StatisticsClass 10 Mathematics NCERT Solutions

Given: The data regarding the number of plants in 20 houses.

To Find: The mean number of plants per house.

Solution: We will use the direct method to find the mean as the numerical values of the frequency ($f_i$) and the class mark ($x_i$) are small.

First, we find the class mark ($x_i$) for each class interval. The class mark is the midpoint of the interval. Class mark ($x_i$) = $\frac{\text{Upper class limit} + \text{Lower class limit}}{2}$

Let's create a table to calculate the mean:

Formula for Mean (Direct Method): $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$

Substituting the values from the table: $\bar{x} = \frac{162}{20} = 8.1$

Method Used and Justification: We used the Direct Method. This method is appropriate here because the values of the class marks ($x_i$) and frequencies ($f_i$) are small, making the calculation of their products ($f_i x_i$) straightforward and not prone to large computational errors.

Final Answer: The mean number of plants per house is 8.1.

Given: The distribution of daily wages of 50 workers.

To Find: The mean daily wages of the workers.

Solution: Since the values of the class marks ($x_i$) will be large, we will use the Step-Deviation Method to simplify the calculations. The class size is uniform, $h = 20$.

Let's choose an assumed mean, $a$. We can take the middle class mark as the assumed mean. The class marks are 510, 530, 550, 570, 590. Let's take $a = 550$.

We will create a table with the necessary columns for the step-deviation method.

From the table, we have: $\Sigma f_i = 50$ $\Sigma f_i u_i = (-24 - 14) + (6 + 20) = -38 + 26 = -12$ Assumed mean, $a = 550$ Class size, $h = 20$

Formula for Mean (Step-Deviation Method): $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

Substituting the values: $\bar{x} = 550 + \left(\frac{-12}{50}\right) \times 20$ $\bar{x} = 550 - \frac{12 \times 20}{50}$ $\bar{x} = 550 - \frac{240}{50}$ $\bar{x} = 550 - 4.8$ $\bar{x} = 545.2$

Final Answer: The mean daily wages of the workers of the factory is ₹ 545.20.

Given: The distribution of daily pocket allowance of children, with a missing frequency $f$. The mean pocket allowance is ₹ 18.

To Find: The value of the missing frequency $f$.

Solution: We will use the direct method. First, let's create a table to organize the calculations.

From the table, we have: $\Sigma f_i = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f$ $\Sigma f_i x_i = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f$

Formula for Mean (Direct Method): $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$

We are given that the mean, $\bar{x} = 18$.

Substituting the values into the formula: $18 = \frac{752 + 20f}{44 + f}$

Now, we solve for $f$: $18(44 + f) = 752 + 20f$ $792 + 18f = 752 + 20f$ $792 - 752 = 20f - 18f$ $40 = 2f$ $f = \frac{40}{2}$ $f = 20$

Final Answer: The missing frequency $f$ is 20.

Given: The distribution of heartbeats per minute for 30 women.

To Find: The mean heartbeats per minute.

Solution: The class marks ($x_i$) are not very large, but to simplify calculations, we can use the Assumed Mean Method. The class size is uniform, $h = 3$.

Let's choose an assumed mean, $a$. The class marks are 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5. Let's take the middle value, $a = 75.5$.

Let's create the table for the assumed mean method:

From the table: $\Sigma f_i = 30$ $\Sigma f_i d_i = (-18 - 24 - 9) + (21 + 24 + 18) = -51 + 63 = 12$ Assumed mean, $a = 75.5$

Formula for Mean (Assumed Mean Method): $\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$

Substituting the values: $\bar{x} = 75.5 + \frac{12}{30}$ $\bar{x} = 75.5 + 0.4$ $\bar{x} = 75.9$

Final Answer: The mean heartbeats per minute for these women is 75.9.

Given: The distribution of mangoes in packing boxes.

To Find: The mean number of mangoes per box.

Solution: The given class intervals are not continuous. We need to make them continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.

We will use the Step-Deviation Method as the values of $f_i$ and $x_i$ are large. The class size is $h = 3$. Let's choose the assumed mean $a = 57$.

Let's create the table for calculation:

From the table: $\Sigma f_i = 400$ $\Sigma f_i u_i = (-30 - 110) + (115 + 50) = -140 + 165 = 25$ Assumed mean, $a = 57$ Class size, $h = 3$

Formula for Mean (Step-Deviation Method): $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

Substituting the values: $\bar{x} = 57 + \left(\frac{25}{400}\right) \times 3$ $\bar{x} = 57 + \frac{1}{16} \times 3$ $\bar{x} = 57 + \frac{3}{16}$ $\bar{x} = 57 + 0.1875$ $\bar{x} = 57.1875$

Method Chosen: We chose the Step-Deviation Method. This method simplifies the calculations significantly when the class marks and frequencies are large, and the class size is uniform. It reduces the chance of calculation errors.

