NCERT Solutions For Class 10 Maths: Statistics

EXERCISE 13.1

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.

Which method did you use for finding the mean, and why?

Solution-
We use the Direct Method since the numerical values of the class marks ($x_i$) and frequencies ($f_i$) are small.

1.Create the table including class marks ($x_i$) and products ($f_i x_i$).

2.Calculate the mean ($\bar{x}$) using the Direct Method formula:
$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$

3.$\bar{x} = \frac{162}{20} = 8.1$

4.The mean number of plants per house is 8.1.

5.Method used: Direct Method, because the calculations involving $x_i$ and $f_i$ were sufficiently small.

Question 2: Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution-
Since the class marks ($x_i$) are large, we use the Step-deviation Method to simplify calculations.

1.Create the table, choosing assumed mean $a = 550$ (class mark of $540-560$) and class size $h = 20$.

2.Calculate the mean ($\bar{x}$) using the Step-deviation Method formula:
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 550 + \left(\frac{-12}{50}\right) \times 20$

4.$\bar{x} = 550 + (-0.24) \times 20$

5.$\bar{x} = 550 - 4.8 = 545.2$

6.The mean daily wages of the workers is $\text{` } 545.20$.

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 $f$.

Solution-
We use the Direct Method to find the missing frequency $f$.

1.Create the table including class marks ($x_i$) and products ($f_i x_i$).

2.Given the mean $\bar{x} = 18$. Use the Direct Method formula:
$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$

3.$18 = \frac{752 + 20f}{44 + f}$

4.$18(44 + f) = 752 + 20f$

5.$792 + 18f = 752 + 20f$

6.$792 - 752 = 20f - 18f$

7.$40 = 2f$

8.$f = 20$.

9.The missing frequency $f$ is 20.

Question 4: Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution-
We use the Assumed Mean Method or Step-deviation Method since the class marks are large, choosing $a = 75.5$ (class mark of $74-77$) and class size $h = 3$. We will use the Step-deviation method for efficiency.

1.Create the table with class marks ($x_i$), deviation ($d_i$), and simplified deviation ($u_i$).

2.Calculate the mean ($\bar{x}$) using the Step-deviation Method formula:
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 75.5 + \left(\frac{4}{30}\right) \times 3$

4.$\bar{x} = 75.5 + \frac{12}{30} = 75.5 + 0.4$

5.$\bar{x} = 75.9$.

6.The mean heartbeats per minute is 75.9.

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.

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

Solution-
Since the classes are not continuous (e.g., $52$ to $53$), but the mean calculation does not strictly require continuity for the $x_i$ values, we find the class marks and use the Step-deviation Method due to the large frequencies.

1.Determine the class marks ($x_i$).

2.We chose assumed mean $a = 57$ and class size $h = 3$ (the difference between consecutive class marks).

3.Calculate the mean ($\bar{x}$) using the Step-deviation Method:
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

4.$\bar{x} = 57 + \left(\frac{25}{400}\right) \times 3$

5.$\bar{x} = 57 + \frac{75}{400} = 57 + 0.1875$

6.$\bar{x} = 57.1875$.

7.The mean number of mangoes is $57.19$ (approx.).

8.Method chosen: Step-deviation Method, because $x_i$ and $f_i$ were numerically large.

Question 6: The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Solution-
We use the Step-deviation Method due to the large class size and large class marks. We choose $a = 225$ (class mark of $200-250$) and class size $h = 50$.

1.Create the table.

2.Calculate the mean ($\bar{x}$) using the Step-deviation Method formula:
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50$

4.$\bar{x} = 225 + (-7 \times 2) = 225 - 14$

5.$\bar{x} = 211$.

6.The mean daily expenditure on food is $\text{` } 211$.

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:

Find the mean concentration of SO 2 in the air.

Solution-
We use the Step-deviation Method due to the small decimal class marks ($x_i$). We choose $a = 0.10$ and class size $h = 0.04$.

