UP Board Solutions Class 11 Maths Chapter 15 Statistics

UP Board Solutions For Class 11 Maths Chapter 15 Statistics (सांख्यिकी)

प्रश्नावली 15.1

प्रश्न 1 व 2 में दिए गए आँकड़ों के लिए माध्य के सापेक्ष विचलन ज्ञात कीजिए:

 

Question 1. 4, 7, 8, 9, 10, 12, 13, 17.
Answer: हल : समांतर माध्य \( \bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10 \) \[ \sum_{i} |x_i - \bar{x}| = |4 - 10| + |7 - 10| + |8 - 10| + |9 - 10| \] \[ + |10 - 10| + |12 - 10| + |13 - 10| + |17 - 10| \] \[ = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24 \] \[ \therefore \text{माध्य के सापेक्ष माध्य विचलन MD} (\bar{x}) = \frac{\Sigma|x_i-\bar{x}|}{n} = \frac{24}{8} = 3. \]In simple words: To find the mean deviation about the mean, first calculate the mean of the given data. Then, find the absolute difference between each data point and the mean. Finally, sum these absolute differences and divide by the total number of data points.

🎯 Exam Tip: Remember to calculate the arithmetic mean accurately before proceeding to find the absolute deviations. A small error in the mean will propagate throughout the calculation.

 

Question 2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Answer: हल : समांतर माध्य, \( \bar{x} = \frac{38+70+48+40+42+55+63+46+54+44}{10} \) \[ = \frac{500}{10} = 50 \] \[ \Sigma|x_i - \bar{x}| = |38 - 50| + |70 - 50| + |48 - 50| + |40 - 50| \] \[ + |42 - 50| + |55 - 50| + |63 - 50| + |46 - 50| + |54 - 50| + |44 - 50| \] \[ = 12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 = 84 \] \[ \therefore \text{MD} (\bar{x}) = \frac{\Sigma|x_i-\bar{x}|}{n} = \frac{84}{10} = 8.4. \]In simple words: First, find the mean of all the numbers. Then, calculate how far each number is from the mean (ignoring negative signs). Add up these distances and divide by the total count of numbers to get the mean deviation.

🎯 Exam Tip: Always double-check your arithmetic when summing large sets of numbers, as a minor calculation error can lead to an incorrect mean and subsequent deviations.

प्रश्न 3 व 4 के आँकड़ों के लिए माध्यिका के सापेक्ष माध्य विचलन ज्ञात कीजिए:

 

Question 3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
Answer: हलः आँकड़ों को आरोही क्रम में लिखने पर 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 \( n = 12 \)
\( \frac{12}{2} = \text{6वाँ पद} = 13 \) और \( \text{7वाँ पद} = 14 \) माध्यिका \( M = \frac{13+14}{2} = \frac{27}{2} = 13.5 \) \[ \Sigma|x_i - M| = |10 - 13.5| + |11 - 13.5| + |11 - 13.5| + |12 - 13.5| \] \[ + |13 - 13.5| + |13 - 13.5| + |14 - 13.5| + |16 - 13.5| \] \[ + |16 - 13.5| + |17 - 13.5| + |17 - 13.5| + |18 - 13.5| \] \[ = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 \] \[ = 28 \] \[ \therefore \text{माध्य विचलन (M)} = \frac{\Sigma|x_i - M|}{n} = \frac{28}{12} = 2.33. \]In simple words: First, arrange the numbers in increasing order and find the median. Then, calculate the absolute difference between each number and the median. Sum these absolute differences and divide by the total count to get the mean deviation about the median.

🎯 Exam Tip: For median calculation, always sort the data first. If the number of observations (n) is even, the median is the average of the two middle terms.

 

Question 4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Answer: हल : दिए हुए आँकड़ों को आरोही क्रम में लिखने पर 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 \( n = 10 \)
\( \frac{10}{2} = \text{5वाँ पद} = 46 \), और \( 5 + 1 = \text{6वाँ पद} = 49 \) माध्यिका \( M = \frac{46+49}{2} = \frac{95}{2} = 47.5 \) \[ \Sigma|x_i - M| = |36 - 47.5| + |42 - 47.5| + |45 - 47.5| + |46 - 47.5| + |46 - 47.5| \] \[ + |49 - 47.5| + |51 - 47.5| + |53 - 47.5| + |60 - 47.5| + |72 - 47.5| \] \[ = 11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5 \] \[ = 70 \] माध्य विचलन \( (\text{M}) = \frac{\Sigma|x_i - M|}{n} = \frac{70}{10} = 7. \)In simple words: Arrange the data in ascending order to find the median. Then, calculate the absolute difference of each data point from the median. Sum these differences and divide by the total count to get the mean deviation from the median.

🎯 Exam Tip: Pay close attention to sorting data when calculating the median. Any misplacement of values will lead to an incorrect median and subsequent errors in mean deviation.

प्रश्न 5 व 6 के आँकड़ों के लिए माध्य के सापेक्ष माध्य विचलन ज्ञात कीजिए:

 

Question 5.

