StatisticsClass 11 Mathematics NCERT Solutions

Given: The data is $4, 7, 8, 9, 10, 12, 13, 17$. Number of observations, $n = 8$.

To Find: The mean deviation about the mean.

Solution: Step 1: Calculate the mean ($\bar{x}$). $\bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$

Step 2: Find the deviations of each observation from the mean, i.e., $x_i - \bar{x}$. The deviations are: $4-10 = -6$ $7-10 = -3$ $8-10 = -2$ $9-10 = -1$ $10-10 = 0$ $12-10 = 2$ $13-10 = 3$ $17-10 = 7$

Step 3: Find the absolute values of the deviations, i.e., $|x_i - \bar{x}|$. The absolute deviations are $6, 3, 2, 1, 0, 2, 3, 7$.

Step 4: Find the mean of the absolute values of the deviations. $\text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{8} |x_i - \bar{x}|}{n}$ $\text{M.D.}(\bar{x}) = \frac{6+3+2+1+0+2+3+7}{8} = \frac{24}{8} = 3$

Final Answer: The mean deviation about the mean is 3.

Given: The data is $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$. Number of observations, $n = 10$.

To Find: The mean deviation about the mean.

Solution: Step 1: Calculate the mean ($\bar{x}$). $\bar{x} = \frac{38+70+48+40+42+55+63+46+54+44}{10} = \frac{500}{10} = 50$

Step 2: Find the absolute deviations of each observation from the mean, i.e., $|x_i - \bar{x}|$. The absolute deviations are: $|38-50| = 12$ $|70-50| = 20$ $|48-50| = 2$ $|40-50| = 10$ $|42-50| = 8$ $|55-50| = 5$ $|63-50| = 13$ $|46-50| = 4$ $|54-50| = 4$ $|44-50| = 6$

Step 3: Find the sum of the absolute deviations. $\sum_{i=1}^{10} |x_i - \bar{x}| = 12+20+2+10+8+5+13+4+4+6 = 84$

Step 4: Calculate the mean deviation about the mean. $\text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{10} |x_i - \bar{x}|}{n} = \frac{84}{10} = 8.4$

Final Answer: The mean deviation about the mean is 8.4.

Given: The data is $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$. Number of observations, $n = 12$.

To Find: The mean deviation about the median.

Solution: Step 1: Arrange the data in ascending order. $10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$

Step 2: Calculate the median (M). Since $n=12$ (even), the median is the mean of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations. $(\frac{12}{2})^{th} = 6^{th}$ observation = 13 $(\frac{12}{2}+1)^{th} = 7^{th}$ observation = 14 $\text{Median (M)} = \frac{13+14}{2} = 13.5$

Step 3: Find the absolute deviations of each observation from the median, i.e., $|x_i - M|$. The absolute deviations are: $|10 - 13.5| = 3.5$ $|11 - 13.5| = 2.5$ $|11 - 13.5| = 2.5$ $|12 - 13.5| = 1.5$ $|13 - 13.5| = 0.5$ $|13 - 13.5| = 0.5$ $|14 - 13.5| = 0.5$ $|16 - 13.5| = 2.5$ $|16 - 13.5| = 2.5$ $|17 - 13.5| = 3.5$ $|17 - 13.5| = 3.5$ $|18 - 13.5| = 4.5$

Step 4: Find the sum of the absolute deviations. $\sum_{i=1}^{12} |x_i - M| = 3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5 = 28$

Step 5: Calculate the mean deviation about the median. $\text{M.D.}(M) = \frac{\sum_{i=1}^{12} |x_i - M|}{n} = \frac{28}{12} \approx 2.33$

Final Answer: The mean deviation about the median is approximately 2.33.

Given: The data is $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$. Number of observations, $n = 10$.

To Find: The mean deviation about the median.

