NCERT Class 10 Maths Statistics
The chapter 14 begins with the concepts related to the measures of the central tendency i.e.: mean, median and mode for grouped data. The formula to determine the class mark, direct method to find the mean, assumed mean method and step-deviation method is discussed under the concept of ‘mean for grouped data’. Next, the formula to find the mode for grouped data is presented. Then the median of grouped data is introduced and cumulative frequency distribution of the less than type and more than type is discussed. The last section deals with the graphical representation of cumulative frequency distribution and the resulting curve is called the cumulative frequency curve, or an ogive.
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Exercise 14.1
Chapter 14 Ex.14.1 Question 1
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in \(20\) houses in a locality. Find the mean number of plants per house.
Number of plants | \(0 - 2\) | \(2 - 4\) | \(4 - 6\) | \(6 - 8\) | \(8 - 10\) | \(10 -12\) | \(12 - 14\) |
Number of houses | \(1\) | \(2\) | \(1\) | \(5\) | \(6\) | \(2\) | \(3\) |
Which method did you use for finding the mean, and why?
Solution
What is known?
The number of plants in \(20\) houses in a locality.
What is unknown?
The mean number of plants per house and the method used for finding the mean.
Reasoning:
We can solve this question by any method of finding mean but here we will use direct method to solve this question because the data given is small.
The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations.
We know that if \(\ x_1, x_2, \ldots, x_n \) are observations with respective frequencies \(f_{1}, f_{2}, \ldots, f_{n}\) then this means observation \(\ x_1 \text { occurs } f_1 \text { times, } x_2 \text { occurs } f_2\) times, and so on.
\(\ x\) is the class mark for each interval, you can find the value of \(\ x\) by using
class mark
\[\begin{align}\left(x_i\right) =\frac{\text { upper limit + lower limit }}{2}\end{align}\]
Now, the sum of the values of all the observations \(=\ f_1 \ x_1+\ f_2 x_2+\ldots+\ f_n \ x_n\) and the number of observations \(=f_1+f_2+\ldots+f_n\).
So, the mean of the data is given by
\[\begin{align}\ \overline x\!=\!\frac{f_{1} x_{1}\!+\!f_{2} x_{2}\!+\!\ \ldots \ldots\!+\!f_{n} x_{n}}{f_{1}\!+\!f_{2}\!+\!\ldots \ldots\!+\!f_{n}}\end{align}\]
\(\begin{align}\overline x=\frac{\sum f_{i} x_{i}}{\Sigma f_{i}}\end{align}\) where \(i\) varies from \(1\) to \(n\)
Steps:
Number of plants | Number of houses\(\ (f_i)\) | \(\ x_i\) | \(\ f_ix_i\) | |
---|---|---|---|---|
\(0 - 2\) | \(1\) | \(1\) | \(1\) | |
\(2 - 4\) | \(2\) | \(3\) | \(6\) | |
\(4 - 6\) | \(1\) | \(5\) | \(5\) | |
\(6 - 8\) | \(5\) | \(7\) | \(35\) | |
\(8 - 10\) | \(6\) | \(9\) | \(54\) | |
\(10 - 12\) | \(2\) | \(11\) | \(22\) | |
\(12 - 14\) | \(3\) | \(13\) | \(39\) | |
\(\Sigma f_i = 20\) | \(\Sigma f_ix_i = 162\) |
From the table it can be observed that,
\[\begin{align}\Sigma f_i=20, \\ \Sigma f_ix_i =162\end{align}\]
\[\begin{align} \text { Mean } \overline x &=\frac{\Sigma f_i x_i}{\Sigma f_i} \\ &=\frac{162}{20} \\ &=8.1 \end{align}\]
Thus, the mean number of plants each house has \(8.1.\)
Here, we have used the direct method because the value of \(x_i\) and \(f_i\) are small.
Chapter 14 Ex.14.1 Question 2
Consider the following distribution of daily wages of \(50\) workers of a factory.
Daily wages (in Rs) | \(500 –520\) | \(520 – 540\) | \(540 –560\) | \(560 – 580 \) | \(580 – 600 \) |
Number of workers | \(12\) | \(14\) | \(8\) | \(6\) | \(10\) |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution
What is known?