Answer: हलः\[ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{350}{25} = 14 \] \[ \text{माध्य विचलन} = \frac{\Sigma f_i |x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32. \]In simple words: For grouped data, calculate the mean by multiplying each value by its frequency, summing these products, and dividing by the total frequency. Then, for each value, find the absolute difference from the mean, multiply by its frequency, sum these products, and divide by the total frequency to get the mean deviation.

🎯 Exam Tip: When dealing with frequency distributions, ensure you calculate the weighted mean (\( \Sigma f_i x_i / \Sigma f_i \)) correctly before computing deviations, as this is the foundational step.

 

Question 6.

Answer: हलः\[ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4000}{80} = 50 \] \[ \text{माध्य विचलन} = \frac{\Sigma f_i |x_i - \bar{x}|}{N} = \frac{1280}{80} = 16. \]In simple words: First, find the mean of the data by multiplying each value by its frequency, summing these, and dividing by the total frequency. Then, for each data point, calculate its absolute difference from the mean, multiply by its frequency, sum these products, and divide by the total frequency to get the mean deviation.

🎯 Exam Tip: When computing mean deviation for frequency distributions, careful organization of the calculation table is crucial to avoid errors in summing \(f_i x_i\) and \(f_i |x_i - \bar{x}|\).

प्रश्न 7 व 8 के आँकड़ों के लिए माध्यिका के सापेक्ष माध्य विचलन ज्ञात कीजिए:

 

Question 7.

Answer: हल :बारंबारता का योग \( N = 26 \)
\( \frac{26}{2} = \text{13वाँ पद} = 7 \) और \( \text{14वाँ पद} = 7 \)
\( \therefore \text{माध्यिका} = \frac{7+7}{2} = 7 \) \[ \therefore \text{माध्यिका के सापेक्ष विचलन (M)} = \frac{\Sigma f_i |x_i - M|}{N} = \frac{84}{26} = 3.23. \]In simple words: First, find the median of the frequency distribution by identifying the value corresponding to the middle position. Then, calculate the absolute difference of each value from the median, multiply by its frequency, sum these products, and divide by the total frequency.

🎯 Exam Tip: For discrete frequency distributions, calculate the cumulative frequency (c.f.) to find the median. The median is the value of \(x_i\) whose c.f. is just greater than or equal to \(N/2\).

 

Question 8.

Answer: हलःमाध्यिका \( M = \frac{29+1}{2} = \text{15वाँ पद} = 30 \) \[ \text{M.D. (M)} = \frac{\Sigma f_i |x_i - M|}{N} = \frac{148}{29} = 5.1. \]In simple words: First, calculate the cumulative frequency and find the median. Then, for each data point, determine its absolute difference from the median, multiply by its frequency, sum these products, and divide by the total frequency to obtain the mean deviation.

🎯 Exam Tip: When \(N\) is odd for a discrete frequency distribution, the median is the value corresponding to the \((N+1)/2\)-th observation. Calculate cumulative frequencies accurately.

प्रश्न 9 व 10 के आँकड़ों के लिए मध्य के सापेक्ष माध्य विचलन ज्ञात कीजिए ।

 

Question 9.

Answer: हल : माना \( a = 350, h = 100, d_i = \frac{x_i - 350}{100} \)\[ \bar{x} = a + \frac{\Sigma f_i d_i}{N} \times h \] \[ = 350 + \frac{4}{50} \times 100 \] \[ = 358 \] \[ \text{माध्य विचलन} = \frac{\Sigma f_i |x_i - \bar{x}|}{N} = \frac{7856}{50} = 157.92. \]In simple words: For grouped data, first find the midpoint of each class. Calculate the mean using these midpoints and their frequencies, often using the step-deviation method for easier calculation. Then, for each midpoint, find its absolute difference from the mean, multiply by its frequency, sum these products, and divide by the total frequency.

🎯 Exam Tip: When using the step-deviation method, ensure accurate calculation of \(d_i\) and the mean. The \(|x_i - \bar{x}|\) column must use the actual mean, not the assumed mean.

 

Question 10.

Answer: हलः \[ d_i = \frac{x_i - 130}{10} \]\[ \text{माध्य} \bar{x} = a + \left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right) \times h \] \[ = 130 + \left(\frac{-47}{100}\right) \times 10 \] \[ = 130 - 4.7 = 125.3 \] \[ \text{माध्य विचलन} = \frac{\Sigma f_i |x_i - \bar{x}|}{N} = \frac{1128.8}{100} = 11.288. \]In simple words: First, calculate the midpoints for each class interval. Then, use an assumed mean and step deviation to find the true mean. After finding the true mean, calculate the absolute difference of each midpoint from the true mean, multiply by its frequency, sum these products, and divide by the total frequency to get the mean deviation.

🎯 Exam Tip: For continuous frequency distributions, correctly calculating midpoints \((x_i)\) and applying the step deviation formula to find the mean \(( \bar{x} )\) are critical. Make sure class intervals are inclusive or exclusive as appropriate.