Solution: Step 1: Arrange the data in ascending order. $36, 42, 45, 46, 46, 49, 51, 53, 60, 72$

Step 2: Calculate the median (M). Since $n=10$ (even), the median is the mean of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations. $(\frac{10}{2})^{th} = 5^{th}$ observation = 46 $(\frac{10}{2}+1)^{th} = 6^{th}$ observation = 49 $\text{Median (M)} = \frac{46+49}{2} = 47.5$

Step 3: Find the absolute deviations of each observation from the median, i.e., $|x_i - M|$. The absolute deviations are: $|36 - 47.5| = 11.5$ $|42 - 47.5| = 5.5$ $|45 - 47.5| = 2.5$ $|46 - 47.5| = 1.5$ $|46 - 47.5| = 1.5$ $|49 - 47.5| = 1.5$ $|51 - 47.5| = 3.5$ $|53 - 47.5| = 5.5$ $|60 - 47.5| = 12.5$ $|72 - 47.5| = 24.5$

Step 4: Find the sum of the absolute deviations. $\sum_{i=1}^{10} |x_i - M| = 11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5 = 70$

Step 5: Calculate the mean deviation about the median. $\text{M.D.}(M) = \frac{\sum_{i=1}^{10} |x_i - M|}{n} = \frac{70}{10} = 7$

Final Answer: The mean deviation about the median is 7.

Given: A discrete frequency distribution.

To Find: The mean deviation about the mean.

Solution: We create a table to perform the calculations.

Step 1: Calculate the mean ($\bar{x}$). $N = \sum f_i = 25$ $\sum f_i x_i = 350$ $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{350}{25} = 14$

Step 2: Calculate the absolute deviations from the mean and the required sum. We extend the table:

| $x_i$ | $f_i$ | $f_i x_i$ | $|x_i - \bar{x}|$ = $|x_i - 14|$ | $f_i |x_i - \bar{x}|$ | |---|---|---|---|---| | 5 | 7 | 35 | 9 | $7 \times 9 = 63$ | | 10 | 4 | 40 | 4 | $4 \times 4 = 16$ | | 15 | 6 | 90 | 1 | $6 \times 1 = 6$ | | 20 | 3 | 60 | 6 | $3 \times 6 = 18$ | | 25 | 5 | 125 | 11 | $5 \times 11 = 55$ | | Total | N=25 | 350 | | $\sum f_i |x_i - \bar{x}| = 158$ |

Step 3: Calculate the mean deviation about the mean. $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32$

Final Answer: The mean deviation about the mean is 6.32.

Given: A discrete frequency distribution.

To Find: The mean deviation about the mean.

Solution: We create a table to perform the calculations.

Step 1: Calculate the mean ($\bar{x}$). $N = \sum f_i = 80$ $\sum f_i x_i = 4000$ $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{4000}{80} = 50$

Step 2: Calculate the absolute deviations from the mean and the required sum. We extend the table:

| $x_i$ | $f_i$ | $f_i x_i$ | $|x_i - \bar{x}|$ = $|x_i - 50|$ | $f_i |x_i - \bar{x}|$ | |---|---|---|---|---| | 10 | 4 | 40 | 40 | $4 \times 40 = 160$ | | 30 | 24 | 720 | 20 | $24 \times 20 = 480$ | | 50 | 28 | 1400 | 0 | $28 \times 0 = 0$ | | 70 | 16 | 1120 | 20 | $16 \times 20 = 320$ | | 90 | 8 | 720 | 40 | $8 \times 40 = 320$ | | Total | N=80 | 4000 | | $\sum f_i |x_i - \bar{x}| = 1280$ |

Step 3: Calculate the mean deviation about the mean. $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{1280}{80} = 16$

Final Answer: The mean deviation about the mean is 16.

Given: A discrete frequency distribution. The observations $x_i$ are already in ascending order.

To Find: The mean deviation about the median.

Solution: Step 1: Find the cumulative frequencies (c.f.) to locate the median.

Total number of observations, $N = \sum f_i = 26$. This is an even number.

Step 2: Calculate the median (M). The median is the mean of the $(\frac{N}{2})^{th}$ and $(\frac{N}{2}+1)^{th}$ observations. $(\frac{26}{2})^{th} = 13^{th}$ observation. $(\frac{26}{2}+1)^{th} = 14^{th}$ observation. From the cumulative frequency table, both the 13th and 14th observations are 7. So, Median (M) = $\frac{7+7}{2} = 7$.