Distribution of daily wages of \(50\) workers of a factory is given-
What is unknown?
The mean daily wages of the workers of the factory.
Reasoning:
We will use Assumed Mean Method to solve this questionbecause the data given is large.
Sometimes when the numerical values of \(x_i\) and \(f_i\) are large, finding the product of \(x_i\) and \(f_i\) becomes tedious.We can do nothing with the \(f_i’s\), but we can change each xi to a smaller number so that our calculations become easy.Now we have to subtract a fixed number from each of these \(x_i’s\).
The first step is to choose one among the \(x_i’s\) as the assumed mean, and denote it by \(‘a’.\) Also, to further reduce our calculation work, we may take ‘a’ to be that \(x_i\) which lies in the centre of \(\begin{align}{x} _1, {x} _ 2, \ldots, {x_n}\end{align}\). So, we can choose \(a.\)
The next step is to find the difference \(d_i\) between a and each of the \(x_i\)’s, that is, the deviation of ‘a’ from each of the \(x_i\)'s i.e., \(d_i = x_i – a\)
The third step is to find the product of \(d_i\) with the corresponding\(f_i\), and take the sum of all the \({f}_{{i}} {d}_{{i}}^{\prime} {s}\)
Now put the values in the below formula
\[\begin{align}\operatorname{Mean} \,\,(\overline{{x}})= {a}+\left(\frac{\Sigma {f}_{{i}} {u}_{{i}}}{\Sigma {f}_{{i}}}\right) \end{align}\]
Steps:
We know that,
class mark (\(x_i\)) \(\begin{align}={\frac{\text { upper limit+lower limit }}{2}} \end{align}\)
Taking assumed mean, a = 550
Daily wages (in Rs) |
No of workers \((f_i)\) |
\((X_i)\) | \( d_i = x_i -150\) | \( f_iu_i\) |
\(500-520\) | \(12\) | \(510\) | \(- 40\) | \(- 480\) |
\(520-540\) | \(14\) | \(530\) | \(- 20\) | \(- 280\) |
\(540-560\) | \(8\) | \(550 (a)\) | \(0\) | \(0\) |
\(560-580\) | \(6\) | \(570\) | \(20\) | \(120\) |
\(580-600\) | \(10\) | \(590\) | \(40\) | \(400\) |
\(\Sigma f_i=50\) | \(\Sigma f_id_i= -240\) |
It can be observed from the table,
\[\begin{align} \Sigma f_{i} &=50 \\ \Sigma f_{i} u_{i} &=-240 \end{align}\]
\[\begin{align}{{{\rm Mean }}\,\,(\overline {{x}} ) } &={{{a}} + \left( {\frac{{{{ }}\Sigma {{f_iu_i}}}}{{\Sigma {{f_i}}}}} \right){{h}}}\\{}&{ = 550 + \left( {\frac{{ - 240}}{{50}}} \right)}\\{}&{ = 550 - \frac{{24}}{5}}\\&=550-4.8\\ {}&{ = 545.2}\end{align}\]
Thus, the mean daily wages of the workers of the factory is \(\rm Rs.545.20\)
Chapter 14 Ex.14.1 Question 3
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs \(18\). Find the missing frequency \( f\).
Daily pocket allowance (in Rs) | \(11 – 13\) | \(13 – 15\) | \(15 – 17\) | \(17 – 19\) | \(19 – 21\) | \(21 – 23\) | \(23 – 25\) |
Number of children | \(7\) | \(6\) | \(9\) | \(13\) | \(f\) | \(5\) | \(4\) |
Solution
What is known?
The mean pocket allowance is Rs \(18\).
The missing frequency \(f\).