 

Question 11. निम्नलिखित आँकड़ों के लिए माध्यिका के सापेक्ष माध्य विचलन ज्ञात कीजिए:

Answer: हल:\( N = 50 \) माध्यिका वर्ग: \(\frac{N}{2} = \frac{50}{2} = 25\). The class 20-30 has c.f. 28, which is just greater than 25. So, median class is 20-30. \( l = 20, f = 14, C = 14, h = 10 \) \[ \text{माध्यिका} = l + \frac{\frac{N}{2} - C}{f} \times h \] \[ = 20 + \frac{25 - 14}{14} \times 10 \] \[ = 20 + \frac{11}{14} \times 10 = 20 + 7.86 = 27.86 \] \[ \text{माध्य विचलन (M)} = \frac{\Sigma f_i |x_i - M|}{\Sigma f_i} = \frac{517.16}{50} = 10.34 \]In simple words: For grouped data, find the median class using cumulative frequencies. Calculate the median using the formula. Then, find the midpoint of each class, calculate its absolute difference from the median, multiply by frequency, sum these, and divide by total frequency.

🎯 Exam Tip: When calculating the median for continuous frequency distributions, remember the formula involves the lower limit of the median class (\(l\)), cumulative frequency of the preceding class (\(C\)), frequency of the median class (\(f\)), and class size (\(h\)).

 

Question 12. नीचे दिए गए 100 व्यक्तियों की आयु के बंटन की माध्यिका आयु के सापेक्ष माध्य विचलन की गणना कीजिए:

Answer: हल : दिए गए आँकड़ों की सतत बारंबारता बंटन में बदलते हुए :माध्यिका वर्ग : 35.5-40.5, \(l = 35.5, h = 5, C = 37, f = 26 \) \[ \text{माध्यिका} = l + \frac{\frac{N}{2} - C}{f} \times h \] \[ = 35.5 + \frac{50 - 37}{26} \times 5 \] \[ = 35.5 + \frac{13}{26} \times 5 = 35.5 + 2.5 = 38 \] \[ \text{माध्य विचलन (M)} = \frac{\Sigma f_i |x_i - M|}{N} = \frac{735}{100} = 7.35 \]In simple words: First, convert the data into continuous classes if needed. Calculate cumulative frequencies to find the median class and then the median. Then, for each class midpoint, calculate its absolute difference from the median, multiply by its frequency, sum these products, and divide by the total frequency.

🎯 Exam Tip: When class intervals are discontinuous (e.g., 16-20, 21-25), convert them into continuous classes (e.g., 15.5-20.5, 20.5-25.5) before calculating the median for accuracy.

प्रश्नावली 15.2

प्रश्न 1 से 5 तक के लिए आँकड़ों के लिए माध्य व प्रसरण ज्ञात कीजिए।

 

Question 1. 6, 7, 10, 12, 13, 4, 8, 12.
Answer: हल: \[ \text{माध्य} \bar{x} = \frac{\Sigma x_i}{n} \] \[ = \frac{6+7+10+12+13+4+8+12}{8} \] \[ = \frac{72}{8} = 9. \]

\[ \text{प्रसरण} = \frac{\Sigma (x_i - \bar{x})^2}{n} = \frac{74}{8} = 9.25. \]In simple words: First, find the mean of the given numbers. Then, for each number, subtract the mean and square the result. Sum these squared differences and divide by the total count to get the variance.

🎯 Exam Tip: Variance measures how spread out the numbers are. Ensure the deviations from the mean are squared correctly, as this is the key difference between mean deviation and variance.

 

Question 2. प्रथम n प्राकृत संख्याएँ।
Answer: हल : पहली n प्राकृत संख्याएँ: 1, 2, 3, ...., n \[ \text{माध्य} \bar{x} = \frac{1+2+3+....+n}{n} = \frac{1}{n} \frac{n(n+1)}{2} \] \[ = \frac{n+1}{2} \] \[ [\text{पहली n प्राकृत संख्याओं का योग} \frac{n(n+1)}{2}] \] \[ \Sigma x_i^2 = 1^2 + 2^2 + 3^2 + ..... + n^2 \] \[ = \frac{n(n+1)(2n+1)}{6} \] \[ \text{प्रसरण} = \frac{\Sigma(x_i - \bar{x})^2}{n} = \frac{1}{n^2} [n \Sigma x_i^2 - (\Sigma x_i)^2] \] \[ = \frac{1}{n^2} \left[ n \frac{n(n+1)(2n+1)}{6} - \left(\frac{n(n+1)}{2}\right)^2 \right] \] \[ = \frac{1}{n^2} \left[ \frac{n^2(n+1)(2n+1)}{6} - \frac{n^2(n+1)^2}{4} \right] \] \[ = \frac{(n+1)}{n^2} \left[ \frac{n^2(2n+1)}{6} - \frac{n^2(n+1)}{4} \right] \] \[ = \frac{n^2(n+1)}{n^2} \left[ \frac{2n+1}{6} - \frac{n+1}{4} \right] \] \[ = (n+1) \left[ \frac{2(2n+1) - 3(n+1)}{12} \right] \] \[ = (n+1) \left[ \frac{4n+2 - 3n-3}{12} \right] \] \[ = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}. \]In simple words: For the first 'n' natural numbers, the mean is \((n+1)/2\). The variance is calculated by taking the sum of squares of these numbers, subtracting the square of the sum of numbers, dividing by 'n', and then dividing by 'n' again to get the average squared deviation. This simplifies to \((n^2-1)/12\).