Final Answer: The mean number of mangoes kept in a packing box is approximately 57.19.

Given: The daily expenditure on food of 25 households.

To Find: The mean daily expenditure on food.

Solution: We will use the Step-Deviation Method because the class marks are large numbers. The class size is uniform, $h = 50$.

Let's choose an assumed mean, $a$. The class marks are 125, 175, 225, 275, 325. Let's take $a = 225$.

Let's create the table for calculation:

From the table: $\Sigma f_i = 25$ $\Sigma f_i u_i = (-8 - 5) + (2 + 4) = -13 + 6 = -7$ Assumed mean, $a = 225$ Class size, $h = 50$

Formula for Mean (Step-Deviation Method): $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

Substituting the values: $\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50$ $\bar{x} = 225 - 7 \times \frac{50}{25}$ $\bar{x} = 225 - 7 \times 2$ $\bar{x} = 225 - 14$ $\bar{x} = 211$

Final Answer: The mean daily expenditure on food is ₹ 211.

Given: The data for the concentration of SO$_2$ in the air for 30 localities.

To Find: The mean concentration of SO$_2$ in the air.

Solution: The class marks are small decimal numbers, so the direct method is suitable for this calculation.

Let's create a table to calculate the mean:

From the table: $\Sigma f_i = 30$ $\Sigma f_i x_i = 2.96$

Formula for Mean (Direct Method): $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$

Substituting the values: $\bar{x} = \frac{2.96}{30}$ $\bar{x} \approx 0.09866...$

Rounding to four decimal places, we get 0.0987.

Final Answer: The mean concentration of SO$_2$ in the air is approximately 0.099 ppm.

Given: The absentee record of 40 students.

To Find: The mean number of days a student was absent.

Solution: The class sizes are not uniform (6, 4, 4, 6, 8, 10, 2). Therefore, the step-deviation method is not directly applicable in its usual form. The direct method is the most reliable method here.

Let's create a table to calculate the mean using the direct method:

From the table: $\Sigma f_i = 40$ $\Sigma f_i x_i = 499$

Formula for Mean (Direct Method): $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$

Substituting the values: $\bar{x} = \frac{499}{40}$ $\bar{x} = 12.475$

Final Answer: The mean number of days a student was absent is 12.475 days.

Given: The literacy rate of 35 cities.

To Find: The mean literacy rate.

Solution: We will use the Step-Deviation Method to find the mean. The class size is uniform, $h = 10$.

Let's choose an assumed mean, $a$. The class marks are 50, 60, 70, 80, 90. Let's take $a = 70$.

Let's create the table for calculation:

From the table: $\Sigma f_i = 35$ $\Sigma f_i u_i = (-6 - 10) + (8 + 6) = -16 + 14 = -2$ Assumed mean, $a = 70$ Class size, $h = 10$

Formula for Mean (Step-Deviation Method): $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$

Substituting the values: $\bar{x} = 70 + \left(\frac{-2}{35}\right) \times 10$ $\bar{x} = 70 - \frac{20}{35}$ $\bar{x} = 70 - \frac{4}{7}$ $\bar{x} \approx 70 - 0.5714$ $\bar{x} \approx 69.4286$

Final Answer: The mean literacy rate is approximately 69.43%.

Given: The ages of patients admitted to a hospital.

To Find: The mode and the mean of the data, and to compare and interpret them.

The class with the maximum frequency is the modal class. From the table, the maximum frequency is 23, which corresponds to the class interval 35-45. So, the modal class is 35-45.

We have: Lower limit of the modal class, $l = 35$ Frequency of the modal class, $f_1 = 23$ Frequency of the class preceding the modal class, $f_0 = 21$ Frequency of the class succeeding the modal class, $f_2 = 14$ Class size, $h = 10$

Formula for Mode: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values: $\text{Mode} = 35 + \left(\frac{23 - 21}{2(23) - 21 - 14}\right) \times 10$ $\text{Mode} = 35 + \left(\frac{2}{46 - 35}\right) \times 10$ $\text{Mode} = 35 + \left(\frac{2}{11}\right) \times 10$ $\text{Mode} = 35 + \frac{20}{11}$ $\text{Mode} = 35 + 1.818...$ $\text{Mode} \approx 36.82$

Part 2: Finding the Mean We will use the step-deviation method. Let the assumed mean $a = 40$.

$\Sigma f_i = 80$, $\Sigma f_i u_i = (-18-22-21) + (14+10) = -61 + 24 = -37$. $a = 40$, $h = 10$.