Reasoning:
We will use assumed mean method to solve this question
\[\begin{align} \operatorname{Mean,}\,\,\,(\overline{{x}})={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \end{align} \]
Solution:
We know that,
Class mark
\[{{x}_{i}}=\,\,\frac{\left[ \begin{align} & \text{Upper}\ \text{class}\ \text{limit}\text{+} \\ & \text{Lower}\ \text{class}\ \text{limit} \\ \end{align} \right]}{2}\]
Taking assumed mean\(, a = 18\)
Steps:
Daily pocket allowance (in Rs) | No of children \( (f_i)\) | \( X_i\) | \( d_i = x_i -a \) | \( f_id_i \) |
\(11 – 13\) | \(7\) | \(12\) | \(-6\) | \(-42\) |
\(13 – 15\) | \(6\) | \(14\) | \(-4\) | \(-24\) |
\(15 – 17\) | \(9\) | \(16\) | \(-2\) | \(-18\) |
\(17 – 19\) | \(13\) | \(18 (a)\) | \(0\) | \(0\) |
\(19 – 21\) | \(f\) | \(20\) | \(2\) | \(2f\) |
\(21 – 23\) | \(5\) | \(22\) | \(4\) | \(20\) |
\(23 – 25\) | \(4\) | \(24\) | \(6\) | \(24\) |
\(\Sigma f_i=40 +f \) | \(\Sigma f_id_i=2f-40\) |
From the table, we obtain
\(\Sigma {f}_{i} = 40 + f\)
\(\Sigma {f}_{i} {d}_{i} = 2f - 40\)
\[\begin{align} \text { Mean, }\,\,(\overline{{x}}) &={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \\ 18 &=18+\left(\frac{-40+2 {f}}{40+f}\right) \\ 18-18 &=\frac{2 {f}-40}{40+f} \\ 2f - 40 &= 0 \\ {f} &=20 \end{align}\]
Hence, the missing frequency \(f \) = \(20\).
Chapter 14 Ex.14.1 Question 4
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart beats per minute | \(65 – 68\) | \(68 – 71\) | \(71 – 74\) | \(74 – 77\) | \(77 – 80\) | \(80 – 83\) | \(83 – 86\) |
Number of women | \(2\) | \(4\) | \(3\) | \(8\) | \(7\) | \(4\) | \(2\) |
Solution
What is known?
The heart beats per minute of \(30 \) women.
The mean heart beats per minute for these women.
Reasoning:
We will use Step-deviation Method to solve this question because the data given is large and will be convenient to apply to all the \(d_i\) have a common factor.
Sometimes when the numerical values of \(x_i\) and \(f_i \) are large, finding the product of \(x_i\) and \(f_i \)becomes tedious. We can do nothing with the \(f_i,\) but we can change each \(x_i\) to a smaller number so that our calculations become easy. Now we have to subtract a fixed number from each of these \(x_i.\)
The first step is to choose one among the \(x_i \)as the assumed mean and denote it by \(‘a’\). Also, to further reduce our calculation work, we may take \(‘a’\) to be that \(x_i\) which lies in the centre of \(x_1, x_2, . . ., x_n.\) So, we can choose \(a.\)
The next step is to find the difference \(‘d_i’\) between a and each of the \(x_i,\) that is, the deviation of \(‘a’\) from each of the \(x_i.\) i.e.,\(d_i = x_i - a\)
The third step is to find \(‘u_i’\) by dividing \(d_i\) and class size \(h\) for each of the \(x_i.\) i.e.,\(u_i = \frac{{d_i}}{h}\)
The next step is to find the product of \(u_i\) with the corresponding \(f_i,\) and take the sum of all the \(f_iu_i\)
The step-deviation method will be convenient to apply if all the \(d_i \)have a common factor.
