
$105-125$	$13$
$125-145$	$20$
$145-165$	$14$
$165-185$	$8$
$185-205$	$4$

Answer

First, we will convert the graph into tabular form given below:

Monthly consumption (in units)	Number of consumers $(f_i)$	Class mark $(x_i)$	$d_i= x_i- 135$	${u_i} = \frac{{{x_i} - 135}}{5}$	$f_iu_i$	Cumulative Frequency
$65-85$ $85-105$ $105-125$ $125-145$ $145-165$ $165-185$ $185-205$	$4$ $5$ $13$ $20$ $14$ $8$ $4$	$75$ $95$ $115$ $135$ $155$ $175$ $195$	$–60$ $–40$ $–20$ $0$ $20$ $40$ $60$	$–3$ $–2$ $–1$ $0$ $1$ $2$ $3$	$–12$ $–10$ $–13$ $0$ $14$ $16$ $12$	$4$ $9$ $22$ $42$ $56$ $64$ $68$
Total	$\sum {{f_i}} = 68$				$\sum {{f_i}{u_i}} = 7$

Let $a = 135$.
Now, $h = 20$
Using the step-deviation method,
Mean, $\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h = 135 + \left( {\frac{7}{{68}}} \right) \times 20$
$ = 135 + \frac{{35}}{{17}} = 135 + 2.05 = 137.05$
Now, $N = 68$
So, $\frac{N}{2} = \frac{{68}}{2} = 34$
This observation lies in class $125-145.$
Therefore, $125-145$ is the median class.
So, $l = 125, CF = 22, f = 20$
$\therefore Median = l + \left( {\frac{{\frac{N}{2} - CF}}{f}} \right) \times h$
$ = 125 + \left( {\frac{{34 - 22}}{{20}}} \right) \times 20 = 125 + 12 = 137$
Mode $=\ 3$ Median $-\ 2$ Mean
$= 3 \times 137 - 2 \times137.05 = 136.9$

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Question 124 Marks

The median of the following data is $525.$ Find the values of $x$ and $y,$ if the total frequency is $100.$

Class interval	Frequency
$0-100$	$2$
$100-200$	$5$
$200-300$	$x$
$300-400$	$12$
$400-500$	$17$
$500-600$	$20$
$600-700$	$y$
$700-800$	$9$
$800-900$	$7$
$900-1000$	$4$

Answer

Class intervals	Frequency (f)	Cumulative frequency (cf/F)
$0-100$	$2$	$2$
$100-200$	$5$	$7$
$200-300$	$x$	$7 + x$
$300-400$	$12$	$19 + x$
$400-500$	$17$	$36 + x$
$500-600$	$20$	$56 + x$
$600-700$	$y$	$56 + x + y$
$700-800$	$9$	$65 + x + y$
$800-900$	$7$	$72 + x + y$
$900-1000$	$4$	$76 + x + y$
		Total $= 76 + x + y$

We have,
$N = \Sigma f _i = 100$
$\Rightarrow 76 + x + y = 100 $
$\Rightarrow x + y = 24$
It is given that the median is $525$. Clearly, it lies in the class $500 - 600$
$\therefore l = 500, h = 100, f = 20, F = 36 + x$ and $N = 100$
Now, Median $= l + \frac { \frac { N } { 2 } - F } { f } \times h$
$\Rightarrow 525 = 500 + \frac { 50 - ( 36 + x ) } { 20 } \times 100$
$\Rightarrow 525 - 500 = (14 - x)5$
$\Rightarrow 25 = 70 - 5x$
$\Rightarrow 5x = 45$
$\Rightarrow x = 9$
Putting $x = 9$ in $x + y = 24,$ we get $y = 15$
Hence, $x = 9$ and $y = 15$

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Question 134 Marks

The mode of frequency distribution given below is 34.5 and number of observations is 165. Find the missing frequencies a and b.

Class	5-14	14-23	23-32	32-41	41-50	50-59
Frequency	5	11	a	53	b	26

Answer

Mode $=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h$
l=32,$f_1$=53,$f_0$=a,$f_2$=b
Class size (h): The class intervals are 5-14, 14-23, 23-32, etc. The width of each class is the difference between the upper and lower limits, which is 14−5=9 or 23−14=9. So, h=9.
Now, substitute these values and the given mode into the formula:
$34.5=32+\left(\frac{53-a}{2(53)-a-b}\right) \times 9$
We can simplify the denominator by using Equation 1 $(a+b=70):$
$2(53)-a-b=106-(a+b)=106-70=36$
Substitute this simplified denominator back into the equation:
$34.5=32+\left(\frac{53-a}{36}\right) \times 9$
Subtract 32 from both sides:
$\begin{array}{l}34.5-32=\left(\frac{53-a}{36}\right) \times 9 \\ 2.5=\left(\frac{53-a}{36}\right) \times 9\end{array}$
Divide both sides by 9:
$\begin{array}{l}\frac{2.5}{9}=\frac{53-a}{36} \\ 2.5 \times 36=9 \times(53-a) \\ 90=9 \times(53-a) \\ 10=53-a \\ a=53-10 \\ a=43\end{array}$
Now that we have the value of a, we can use Equation 1 to find b.
$\begin{array}{l}a+b=70 \\ 43+b=70 \\ b=70-43 \\ b=27\end{array}$
The missing frequencies are a = 43 and b = 27.

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