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Maths 4 Marks Questions

Statistics · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 15 questions.

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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  
i. $N=100$
Mean $=\frac{\Sigma f_i x_i}{N}=\frac{832}{100}=8.32$
ii. $\frac{N}{2}=\frac{100}{2}=50$
The cumulative frequency just greater than $\frac{N}{2}$ is 76 , then the median class is $7-10$ such that
$I=7, h=10-7=3, f=40, F=36$
$\text { Median }=I+\frac{\frac{N}{2}-F}{f} \times h$
$=7+\frac{50-36}{40} \times 3$
$=7+\frac{42}{40}=7+1.05=8.05$
iii. 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 $=1+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h =35+\left(\frac{50-45}{33}\right) \times 5$
$= 35 + \frac{{25}}{{33}} = 35 + 0.76 = 35.76$ 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 $\sum {{f_i}} = 68$       $\sum {{f_i}{u_i}} = 7$  
  1. 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$
  2. 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$
  3. Mode = $3$ Median $- 2$ Mean$=3 \times 137-2 \times 137.05=136.9$
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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 following table gives the literacy rate $($in percentage$)$ of $35$ cities. Find the mean literacy rate.
Literacy rate $($in $\%)$ $45-55$ $55-65$ $65-75$ $75-85$ $85-95$
Number of cities $3$ $10$ $11$ $8$ $3$
Answer
Take $a = 70, h = 10$
Literacy rate $($in $\%)$ Number of cities $(f_i)$ Class mark $(x_i)$ $d_i = x_i –70$   ${u_i} = \frac{{{x_i} - 70}}{{10}}$ $f_iu_i$
$45-55$
$55-65$
$65-75$
$75-85$
$85-95$		 $3$
$10$
$11$
$8$
$3$		 $50$
$60$
$70$
$80$
$90$		 –$20$
$–10$
$0$
$10$
$20$		 $–2$
$–1$
$0$
$1$
$2$		 $–6$
$–10$
$0$
$8$
$6$
Total $\sum f_i= 35$       $\sum f_iu_i = -2$
sing the step$-$deviation method,
$\overline x   = a +  \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times  h = 70 +  \left( {\frac{{ - 2}}{{35}}} \right) \times  10$
$= 70 - \frac{4}{7} = 70 - 0.57 = 69.43\%$
Hence, the mean literacy rate is $69.43\%$
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Question 144 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 teachers15 - 2525 - 3535 - 4545 - 5555 - 6565 - 7575 - 85
Number of states/U.T.61174421
### 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 .
Percentage of female teachers15 - 2525 - 3535 - 4545 - 5555 - 6565 - 7575 - 85
Number of states/U.T.61174421
Answer
Let, $a = 50$
C.INumber of states/ U.T. (fi)xidi = xi - 50fidi
15 - 25620-30-180
25 - 351130-20-220
35 - 45740-10-70
45 - 5545000
55 - 654601040
65 - 752702040
75 - 851803030
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 154 Marks
Heart Rate: Heart rate is one of the body's "Vital Signs" of health. It measures the number of times the heart beats or contracts per minute. While a normal heart rate does not guarantee that a person is free from health problems, it is useful benchmark for identifying many health problems. 
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30 women were examined by AIIMS doctors and the number of heart beats per minute was recorded and the Summary was given as follows:
Number of Heart Beats per minuteNumber of Women
65-682
68-714
71-743
74-778
77-807
80-834
83-862
Answer the following from the above information: 
(i) How many women have heart beat in range of 68-77.
(ii) What is the median class of heart beats per minute for these women? 
(iii) Find the mode for the heart beat per minute for these women. 
Answer
(i) Number of women with a heart rate in the range of 68-77.
Number of women in the 68-71 range = 4
Number of women in the 71-74 range = 3
Number of women in the 74-77 range = 8
Total number of women = 4+3+8=15
There are 15 women with a heart rate in the range of 68-77.
(ii) The median class of heart beats per minute.
First, we find the cumulative frequencies (cf) to identify the median class. The total number of observations, $n=30$.
The median class is the one containing the $\frac{n}{2}$ or $\frac{30}{2}=15$ th observation.
Number of Heart Beats per minuteNumber of Women (f)Cumulative Frequency (cf)
65-6822
68-7146
71-7439
74-77817
77-80724
80-83428
83-86230
Therefore, the median class is 74-77.
(iii) The mode for the heart beat per minute.
The mode is the value with the highest frequency. The class with the highest frequency is the modal class.
From the table, the highest frequency is 8, which corresponds to the class 74-77.
Using the mode formula:
Mode $=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h$
l = lower limit of the modal class = 74
$f_1=$ frequency of the modal class $=8$
$\begin{array}{l}f_0=\text { frequency of the preceding class }=3 \\ f_2=\text { frequency of the succeeding class }=7 \\ h=\text { class size }=77-74=3\end{array}$
Substitute these values into the formula:
Mode $=74+\left(\frac{8-3}{2(8)-3-7}\right) \times 3$
=$74+\left(\frac{5}{16-10}\right) \times 3$
$\begin{array}{l}=74+\left(\frac{5}{6}\right) \times 3 \\ =74+\frac{15}{6} \\ =74+2.5\end{array}$
Mode = 76.5
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