Now put the values in the below formula
Mean, \(\overline x = a + (\frac{{\Sigma f_i u_i}}{\Sigma f_i}) \times h\)
Steps:
We know that,
Class mark, \( x_i = \frac{{\text{Upper class limit + Lower class limit}}}{2}\)
Class size, \(h=3\)
Taking assumed mean,\( a= 75.5\)
Number of heart beats per minute | No of women \( (f_i)\) | \( X_i\) | \( d_i = x_i -a \) | \(\begin{align} {u_i}=\frac{ {x_i}-{a}}{h}\end{align}\) | \( f_iu_i \) |
\(65-68\) | \(2\) | \(66.5\) | \(-9\) | \(-3\) | \(-6\) |
\(68-71\) | \(4\) | \(69.5\) | \(-6\) | \(-2\) | \(-8\) |
\(71 -74\) | \(3\) | \(72.5\) | \(-3\) | \(-1\) | \(-3\) |
\(74-77\) | \(8\) | \(75.5(a)\) | \(0\) | \(0\) | \(0\) |
\(77 – 80\) | \(7\) | \(78.5\) | \(3\) | \(1\) | \(7\) |
\(80 – 83\) | \(4\) | \(81.5\) | \(6\) | \(2\) | \(8\) |
\(83 – 86\) | \(2\) | \(84.5\) | \(9\) | \(3\) | \(6\) |
\(\Sigma f_i=30 \) | \(\Sigma f_iu_i=4\) |
From the table, we obtain
\(\Sigma f_i=30\)
\(\Sigma f_i u_i=4\)
\[\begin{align} \operatorname{Mean}\,\,(\overline{{x}}) &={a}+\left(\frac{\Sigma {f}_{{i}} {u}_{{i}}}{\Sigma {f}_{{i}}}\right) {h} \\ \overline{{x}} &=75.5+\left(\frac{4}{30}\right) 3 \\ \overline{{x}} &=75.5-\frac{{2}}{5} \\\overline{{x}} &=75.5-0.4 \\ \overline {{x}}&=75.9 \end{align}\]
Hence, the mean heartbeat per minute for these women is \(75.9\)
Chapter 14 Ex.14.1 Question 5
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?
Solution
What is known?
The distribution of mangoes according to the number of boxes.
The mean number of mangoes kept in a packing box.
Reasoning:
We solve this question by step deviation method.
Hence, the given class interval is not continuous. First we have to make it continuous. There is a gap of \(1\) between two class interval. Therefore, \(0.5\) has to be added to the upper class limit and \(0.5\) has to be subtracted from the lower class limit of each interval.
\[\text{Mean,}\; \overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\]
Solution:
We know that,
Class mark,
\({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Class size, \(h = 3\)
Taking assumed mean, \(a = 57\)
Class interval | No of house hold \((f_i)\) | \( x_i\) | \( d_i = x_i -a \) | \(\begin{align}{u_i}=\frac{ d_i}{h}\end{align}\) | \( F_iu_i \) |
\(49.5-52.5\) | \(15\) | \(51\) | \(-6\) | \(-2\) | \(-30\) |
\(52.5-55.5\) | \(110\) | \(54\) | \(-3\) | \(1\) | \(-110\) |
\(55.5-58.5\) | \(135\) | \(57(a)\) | \(0\) | \(0\) | \(0\) |
\(58.5-61.5\) | \(115\) | \(60\) | \(3\) | \(1\) | \(115\) |
\(61.5-64.5\) | \(25\) | \(63\) | \(6\) | \(2\) | \(50\) |
\(\Sigma f_i=400 \) | \(\Sigma f_iu_i=25\) |
From the table, we obtain
\[\begin{align}\sum {{f_i} = 400} \\\sum {{f_i}{u_i}} = 25\end{align}\]
Mean, \(\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\)
\[\begin{align}& = 57 + \left( {\frac{{25}}{{400}}} \right) \times 3\\ &= 57 + \frac{1}{{16}} \times 3\\ &= 57 + \frac{3}{{16}}\\ &= 57 + 0.19\\ &= 57.19\end{align}\]
The mean number of mangoes kept in a packing box are \(57\).\(19\)
Chapter 14 Ex.14.1 Question 6
The table below shows the daily expenditure on food of \(25\) households in a locality.
Daily expenditure (in Rs) | \(100 – 150\) | \(150 – 200\) | \(200 – 250\) | \(250 – 300\) | \(300 – 350\) |
Number of households | \(4\) | \(5\) | \(12\) | \(2\) | \(2\) |
Find the mean daily expenditure on food by a suitable method.
Solution
What is known?
The daily expenditure on food of \(25\) households in a locality.
The mean daily expenditure on food.
Reasoning:
We will solve this question by step deviation method.