1.Create the table.

2.Calculate the mean ($\bar{x}$) using the Step-deviation Method formula:
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 0.10 + \left(\frac{-1}{30}\right) \times 0.04$

4.$\bar{x} = 0.10 - \frac{0.04}{30} \approx 0.10 - 0.00133$

5.$\bar{x} = 0.09867$.

6.The mean concentration of SO$_2$ in the air is $0.099$ ppm (approx.).

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.

Solution-
Since the class size varies ($h$ is unequal), we use the Assumed Mean Method. We choose $a = 17$ (class mark of $14-20$).

1.Create the table with class marks ($x_i$) and deviation ($d_i = x_i - 17$).

2.Calculate the mean ($\bar{x}$) using the Assumed Mean Method formula:
$\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$

3.$\bar{x} = 17 + \frac{-181}{40}$

4.$\bar{x} = 17 - 4.525 = 12.475$.

5.The mean number of days a student was absent is $12.48$ days (approx.).

Question 9: The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution-
We use the Step-deviation Method with $a = 70$ (class mark of $65-75$) and class size $h = 10$.

1.Create the table.

2.Calculate the mean ($\bar{x}$) using the Step-deviation Method formula:
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 70 + \left(\frac{-2}{35}\right) \times 10$

4.$\bar{x} = 70 - \frac{20}{35} = 70 - 0.5714$ (approx.)

5.$\bar{x} = 69.4286$.

6.The mean literacy rate is $69.43$ percent (approx.).

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EXERCISE 13.2

Question 1: The following table shows the ages of the patients admitted in a hospital during a year:

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

Solution-
(i) Finding the Mode:
1.The maximum frequency ($f_1$) is 23, corresponding to the modal class $35 - 45$.

2.Identify the required values:
$l$ (lower limit of modal class) $= 35$
$h$ (class size) $= 10$
$f_1$ (frequency of modal class) $= 23$
$f_0$ (frequency of preceding class) $= 21$
$f_2$ (frequency of succeeding class) $= 14$

3.Calculate the Mode using the formula:
Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

4.Mode $= 35 + \frac{23 - 21}{2(23) - 21 - 14} \times 10$

5.Mode $= 35 + \frac{2}{46 - 35} \times 10 = 35 + \frac{20}{11}$

6.Mode $\approx 35 + 1.818 = 36.82$ years.

(ii) Finding the Mean (using Step-deviation Method with $a=40$, $h=10$):
1.Create the table.

2.Calculate the mean ($\bar{x}$):
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 40 + \left(\frac{-37}{80}\right) \times 10 = 40 - 4.625$

4.$\bar{x} = 35.375 \approx 35.38$ years.

(iii) Comparison and Interpretation:
1.The Mode is $36.82$ years, and the Mean is $35.38$ years.

2.Interpretation: The mode signifies that the age group $35 - 45$ years has the maximum number of patients admitted (specifically, 36.82 years is the most frequent age value). The mean signifies that the average age of patients admitted is $35.38$ years.

Question 2: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Solution-
1.The maximum frequency ($f_1$) is 61, corresponding to the modal class $60 - 80$.

2.Identify the required values:
$l$ (lower limit of modal class) $= 60$
$h$ (class size) $= 20$
$f_1$ (frequency of modal class) $= 61$
$f_0$ (frequency of preceding class) $= 52$
$f_2$ (frequency of succeeding class) $= 38$

3.Calculate the Mode using the formula:
Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

4.Mode $= 60 + \frac{61 - 52}{2(61) - 52 - 38} \times 20$

5.Mode $= 60 + \frac{9}{122 - 90} \times 20 = 60 + \frac{9 \times 20}{32} = 60 + \frac{180}{32}$

6.Mode $= 60 + 5.625 = 65.625$ hours.

7.The modal lifetimes of the components is 65.625 hours.