Step 3: Calculate the absolute deviations from the median and the required sum. We create a new table for this calculation:

| $x_i$ | $f_i$ | $|x_i - M|$ = $|x_i - 7|$ | $f_i |x_i - M|$ | |---|---|---|---| | 5 | 8 | 2 | $8 \times 2 = 16$ | | 7 | 6 | 0 | $6 \times 0 = 0$ | | 9 | 2 | 2 | $2 \times 2 = 4$ | | 10 | 2 | 3 | $2 \times 3 = 6$ | | 12 | 2 | 5 | $2 \times 5 = 10$ | | 15 | 6 | 8 | $6 \times 8 = 48$ | | Total | N=26 | | $\sum f_i |x_i - M| = 84$ |

Step 4: Calculate the mean deviation about the median. $\text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N} = \frac{84}{26} \approx 3.23$

Final Answer: The mean deviation about the median is approximately 3.23.

Given: A discrete frequency distribution. The observations $x_i$ are already in ascending order.

To Find: The mean deviation about the median.

Solution: Step 1: Find the cumulative frequencies (c.f.) to locate the median.

Total number of observations, $N = \sum f_i = 29$. This is an odd number.

Step 2: Calculate the median (M). The median is the $(\frac{N+1}{2})^{th}$ observation. $(\frac{29+1}{2})^{th} = 15^{th}$ observation. From the cumulative frequency table, the cumulative frequency just greater than 14 is 21, which corresponds to $x_i = 30$. However, the 15th observation is the first one in the group with $x_i=30$. The 14th observation is 27. The 15th observation corresponds to $x_i=30$. Correction: The 9th to 14th observations are 27. The 15th to 21st observations are 30. So the 15th observation is 30. Median (M) = 30.

Step 3: Calculate the absolute deviations from the median and the required sum. We create a new table for this calculation:

| $x_i$ | $f_i$ | $|x_i - M|$ = $|x_i - 30|$ | $f_i |x_i - M|$ | |---|---|---|---| | 15 | 3 | 15 | $3 \times 15 = 45$ | | 21 | 5 | 9 | $5 \times 9 = 45$ | | 27 | 6 | 3 | $6 \times 3 = 18$ | | 30 | 7 | 0 | $7 \times 0 = 0$ | | 35 | 8 | 5 | $8 \times 5 = 40$ | | Total | N=29 | | $\sum f_i |x_i - M| = 148$ |

Step 4: Calculate the mean deviation about the median. $\text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N} = \frac{148}{29} \approx 5.1$

Final Answer: The mean deviation about the median is approximately 5.1.

Given: A continuous frequency distribution.

To Find: The mean deviation about the mean.

Solution: We create a table to perform the calculations. We first find the mid-point ($x_i$) for each class.

Step 1: Calculate the mean ($\bar{x}$). $N = \sum f_i = 50$ $\sum f_i x_i = 17900$ $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{17900}{50} = 358$

Step 2: Calculate the absolute deviations from the mean and the required sum. We extend the table:

| $x_i$ | $f_i$ | $|x_i - \bar{x}|$ = $|x_i - 358|$ | $f_i |x_i - \bar{x}|$ | |---|---|---|---| | 50 | 4 | 308 | 1232 | | 150 | 8 | 208 | 1664 | | 250 | 9 | 108 | 972 | | 350 | 10 | 8 | 80 | | 450 | 7 | 92 | 644 | | 550 | 5 | 192 | 960 | | 650 | 4 | 292 | 1168 | | 750 | 3 | 392 | 1176 | | Total | N=50 | | $\sum f_i |x_i - \bar{x}| = 7896$ |

Step 3: Calculate the mean deviation about the mean. $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{7896}{50} = 157.92$

Final Answer: The mean deviation about the mean is 157.92.

Given: A continuous frequency distribution.

To Find: The mean deviation about the mean.

Solution: We create a table to perform the calculations. We find the mid-point ($x_i$) for each class.

Step 1: Calculate the mean ($\bar{x}$). $N = \sum f_i = 100$ $\sum f_i x_i = 12530$ $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{12530}{100} = 125.3$

Step 2: Ca