Mean, \(\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\)
Steps:
We know that,
Class mark, \({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Class size, \(h = 50\)
Taking assumed mean, \(a = 225\)
Daily expenditure in Rs. |
Number of house holds \((f_i)\) | \( x_i\) | \( d_i = x_i -a \) | \(\begin{align}{u_i}=\frac{ d_i}{h}\end{align}\) | \(f_iu_i \) |
\(100 –150\) | \(4\) | \(125\) | \(-100\) | \(-2\) | \(-8\) |
\(150 – 200\) | \(5\) | \(175\) | \(-50\) | \(-1\) | \(-5\) |
\(200 – 250\) | \(12\) | \(225 (a)\) | \(0\) | \(0\) | \(0\) |
\(250 – 300\) | \(2\) | \(275\) | \(50\) | \(1\) | \(2\) |
\(300 –350\) | \(2\) | \(325\) | \(100\) | \(2\) | \(4\) |
\(\Sigma f_i=25 \) | \(\Sigma f_iu_i=-7\) |
From the table, we obtain
\[\begin{array}{l}\sum {{f_i} = 25} \\\sum {{f_i}{u_i}} = - 7\end{array}\]
\[\begin{align}{\text { Mean }(\overline{{x}})}&={{a}+\left(\frac{\sum { f_iu_i }}{\sum {f} _{i}}\right) {h}} \\ {\overline{{x}}}&={225+\left(\frac{-7}{25}\right) 50} \\ {\overline{{x}}}&={225-14} \\ {\overline{{x}}}&={211}\end{align}\]
Thus, the mean daily expenditure on food is Rs \(211\).
Chapter 14 Ex.14.1 Question 7
To find out the concentration of \(SO_2\) in the air (in parts per million, i.e., ppm), the data was collected for \(30\) localities in a certain city and is presented below:
Concentration of \(SO_2\) (in ppm) | Frequency |
\(0.00 - 0.04\) | \(4\) |
\(0.04 - 0.08\) | \(9\) |
\(0.08 - 0.12\) | \(9\) |
\(0.12 - 0.16\) | \(2\) |
\(0.16 - 0.20\) | \(4\) |
\(0.20 - 0.24\) | \(2\) |
Find the mean concentration of \(SO_2\) in the air.
Solution
What is known?
The Concentration of \(SO_2\) in the air (in parts per million, i.e., ppm) for \(30\) localities in a certain city:
What is unknown?
The mean concentration of \(SO_2 \)in the air
Reasoning:
We solve this question by step deviation method.
Mean, \(\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\)
Steps:
We know that,
Class mark,\({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Class size\(, h = 0.04\)
Taking assumed mean\(,a = 0.14\)
Content of SO_{2} | Frequency \((f_i)\) | \( x_i\) | \( d_i = x_i -a \) | \(\begin{align}{u_i}=\frac{ d_i}{h}\end{align}\) | \( f_iu_i \) |
\(0.00-0.04\) | \(4\) | \(0.02\) | \(-0.12\) | \(-3\) | \(-12\) |
\(0.04-0.08\) | \(9\) | \(0.06\) | \(-0.08\) | \(-2\) | \(-18\) |
\(0.08-0.12\) | \(9\) | \(0.10\) | \(-0.04\) | \(-1\) | \(-9\) |
\(0.12-0.16\) | \(2\) | \(0.14(a)\) | \(0\) | \(0\) | \(0\) |
\(0.16-0.20\) | \(4\) | \(0.18\) | \(0.04\) | \(1\) | \(4\) |
\(0.20-0.24\) | \(2\) | \(0.22\) | \(0.08\) | \(2\) | \(4\) |
\(\Sigma f_i=30 \) | \(\Sigma f_iu_i=-31\) |
From the table, we obtain
\[\begin{array}{l}\sum {{f_i} = 30} \\\sum {{f_i}{u_i}} = - 31\end{array}\]
\[\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\sum {f}_{{i}} {u}_{{i}}}{{f}_{{i}}}\right) {h} \\ \overline{{x}} &=0.14+\left(\frac{-31}{30}\right) 0.04 \\ \overline{{x}} &=0.14-| 0.04133 \\ \overline{{x}} &=0.099 \end{align}\]
The mean concentration of \(SO_2\) in the air is \(0.099\).