Question 3: 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 :

Solution-
(i) Finding the Mode:
1.The maximum frequency ($f_1$) is 40, corresponding to the modal class $1500 - 2000$.

2.Identify the required values:
$l = 1500$
$h = 500$
$f_1 = 40$
$f_0 = 24$
$f_2 = 33$

3.Calculate the Mode:
Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

4.Mode $= 1500 + \frac{40 - 24}{2(40) - 24 - 33} \times 500$

5.Mode $= 1500 + \frac{16}{80 - 57} \times 500 = 1500 + \frac{16 \times 500}{23}$

6.Mode $\approx 1500 + 347.83 = 1847.83$.

7.The modal monthly expenditure is $\text{` } 1847.83$.

(ii) Finding the Mean (using Step-deviation Method with $a=2750$, $h=500$):
1.Create the table.

2.Calculate the mean ($\bar{x}$):
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 2750 + \left(\frac{-35}{200}\right) \times 500 = 2750 - 35 \times 2.5$

4.$\bar{x} = 2750 - 87.5 = 2662.5$.

5.The mean monthly expenditure is $\text{` } 2662.50$.

Question 4: The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution-
(i) Finding the Mode:
1.The maximum frequency ($f_1$) is 10, corresponding to the modal class $30 - 35$.

2.Identify the required values:
$l = 30$
$h = 5$
$f_1 = 10$
$f_0 = 9$
$f_2 = 3$

3.Calculate the Mode:
Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

4.Mode $= 30 + \frac{10 - 9}{2(10) - 9 - 3} \times 5$

5.Mode $= 30 + \frac{1}{20 - 12} \times 5 = 30 + \frac{5}{8}$

6.Mode $= 30 + 0.625 = 30.625$.

7.The mode is 30.6.

(ii) Finding the Mean (using Step-deviation Method with $a=32.5$, $h=5$):
1.Create the table.

2.Calculate the mean ($\bar{x}$):
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 32.5 + \left(\frac{-23}{35}\right) \times 5 = 32.5 - \frac{23}{7}$

4.$\bar{x} \approx 32.5 - 3.2857 = 29.214$.

5.The mean is 29.2.

(iii) Interpretation:
1.The Mode (30.6) suggests that the most frequent teacher-student ratio across the states/U.T. is approximately 30.6 students per teacher.

2.The Mean (29.2) suggests that the average teacher-student ratio across the states/U.T. is 29.2 students per teacher.

Question 5: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Solution-
1.The maximum frequency ($f_1$) is 18, corresponding to the modal class $4000 - 5000$.

2.Identify the required values:
$l = 4000$
$h = 1000$
$f_1 = 18$
$f_0 = 4$
$f_2 = 9$

3.Calculate the Mode using the formula:
Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

4.Mode $= 4000 + \frac{18 - 4}{2(18) - 4 - 9} \times 1000$

5.Mode $= 4000 + \frac{14}{36 - 13} \times 1000 = 4000 + \frac{14000}{23}$

6.Mode $\approx 4000 + 608.696 = 4608.70$.

7.The mode of the data is 4608.7 runs (approx.).

Question 6: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Solution-
1.The maximum frequency ($f_1$) is 20, corresponding to the modal class $40 - 50$.

2.Identify the required values:
$l = 40$
$h = 10$
$f_1 = 20$
$f_0 = 12$
$f_2 = 11$

3.Calculate the Mode using the formula:
Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

4.Mode $= 40 + \frac{20 - 12}{2(20) - 12 - 11} \times 10$

5.Mode $= 40 + \frac{8}{40 - 23} \times 10 = 40 + \frac{80}{17}$

6.Mode $\approx 40 + 4.7059 = 44.71$.

7.The mode of the data is 44.7 cars (approx.).

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EXERCISE 13.3

Question 1: 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.

Solution-
(i) Finding the Median:
1.Total number of observations $n = 68$. Calculate $n/2 = 34$.