🎯 Exam Tip: Remember the formulas for the sum of the first n natural numbers and the sum of their squares. These are crucial shortcuts for solving problems involving sequences of natural numbers efficiently.

 

Question 3. 3 के प्रथम 10 गुणज ।
Answer: हलः प्रथम दस 3 के गुणज : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Let \( A = 15, h = 3 \). Then \( y_i = \frac{x_i - A}{h} = \frac{x_i - 15}{3} \)

\[ \text{माध्य} \bar{x} = A + \frac{\Sigma y_i}{n} \times h \] \[ = 15 + \frac{5}{10} \times 3 \] \[ = 15 + 1.5 = 16.5. \] \[ \text{प्रसरण}, \sigma^2 = \frac{h^2}{n^2} [n \Sigma y_i^2 - (\Sigma y_i)^2] \] \[ = \frac{3^2}{10^2} [10 \times 85 - (5)^2] \] \[ = \frac{9}{100} [850 - 25] \] \[ = \frac{9 \times 825}{100} \] \[ = \frac{7425}{100} = 74.25 \] अतः माध्य = 16.5, प्रसरण = 74.25.In simple words: The first 10 multiples of 3 are 3, 6, ..., 30. Using a shortcut method (step deviation) by assuming a mean and finding coded values, the true mean is found. The variance is then calculated using the formula that incorporates the assumed mean, class width, and the sum of coded values and their squares.

🎯 Exam Tip: For arithmetic progressions or data with a common difference, the step-deviation method simplifies calculations significantly. Ensure proper application of the variance formula involving \(h\), \(n\), \(\Sigma y_i\), and \(\Sigma y_i^2\).

 

Question 4.

Answer: हल:\[ \text{माध्य} \bar{x} = \frac{760}{40} = 19 \] \[ \text{प्रसरण} \sigma^2 = \frac{\Sigma f_i (x_i - \bar{x})^2}{N} \] \[ = \frac{1736}{40} = 43.4 \] अतः माध्य = 19, प्रसरण = 43.4.In simple words: First, calculate the mean by multiplying each value by its frequency, summing these products, and dividing by the total frequency. Then, find the difference between each value and the mean, square it, multiply by its frequency, sum these results, and divide by the total frequency to get the variance.

🎯 Exam Tip: For frequency distributions, variance calculation requires careful computation of \(f_i x_i\) for the mean, and then \(f_i (x_i - \bar{x})^2\) for the sum of squared deviations. Organize your table clearly to avoid mistakes.

Question 5.

Answer:
हलः \[ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{350}{25} = 14 \] माध्य विचलन = \[ \frac{\Sigma f_i|x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32. \]
In simple words: First, calculate the mean (\(\bar{x}\)) of the given data using the formula \(\Sigma f_i x_i / \Sigma f_i\). Then, find the absolute deviation of each data point from the mean, multiply by its frequency, sum these values, and divide by the total frequency to get the mean deviation.

🎯 Exam Tip: Remember to use the absolute values of deviations ( \(|x_i - \bar{x}|\) ) when calculating mean deviation. A common mistake is to overlook the absolute value, which would lead to an incorrect sum of deviations. Ensure all calculations are accurate, especially when dealing with multiple columns.

 

Question 6.

Answer:
हलः
\[ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4000}{80} = 50 \] माध्य विचलन = \[ \frac{\Sigma f_i|x_i - \bar{x}|}{N} = \frac{1280}{80} = 16. \]
In simple words: To find the mean deviation, first calculate the mean (\(\bar{x}\)) of the grouped data. Then, for each class, find the absolute difference between its midpoint (\(x_i\)) and the mean, multiply by its frequency (\(f_i\)), sum these products, and finally divide by the total number of observations (N).

🎯 Exam Tip: When calculating mean deviation for grouped data, ensure the correct midpoint (\(x_i\)) is used for each class interval. Double-check your summation of \(f_i x_i\) and \(f_i|x_i - \bar{x}|\) as these are critical steps for accuracy. Precise calculation of the mean is foundational for the rest of the problem.

 

Question 7.

Answer:
हल :बारंबारता का योग = 26
\[ \frac{26}{2} = 13 \] वाँ पद = 7 और 14वाँ पद = 7
\[ \text{माध्यिका} = \frac{7+7}{2} = 7 \]
माध्यिका के सापेक्ष विचलन (M) = \[ \frac{\Sigma f_i |x_i - M|}{N} = \frac{84}{26} = 3.23. \]
In simple words: First, arrange the data and calculate the cumulative frequency to find the median (M). Since N=26 is even, the median is the average of the 13th and 14th terms. Then, calculate the absolute deviation of each \(x_i\) from the median, multiply by its frequency, sum these products, and divide by the total frequency to get the mean deviation about the median.

🎯 Exam Tip: When finding the median for an even number of observations in grouped data, correctly identify the two middle terms from the cumulative frequency. A small error in finding the median will propagate through all subsequent calculations. Ensure \(|x_i - M|\) is correctly calculated for each row.

 

Question 8.