Chapter 14 Ex.14.1 Question 8
A class teacher has the following absentee record of \(40\) students of a class for the whole term. Find the mean number of days a student was absent.
Number of days | \(0 – 6\) | \(6 – 10\) | \(10 – 14\) | \(14 – 20\) | \(20 – 28\) | \(28 – 38\) | \(38 – 40\) |
Number of students | \(11\) | \(10\) | \(7\) | \(4\) | \(4\) | \(3\) | \(1\) |
Solution
What is known?
The Absentee record of \(40\) students of a class for the whole term.
The mean number of days a student was absent.
Reasoning:
We solve this question by assumed mean method.
Mean, \(\overline x = a + \left( {\frac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}} \right)\)
Steps:
We know that,
Class mark,\({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Taking assumed mean,\(a=17\)
Number of days |
Number of students \(f_i\) |
\(x_i\) | \( d_i = x_i -a \) | \( f_id_i \) |
\(0 – 6\) | \(11\) | \(3\) | \(-14\) | \(-154\) |
\(6 – 10\) | \(10 \) | \(8\) | \(-9\) | \(-90\) |
\(10 – 14\) | \(7 \) | \(12\) | \(-5\) | \(-35\) |
\(14 – 20\) | \(4\) | \(17(a)\) | \(0\) | \(0\) |
\(20 – 28\) | \(4\) | \(24\) | \(7\) | \(28\) |
\(28 – 38\) | \(3\) | \(33\) | \(18\) | \(48\) |
\(38 – 40\) | \(1\) | \(39\) | \(22\) | \(22\) |
\(\Sigma f_i=40 \) | \(\Sigma f_id_i=-181\) |
From the table,we obtain
\[\begin{array}{l}\sum {{f_i} = 40} \\\sum {{f_i}{d_i}} = - 181\end{array}\]
\[\begin{align} \operatorname{Mean}(\overline{{x}}) &={a}+\left(\frac{\Sigma f_{1} d_{i}}{\Sigma {f}_{{i}}}\right) \\ \overline{{x}} &=17+\left(\frac{-181}{40}\right) \\ \overline{{x}} &=17-\frac{181}{40} \\ \overline{{x}} &=17-4.525 \\ \overline{{x}} &=12.475 \\ \overline{{x}} & =12.48 \end{align}\]
Thus, the mean number of days a student was absent is \(12.48\).
Chapter 14 Ex.14.1 Question 9
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 | \(3\) | \(10\) | \(11\) | \(8\) | \(3\) |
Solution
What is known?
The Literacy rate (in percentage) of \(35\) cities
The Mean literacy rate.
Reasoning:
We solve this question by assumed mean method.
Mean, \(\overline x = a + \left( {\frac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}} \right)\)
Steps:
We know that,
Class mark, \({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Taking assumed mean\(, a=70\)
Literacy rate | No of cities\((f_i)\) | \( X_i\) | \( d_i = x_i -a \) | \( f_id_i \) |
\(45 – 55\) | \(3\) | \(50\) | \(-20\) | \(-60\) |
\(55 – 65\) | \(10\) | \(60\) | \(-10\) | \(-100\) |
\(65 – 75\) | \(11\) | \(70(a)\) | \(0\) | \(0\) |
\(75 – 85\) | \(8\) | \(80\) | \(10\) | \(80\) |
\(85 – 95\) | \(3\) | \(90\) | \(20\) | \(60\) |
\(\Sigma f_i=35 \) | \(\Sigma f_id_i=-20\) |
From the table,we obtain
\[\begin{array}{l}\sum {{f_i} = 35} \\\sum {{f_i}{d_i}} = - 20\end{array}\]
\[\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\Sigma f_{i} d_{i}}{\Sigma {f}_{{i}}}\right) \\ \overline{{x}} &=70+\left(\frac{-20}{35}\right) \\ \overline{{x}} &=70-\frac{4}{7} \\ \overline{{x}} &=70-0.57 \\ \overline{{x}} &=69.43 \end{align}\]
Thus, the mean literacy rate is \(69.43\)%