Formula for Mean: $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$ $\bar{x} = 40 + \left(\frac{-37}{80}\right) \times 10$ $\bar{x} = 40 - \frac{37}{8}$ $\bar{x} = 40 - 4.625$ $\bar{x} = 35.375$

Comparison and Interpretation: The mode age of the patients is approximately 36.82 years, while the mean age is approximately 35.38 years. The mode indicates that the maximum number of patients admitted to the hospital are of age 36.82 years. The mean indicates that the average age of a patient admitted to the hospital is 35.38 years.

Final Answer: Mode = 36.82 years. Mean = 35.38 years. The maximum number of patients are around 36.82 years old, whereas the average age of all patients is 35.38 years.

Given: The observed lifetimes of 225 electrical components.

To Find: The modal lifetimes of the components.

Solution: To find the mode, we first identify the modal class, which is the class with the highest frequency. From the table, the maximum frequency is 61. The corresponding class interval is 60-80. Therefore, the modal class is 60-80.

We have: Lower limit of the modal class, $l = 60$ Frequency of the modal class, $f_1 = 61$ Frequency of the class preceding the modal class, $f_0 = 52$ Frequency of the class succeeding the modal class, $f_2 = 38$ Class size, $h = 20$

Formula for Mode: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values: $\text{Mode} = 60 + \left(\frac{61 - 52}{2(61) - 52 - 38}\right) \times 20$ $\text{Mode} = 60 + \left(\frac{9}{122 - 90}\right) \times 20$ $\text{Mode} = 60 + \left(\frac{9}{32}\right) \times 20$ $\text{Mode} = 60 + \frac{180}{32}$ $\text{Mode} = 60 + \frac{45}{8}$ $\text{Mode} = 60 + 5.625$ $\text{Mode} = 65.625$

Final Answer: The modal lifetime of the components is 65.625 hours.

Given: The distribution of total monthly household expenditure of 200 families.

To Find: The modal and mean monthly expenditure.

Part 1: Finding the Mode The maximum frequency is 40, which corresponds to the class interval 1500-2000. So, the modal class is 1500-2000.

We have: $l = 1500$ $f_1 = 40$ $f_0 = 24$ $f_2 = 33$ $h = 500$

Formula for Mode: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$ $\text{Mode} = 1500 + \left(\frac{40 - 24}{2(40) - 24 - 33}\right) \times 500$ $\text{Mode} = 1500 + \left(\frac{16}{80 - 57}\right) \times 500$ $\text{Mode} = 1500 + \left(\frac{16}{23}\right) \times 500$ $\text{Mode} = 1500 + \frac{8000}{23}$ $\text{Mode} = 1500 + 347.826...$ $\text{Mode} \approx 1847.83$

Part 2: Finding the Mean We will use the step-deviation method. Let the assumed mean $a = 2750$. $h = 500$.

$\Sigma f_i = 200$, $\Sigma f_i u_i = (-72-80-33) + (30+44+48+28) = -185 + 150 = -35$. $a = 2750$, $h = 500$.

Formula for Mean: $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$ $\bar{x} = 2750 + \left(\frac{-35}{200}\right) \times 500$ $\bar{x} = 2750 - \frac{35 \times 5}{2}$ $\bar{x} = 2750 - \frac{175}{2}$ $\bar{x} = 2750 - 87.5$ $\bar{x} = 2662.5$

Final Answer: The modal monthly expenditure is ₹ 1847.83. The mean monthly expenditure is ₹ 2662.50.

Given: State-wise teacher-student ratio.

To Find: The mode and mean of the data and interpret them.

Part 1: Finding the Mode The maximum frequency is 10, corresponding to the class interval 30-35. So, the modal class is 30-35.

We have: $l = 30$ $f_1 = 10$ $f_0 = 9$ $f_2 = 3$ $h = 5$

Formula for Mode: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$ $\text{Mode} = 30 + \left(\frac{10 - 9}{2(10) - 9 - 3}\right) \times 5$ $\text{Mode} = 30 + \left(\frac{1}{20 - 12}\right) \times 5$ $\text{Mode} = 30 + \frac{5}{8}$ $\text{Mode} = 30 + 0.625 = 30.625$

Part 2: Finding the Mean We will use the assumed mean method. Let $a = 32.5$.

$\Sigma f_i = 35$, $\Sigma f_i d_i = (-45-80-45) + (15+40) = -170 + 55 = -115$. $a = 32.5$.

Formula for Mean: $\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$ $\bar{x} = 32.5 + \frac{-115}{35}$ $\bar{x} = 32.5 - \frac{23}{7}$ $\bar{x} \approx 32.5 - 3.2857$ $\bar{x} \approx 29.2143$

Interpretation: The mode is approximately 30.6. This means that most states/U.T. have a teacher-student ratio of about 30.6 students per teacher. The mean is approximately 29.2. This means that on average, there are about 29.2 students per teacher across all states/U.T.