2.Create the table including cumulative frequency ($cf$).

3.The median class is $125 - 145$, as its $cf$ (42) is greater than and nearest to $n/2 = 34$.

4.Identify the required values for the median formula:
$l = 125$
$n/2 = 34$
$cf$ (preceding class) $= 22$
$f$ (median class) $= 20$
$h$ (class size) $= 20$

5.Calculate the Median using the formula:
Median $= l + \frac{n/2 - cf}{f} \times h$

6.Median $= 125 + \frac{34 - 22}{20} \times 20 = 125 + 12$

7.Median $= 137$.

(ii) Finding the Mode:
1.The modal class is $125 - 145$, as it has the maximum frequency ($f_1 = 20$).

2.Identify values: $l=125$, $h=20$, $f_1=20$, $f_0=13$, $f_2=14$.

3.Calculate the Mode:
Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

4.Mode $= 125 + \frac{20 - 13}{2(20) - 13 - 14} \times 20$

5.Mode $= 125 + \frac{7}{40 - 27} \times 20 = 125 + \frac{7 \times 20}{13}$

6.Mode $\approx 125 + 10.769 = 135.77$.

(iii) Finding the Mean (using Step-deviation Method with $a=135$, $h=20$):
1.Create the table.

2.Calculate the mean ($\bar{x}$):
$\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.$\bar{x} = 135 + \left(\frac{7}{68}\right) \times 20 \approx 135 + 2.059$

4.$\bar{x} = 137.06$.

(iv) Comparison:
1.Mean $= 137.06$ units, Median $= 137$ units, Mode $= 135.77$ units. The values are close, indicating a fairly symmetrical distribution. They all lie around the center of the data.

**Question 2: If the median of the distribution given below is 28.5, find the values of $x$ and $y$.

Solution-
1.Total frequency $n = 60$. Therefore, $n/2 = 30$.

2.The median is given as $28.5$, which lies in the class interval $20 - 30$. This is the median class.

3.Create the cumulative frequency table.

4.Using the total frequency:
$45 + x + y = 60$
$x + y = 15$ (1)

5.Apply the Median formula using values from the median class $20-30$: $l=20$, $h=10$, $f=20$, $cf = 5 + x$, Median $= 28.5$.
Median $= l + \frac{n/2 - cf}{f} \times h$

6.$28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10$

7.$28.5 - 20 = \frac{25 - x}{2}$

8.$8.5 = \frac{25 - x}{2}$

9.$17 = 25 - x$

10.$x = 25 - 17 = 8$.

11.Substitute $x = 8$ into Equation (1):
$8 + y = 15$

12.$y = 7$.

13.The values of $x$ and $y$ are 8 and 7, respectively.

Question 3: A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Solution-
1.The given data is a cumulative frequency distribution of the less than type. We convert it to a frequency distribution table with continuous class intervals. $n = 100$.

2.The classes start from $18$ and have a class size of 5.

3.Calculate $n/2 = 100/2 = 50$.

4.The median class is $35 - 40$, as its $cf$ (78) is greater than and nearest to 50.

5.Identify the required values for the median formula:
$l = 35$
$n/2 = 50$
$cf$ (preceding class) $= 45$
$f$ (median class) $= 33$
$h$ (class size) $= 5$

6.Calculate the Median:
Median $= l + \frac{n/2 - cf}{f} \times h$

7.Median $= 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{5 \times 5}{33}$

8.Median $\approx 35 + 0.7576 = 35.7576$.

9.The median age is $35.76$ years (approx.).

Question 4: The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Find the median length of the leaves. (Hint : 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.)

Solution-
1.The classes are discontinuous (non-inclusive). We must convert them to continuous classes (e.g., $118 - 0.5 = 117.5$ to $126 + 0.5 = 126.5$). Total observations $n = 40$. $n/2 = 20$.

2.Create the table with continuous classes and cumulative frequency.