Answer:
हलः
\[ \text{माध्यिका} = \frac{29+1}{2} = 15 \] वाँ पद = 30
M.D. (M) = \[ \frac{\Sigma f_i |x_i - M|}{N} = \frac{148}{29} = 5.1. \]
In simple words: First, find the median (M) by calculating the cumulative frequency and identifying the middle term. For N=29, the median is the 15th term. Then, calculate the absolute deviation of each \(x_i\) from the median, multiply by its frequency, sum these products, and divide by the total frequency to get the mean deviation about the median.

🎯 Exam Tip: Accurately identifying the median is paramount for mean deviation calculations. For an odd number of observations (N), the median is the \((N+1)/2\)-th term. Be careful with calculations of \(|x_i - M|\) and their subsequent products with frequency \(f_i\).

 

Question 9.

आय
0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
प्रतिदिन
व्यक्तियों
की संख्या
4
8
9
10
7
5
4
3

Answer:
हल : माना a = 350, h = 100, \(d_i = \frac{x_i - 350}{100}\)\[ \bar{x} = a + \frac{\Sigma f_i d_i}{N} \times h \] \[ = 350 + \frac{4}{50} \times 100 \] \[ = 358 \] माध्य विचलन = \[ \frac{\Sigma f_i|x_i - \bar{x}|}{N} = \frac{7856}{50} = 157.92. \]
In simple words: First, calculate the mean (\(\bar{x}\)) of the grouped data, using the assumed mean method for convenience. Then, for each class, find its midpoint, calculate the absolute deviation from the mean, multiply by the frequency, sum all these values, and finally divide by the total frequency to get the mean deviation.

🎯 Exam Tip: The step-deviation method (using \(d_i\)) is highly efficient for calculating the mean of grouped data, especially with large class intervals. Ensure the class midpoints (\(x_i\)) are accurately determined and that \(|x_i - \bar{x}|\) is correctly calculated for each class before summing. Double-check all arithmetic operations.

 

Question 10.

ऊंचाई (सेमी में) 95-105 105-115 115-125 125-135 135-145 145-155
लड़कों की संख्या
9
13
26
30
12
10

Answer:
हलः\[ \text{माध्य } \bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} \times h \] \[ = 130 + \left(\frac{-47}{100}\right) \times 10 \] \[ = 130 - 4.7 = 125.3 \] माध्य विचलन = \[ \frac{\Sigma f_i|x_i - \bar{x}|}{N} = \frac{1128.8}{100} = 11.288. \]
In simple words: First, calculate the mean (\(\bar{x}\)) of the grouped data, preferably using the step-deviation method with an assumed mean. Then, for each class, find its midpoint, compute the absolute difference from the mean, multiply by its frequency, sum all these values, and finally divide by the total frequency to get the mean deviation.

🎯 Exam Tip: For problems involving grouped data and large numbers, the assumed mean method or step-deviation method is crucial for simplifying calculations. Pay close attention to the signs in the \(f_i d_i\) column and ensure the mean (\(\bar{x}\)) is computed correctly before proceeding to calculate the absolute deviations and the final mean deviation.

 

Question 11.

निम्नलिखित आँकड़ों के लिए माध्यिका के सापेक्ष माध्य विचलन ज्ञात कीजिए:
अंक
0-10 10-20 20-30 30-40 40-50 50-60
लड़कियों
की संख्या
6
8
14
16
4
2

Answer:
हल:
माध्यिका = \(l + \frac{\frac{N}{2} - C}{f} \times h\) \[ = 20 + \frac{25 - 14}{14} \times 10 = 20 + \frac{110}{14} \] \[ = 20 + 7.86 = 27.86 \] माध्य विचलन (M) = \[ \frac{\Sigma f_i |x_i - M|}{\Sigma f_i} = \frac{517.16}{50} = 10.34 \]
In simple words: First, determine the median class from the cumulative frequency and then calculate the median (M) using the formula for grouped data. Next, calculate the midpoint (\(x_i\)) for each class, find the absolute deviation \(|x_i - M|\), multiply by the respective frequency \(f_i\), sum these products, and divide by the total frequency to get the mean deviation about the median.

🎯 Exam Tip: The crucial initial step is to correctly identify the median class and calculate the median (M). Be careful with the formula for the median of grouped data, ensuring correct values for \(l, N, C, f,\) and \(h\). Precision in calculating the median impacts all subsequent deviations. Also, pay attention to decimal places for accuracy.

 

Question 12.

नीचे दिए गए 100 व्यक्तियों की आयु के बंटन की माध्यिका आयु के सापेक्ष माध्य विचलन की गणना कीजिए:
आयु 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55
संख्या 5
6
12
14
26
12
16
9

Answer:
हल : दिए गए आँकड़ों की सतत बारंबारता बंटन में बदलते हुए :माध्यिका वर्ग : 35.5-40.5, \(l = 35.5, h = 5, C = 37, f = 26\)
माध्यिका = \(l + \frac{\frac{N}{2} - C}{f} \times h\) \[ = 35.5 + \left(\frac{50 - 37}{26}\right) \times 5 \] \[ = 35.5 + \frac{13}{26} \times 5 = 35.5 + 2.5 = 38 \] माध्य विचलन (M) = \[ \frac{\Sigma f_i|x_i - M|}{N} = \frac{735}{100} = 7.35 \]
In simple words: First, convert the given data into a continuous frequency distribution. Then, determine the median class and calculate the median (M) using the formula for grouped data. Finally, for each class, find its midpoint, compute the absolute deviation from the median, multiply by its frequency, sum all these products, and divide by the total frequency to find the mean deviation.