Final Answer: Mode $\approx 30.6$ Mean $\approx 29.2$ Interpretation: Most states have a student-teacher ratio of 30.6, while the average ratio for the country is 29.2.

Given: The distribution of runs scored by top batsmen.

To Find: The mode of the data.

Solution: The maximum frequency is 18, which corresponds to the class interval 4000-5000. So, the modal class is 4000-5000.

We have: Lower limit of the modal class, $l = 4000$ Frequency of the modal class, $f_1 = 18$ Frequency of the class preceding the modal class, $f_0 = 4$ Frequency of the class succeeding the modal class, $f_2 = 9$ Class size, $h = 1000$

Formula for Mode: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values: $\text{Mode} = 4000 + \left(\frac{18 - 4}{2(18) - 4 - 9}\right) \times 1000$ $\text{Mode} = 4000 + \left(\frac{14}{36 - 13}\right) \times 1000$ $\text{Mode} = 4000 + \left(\frac{14}{23}\right) \times 1000$ $\text{Mode} = 4000 + \frac{14000}{23}$ $\text{Mode} = 4000 + 608.695...$ $\text{Mode} \approx 4608.7$

Final Answer: The mode of the data is approximately 4608.7 runs.

Given: The number of cars passing through a spot for 100 periods.

To Find: The mode of the data.

Solution: The maximum frequency is 20, which corresponds to the class interval 40-50. So, the modal class is 40-50.

We have: Lower limit of the modal class, $l = 40$ Frequency of the modal class, $f_1 = 20$ Frequency of the class preceding the modal class, $f_0 = 12$ Frequency of the class succeeding the modal class, $f_2 = 11$ Class size, $h = 10$

Formula for Mode: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values: $\text{Mode} = 40 + \left(\frac{20 - 12}{2(20) - 12 - 11}\right) \times 10$ $\text{Mode} = 40 + \left(\frac{8}{40 - 23}\right) \times 10$ $\text{Mode} = 40 + \left(\frac{8}{17}\right) \times 10$ $\text{Mode} = 40 + \frac{80}{17}$ $\text{Mode} = 40 + 4.705...$ $\text{Mode} \approx 44.7$

Final Answer: The mode of the data is approximately 44.7 cars.

Given: The monthly electricity consumption of 68 consumers.

To Find: The median, mean, and mode of the data and compare them.

First, let's create a comprehensive table for all calculations. Total number of consumers, $n = \Sigma f_i = 68$. Class size, $h = 20$. Let's choose assumed mean $a = 135$.

1. Calculation of Mean Using the step-deviation method: $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$ $\bar{x} = 135 + \left(\frac{7}{68}\right) \times 20$ $\bar{x} = 135 + \frac{140}{68} = 135 + \frac{35}{17}$ $\bar{x} \approx 135 + 2.0588 \approx 137.06$

2. Calculation of Mode The maximum frequency is 20, so the modal class is 125-145. $l = 125$, $f_1 = 20$, $f_0 = 13$, $f_2 = 14$, $h = 20$. $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$ $\text{Mode} = 125 + \left(\frac{20 - 13}{2(20) - 13 - 14}\right) \times 20$ $\text{Mode} = 125 + \left(\frac{7}{40 - 27}\right) \times 20$ $\text{Mode} = 125 + \frac{140}{13} \approx 125 + 10.769 \approx 135.77$

3. Calculation of Median Here, $n = 68$, so $\frac{n}{2} = 34$. The cumulative frequency just greater than 34 is 42, which corresponds to the class 125-145. This is the median class. $l = 125$, $\frac{n}{2} = 34$, $cf = 22$ (cumulative frequency of the preceding class), $f = 20$, $h = 20$. $\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$ $\text{Median} = 125 + \left(\frac{34 - 22}{20}\right) \times 20$ $\text{Median} = 125 + \frac{12}{20} \times 20 = 125 + 12 = 137$

Comparison: The three measures of central tendency are close to each other: Mean $\approx 137.06$ Median $= 137$ Mode $\approx 135.77$ This indicates that the distribution is nearly symmetrical.

Final Answer: Mean = 137.06 units Median = 137 units Mode = 135.77 units The values are very close, indicating a fairly symmetric distribution of electricity consumption.

Given: A frequency distribution with median = 28.5 and total frequency = 60.

To Find: The values of the missing frequencies $x$ and $y$.

Solution: First, we create the cumulative frequency table.

From the table, the sum of frequencies is $45 + x + y$. We are given that the total frequency is 60. So, $45 + x + y = 60$ $x + y = 60 - 45$ $x + y = 15$ ---(1)