3.The median class is $144.5 - 153.5$, as its $cf$ (29) is greater than and nearest to $n/2 = 20$.

4.Identify the required values:
$l = 144.5$
$n/2 = 20$
$cf$ (preceding class) $= 17$
$f$ (median class) $= 12$
$h$ (class size) $= 9$

5.Calculate the Median:
Median $= l + \frac{n/2 - cf}{f} \times h$

6.Median $= 144.5 + \frac{20 - 17}{12} \times 9 = 144.5 + \frac{3 \times 9}{12}$

7.Median $= 144.5 + \frac{27}{12} = 144.5 + 2.25$

8.Median $= 146.75$ mm.

Question 5: The following table gives the distribution of the life time of 400 neon lamps :

Find the median life time of a lamp.

Solution-
1.Total observations $n = 400$. $n/2 = 200$.

2.Create the table including cumulative frequency.

3.The median class is $3000 - 3500$, as its $cf$ (216) is greater than and nearest to $n/2 = 200$.

4.Identify the required values:
$l = 3000$
$n/2 = 200$
$cf$ (preceding class) $= 130$
$f$ (median class) $= 86$
$h$ (class size) $= 500$

5.Calculate the Median:
Median $= l + \frac{n/2 - cf}{f} \times h$

6.Median $= 3000 + \frac{200 - 130}{86} \times 500 = 3000 + \frac{70 \times 500}{86}$

7.Median $= 3000 + \frac{35000}{86} \approx 3000 + 406.98$

8.Median $= 3406.98$ hours.

Question 6: 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:

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.

Solution-
1.Total observations $n = 100$. $n/2 = 50$. Class size $h=3$.

2.Create the table for Mean and Median calculations.

(i) Finding the Median:
1.The median class is $7 - 10$ ($cf=76$).
$l = 7$, $h = 3$, $f = 40$, $cf = 36$.

2.Median $= l + \frac{n/2 - cf}{f} \times h$

3.Median $= 7 + \frac{50 - 36}{40} \times 3 = 7 + \frac{14 \times 3}{40}$

4.Median $= 7 + 1.05 = 8.05$.

5.The median number of letters is 8.05.

(ii) Finding the Mean (using Step-deviation Method):
1.Assumed mean $a = 8.5$, $h = 3$, $\Sigma f_i = 100$, $\Sigma f_i u_i = -6$.

2.Mean $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

3.Mean $= 8.5 + \left(\frac{-6}{100}\right) \times 3 = 8.5 - 0.18$

4.Mean $= 8.32$.

5.The mean number of letters is 8.32.

(iii) Finding the Mode:
1.The modal class is $7 - 10$ (maximum frequency $f_1 = 40$).
$l=7$, $h=3$, $f_1=40$, $f_0=30$, $f_2=16$.

2.Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

3.Mode $= 7 + \frac{40 - 30}{2(40) - 30 - 16} \times 3 = 7 + \frac{10}{80 - 46} \times 3$

4.Mode $= 7 + \frac{30}{34} \approx 7 + 0.882$

5.Mode $= 7.88$.

6.The modal size of the surnames is 7.88 letters (approx.).

Question 7: The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution-
1.Total students $n = 30$. $n/2 = 15$.

2.Create the cumulative frequency table.

3.The median class is $55 - 60$, as its $cf$ (19) is greater than and nearest to $n/2 = 15$.

4.Identify the required values:
$l = 55$
$n/2 = 15$
$cf$ (preceding class) $= 13$
$f$ (median class) $= 6$
$h$ (class size) $= 5$

5.Calculate the Median:
Median $= l + \frac{n/2 - cf}{f} \times h$

6.Median $= 55 + \frac{15 - 13}{6} \times 5 = 55 + \frac{2 \times 5}{6}$

7.Median $= 55 + \frac{10}{6} \approx 55 + 1.6667$

8.Median $= 56.67$ kg (approx.).