🎯 Exam Tip: When dealing with discontinuous class intervals, the first step is always to convert them to continuous intervals. This involves adjusting the lower and upper limits by 0.5. After this, accurately determine the median class and the median. Any error in these initial steps will lead to incorrect results. Also, ensure the value of 'h' (class size) is correct after converting to continuous intervals.

 

Exercise 15.2

Question 1.

6, 7, 10, 12, 13, 4, 8, 12.

Answer:
हलः \[ \text{माध्य } \bar{x} = \frac{\Sigma x_i}{n} \] \[ = \frac{6+7+10+12+13+4+8+12}{8} \] \[ = \frac{72}{8} = 9. \]प्रसरण = \[ \frac{\Sigma (x_i - \bar{x})^2}{n} = \frac{74}{8} = 9.25. \]
In simple words: First, calculate the mean (\(\bar{x}\)) of the given data points. Then, for each data point, subtract the mean and square the result. Sum all these squared differences and divide by the total number of data points to find the variance.

🎯 Exam Tip: Accurately calculating the mean is the first critical step. Subsequently, be careful with the subtractions \( (x_i - \bar{x}) \) and squaring them. A small error in these intermediate calculations will lead to an incorrect variance. Double-check all sums for accuracy.

 

Question 2.

प्रथम n प्राकृत संख्याएँ।

Answer:
हल : पहली n प्राकृत संख्याएँ: 1, 2, 3, ...., n
माध्य \( \bar{x} = \frac{1+2+3+....+ n}{n} = \frac{1}{n} \frac{n(n+1)}{2} = \frac{n+1}{2} \)
[पहली n प्राकृत संख्याओं का योग \( \frac{n(n+1)}{2} \)]
\( \Sigma x_i^2 = 1^2 + 2^2 + 3^2 + ..... + n^2 = \frac{n(n+1)(2n+1)}{6} \)
प्रसरण = \( \frac{\Sigma(x_i-\bar{x})^2}{n} = \frac{1}{n^2} [n\Sigma x_i^2 - (\Sigma x_i)^2] \) \[ = \frac{1}{n^2} \left[n \frac{n(n+1)(2n+1)}{6} - \frac{n^2(n+1)^2}{4}\right] \] \[ = \frac{1}{12} [2(n+1)(2n+1) - 3(n+1)^2] \] \[ = \frac{n+1}{12} [2(2n+1) - 3(n+1)] \] \[ = \frac{n+1}{12} [4n+2 - 3n - 3] \] \[ = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}. \]
In simple words: To find the variance of the first n natural numbers, first use the formulas for the sum of the first n natural numbers and the sum of their squares. Then, apply the alternative formula for variance, \( \frac{1}{n^2} [n\Sigma x_i^2 - (\Sigma x_i)^2] \), and simplify the expression to arrive at \( (n^2 - 1)/12 \).

🎯 Exam Tip: Memorizing the formulas for the sum of the first 'n' natural numbers and the sum of their squares is crucial for solving this type of problem quickly. When deriving the variance formula, be extremely careful with algebraic manipulations and factorization to avoid errors.

 

Question 3.

3 के प्रथम 10 गुणज ।

Answer:
हलः
प्रथम दस 3 के गुणज : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30माध्य \( \bar{x} = A + \frac{\Sigma y_i}{n} \times h \) \[ = 15 + \frac{5}{10} \times 3 \] \[ = 15 + 1.5 = 16.5. \] प्रसरण, \( \sigma^2 = \frac{h^2}{n^2} [n \Sigma y_i^2 - (\Sigma y_i)^2] \) \[ = \frac{9}{100} [10 \times 85 - 25] \] \[ = \frac{9}{100} [850 - 25] \] \[ = \frac{9 \times 825}{100} = \frac{7425}{100} = 74.25 \] अतः माध्य = 16.5, प्रसरण = 74.25.
In simple words: List the first 10 multiples of 3. Use the step-deviation method by assuming a mean (e.g., 15) and a common factor (3) to simplify calculations for \(y_i\). Calculate the mean using \(y_i\) and convert it back to \(\bar{x}\). Then, use the variance formula involving \(y_i\) to find the variance.

🎯 Exam Tip: The step-deviation method is highly recommended for problems with arithmetic progressions or multiples, as it significantly simplifies calculations. Choosing an assumed mean and class width that are easy to work with (like 15 and 3 here) will reduce computational errors. Ensure you apply the \(h^2/n^2\) factor correctly in the variance formula.

 

Question 4.

Answer:
हलःमाध्य \( \bar{x} = \frac{760}{40} = 19 \)
प्रसरण \( \sigma^2 = \frac{\Sigma f_i(x_i - \bar{x})^2}{N} \) \[ = \frac{1736}{40} = 43.4 \] अतः माध्य = 19, प्रसरण = 43.4.
In simple words: First, calculate the mean (\(\bar{x}\)) of the grouped data. Then, for each data point, find the difference from the mean, square it, multiply by its frequency, sum all these products, and finally divide by the total frequency to find the variance.

🎯 Exam Tip: For calculating variance, ensuring that \(x_i\) represents the correct data point and that the mean (\(\bar{x}\)) is accurate is crucial. Pay close attention to the squaring operation \((x_i - \bar{x})^2\) and the subsequent multiplication by \(f_i\). Any miscalculation in these steps will lead to an incorrect variance.

 

Question 5.

Answer:
हल : मान लीजिए कल्पित माध्य A = 98,.. \(y_i = x_i - 98\)माध्य \( \bar{x} = A + \frac{\Sigma f_i y_i}{N} = 98 + \frac{44}{22} \) \[ = 98 + 2 = 100. \] प्रसरण \( \sigma^2 = \frac{1}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{1}{(22)^2} [22 \times 728 - 44 \times 44] \] \[ = \frac{1}{22} [728 - 88] \] \[ = \frac{640}{22} = \frac{320}{11} = 29.09. \]
In simple words: First, calculate the mean using the assumed mean method (A=98) to simplify \(y_i = x_i - A\). Then, calculate the variance using the formula involving \(y_i\), which is \( \frac{1}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \).

🎯 Exam Tip: The assumed mean method simplifies calculations significantly for larger data sets. Ensure the correct formula for variance with assumed mean is used. Double-check the summation of \(f_i y_i\) and \(f_i y_i^2\) as these values are critical for the final variance calculation. Be careful with the squaring term \((\Sigma f_i y_i)^2\).

 

Question 6.

लघु विधि द्वारा माध्य वे मानक विचलन ज्ञात कीजिए:

Answer:
हल : मान लीजिए कल्पित माध्य A = 64
तथा \(y_i = x_i - 64\)माध्य, \( \bar{x} = A + \frac{\Sigma f_i y_i}{N} \) \[ = 64 + 0 = 64. \] प्रसरण, \( \sigma^2 = \frac{1}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{1}{(100)^2} [100 \times 286 - 0] \] \[ = \frac{286}{100} = 2.86 \] मानक विचलन, \( \sigma = \sqrt{2.86} = 1.69. \)
In simple words: First, calculate the mean using the assumed mean method, where \(y_i = x_i - A\). Since \(\Sigma f_i y_i\) is 0, the mean is simply the assumed mean. Then, calculate the variance using the formula \( \frac{1}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \). Finally, the standard deviation is the square root of the variance.

🎯 Exam Tip: When using the assumed mean method, choose A strategically (often a central value of \(x_i\)) to make \(y_i\) values smaller, simplifying calculations. If \(\Sigma f_i y_i\) happens to be zero, the mean is equal to the assumed mean, which is a great shortcut. Remember that standard deviation is always the positive square root of variance.

 

Question 7.

वर्ग :
0-30 30-60 60-90 90-120 120-150 150-180 180-210
बारंबारता :
2
3
5
10
3
5
2

Answer:
हल : माना कल्पित माध्य A = 105, वर्ग अंतराल h = 30
\(y_i = \frac{x_i - A}{h} = \frac{x_i - 105}{30}\)माध्य \( \bar{x} = A + \left(\frac{\Sigma f_i y_i}{N}\right) \times h \) \[ = 105 + \frac{2}{30} \times 30 = 107. \] प्रसरण \( \sigma^2 = \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{30 \times 30}{30 \times 30} [30 \times 76 - 2^2] \] \[ = 2280 - 4 = 2276. \]
In simple words: First, calculate the mean using the step-deviation method with an assumed mean and class interval. Then, use the step-deviation formula for variance, \( \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \), plugging in the calculated values to find the variance.

🎯 Exam Tip: The step-deviation method is particularly useful for grouped frequency distributions with equal class intervals. Correctly calculating \(y_i\), \(f_i y_i\), and \(f_i y_i^2\) is key. Ensure \(h\) is the exact class size and \(N\) is the total frequency. Be meticulous in algebraic simplification to avoid errors in the final variance value.

 

Question 8.

वर्ग :
0-10 10-20 20-30 30-40 40-50
बारंबारता :
5
8
15
16
6

Answer:
हल : माना कल्पित माध्य A = 25, वर्ग अंतराल = 10
\(y_i = \frac{x_i - A}{h} = \frac{x_i - 25}{10}\)माध्य, \( \bar{x} = A + \left(\frac{\Sigma f_i y_i}{N}\right) \times h \) \[ = 25 + \left(\frac{10}{50}\right) \times 10 = 25 + 2 = 27. \] प्रसरण, \( \sigma^2 = \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{100}{2500} [50 \times 68 - 100] \] \[ = \frac{50}{25} [68 - 2] = 2 \times 66 = 132. \]
In simple words: First, calculate the mean using the step-deviation method, assuming a mean of 25 and a class interval of 10. Then, compute the variance using the step-deviation formula, substituting the values obtained from the frequency table.

🎯 Exam Tip: For grouped data, selecting an appropriate assumed mean (A) and class interval (h) can simplify calculations for \(y_i\). Ensure that the value \(h^2/N^2\) is correctly applied in the variance formula, and double-check the calculations for \(N \Sigma f_i y_i^2\) and \((\Sigma f_i y_i)^2\). Accuracy in arithmetic is crucial.

 

Question 9.

लघु विधि द्वारा माध्य, प्रसरण व मानक विचलन ज्ञात कीजिए ।

Answer:
हल : A = 92.5, h = 5, \(y_i = \frac{x_i - 92.5}{5}\)माध्य, \( \bar{x} = A + \left(\frac{\Sigma f_i y_i}{N}\right) \times h \) \[ = 92.5 + \frac{6}{60} \times 5 \] \[ = 92.5 + 0.5 = 93. \] प्रसरण, \( \sigma^2 = \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{25}{3600} [60 \times 254 - 36] \] \[ = \frac{12}{144} [5 \times 254 - 3] \] \[ = \frac{1}{12} [1270 - 3] \] \[ = \frac{1267}{12} = 105.58. \] मानक विचलन, \( \sigma = \sqrt{105.58} = 10.28. \)
In simple words: Use the step-deviation method by selecting an assumed mean (A) and class interval (h) to calculate \(y_i\). Compute the mean using the formula involving \(y_i\), then find the variance using the corresponding step-deviation formula. Finally, take the square root of the variance to get the standard deviation.

🎯 Exam Tip: Carefully determine the class midpoints (\(x_i\)), assumed mean (A), and class interval (h). The calculations for \(f_i y_i\) and \(f_i y_i^2\) are critical. Double-check all summations and be precise with arithmetic, especially when dealing with decimals in the mean and variance. Remember to take the positive square root for standard deviation.

 

Question 10.

एक डिजाइन में बनाए गए वृत्तों के व्यास (मिमी में) नीचे दिए गए हैं।
व्यास
33-36 37-40 41-44 45-48 49-52
वृत्तों की संख्या
15
17
21
22
25
वृत्तों के व्यासों का मानक विचलन के माध्य व्यास ज्ञात कीजिए ।

Answer:
हलः
दिए हुए असतत आँकड़ों को सतत बारंबारता बंटन में बदलने के लिए अंतराल इस प्रकार
हैं।
32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5-48.5, 48.5-52.5
माना A = 42.5, h = 4, \(y_i = \frac{x_i - 42.5}{4}\)माध्य, \( \bar{x} = A + \left(\frac{\Sigma f_i y_i}{N}\right) \times h \) \[ = 42.5 + \frac{25}{100} \times 4 \] \[ = 42.5 + 1 = 43.5. \] प्रसरण \( \sigma^2 = \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{16}{(100)^2} [100 \times 199 - (25)^2] \] \[ = \frac{16 \times 25}{100 \times 100} [4 \times 199 - 25] \] \[ = \frac{1}{25} [796 - 25] \] \[ = \frac{771}{25} = 30.84 \] मानक विचलन \( \sigma = \sqrt{30.84} = 5.56. \)
In simple words: First, convert the discontinuous class intervals to continuous ones. Then, use the step-deviation method to calculate the mean (\(\bar{x}\)) with an assumed mean and class interval. Next, find the variance using the step-deviation formula, and finally, take the square root of the variance to obtain the standard deviation.

🎯 Exam Tip: Always convert discontinuous class intervals into continuous ones before starting any calculations for mean, variance, or standard deviation. This ensures correct midpoints and class intervals. Be extremely careful with calculations, especially squaring and large numbers, as accuracy is key for both mean and standard deviation.

 

Exercise 15.3

Question 1.

निम्नलिखित आँकड़ों से बताइए कि A या B में से किसमें अधिक बिखराव है।
अंक
10-20 20-30 30-40 40-50 50-60 60-70 70-80
समूह A 9
17
32
33
40
10
9
समूह B 10
20
30
25
43
15
7

Answer:
हलः
माना कल्पित माध्य A = 45, h = 10.
\(y_i = \frac{x_i - 45}{10}\)
समूह A के लिए :
\( \bar{x} = A + \left(\frac{\Sigma f_i y_i}{N}\right) \times h \) \[ = 45 + \frac{-6}{150} \times 10 = 45 - \frac{2}{5} \] \[ = 45 - 0.4 = 44.6 \] \( \sigma^2 = \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{100}{22500} [150 \times 342 - 36] \] \[ = \frac{36}{225} [25 \times 57 - 1] \] \[ = \frac{4}{25} [1425 - 1] \] \[ = \frac{4 \times 1224}{25} = 227.84 \] \( \sigma = 15.09 \)
विचरण गुणांक, C.V. = \( \frac{\sigma}{\bar{x}} \times 100 \) \[ = \frac{15.09}{44.6} \times 100 = 33.83. \]
समूह B के लिए :
\( \bar{x} = A + \left(\frac{\Sigma f_i y_i}{N}\right) \times h \) \[ = 45 + \frac{-6}{150} \times 10 = 45 - \frac{2}{5} \] \[ = 45 - 0.4 = 44.6 \] \( \sigma^2 = \frac{h^2}{N^2} [N \Sigma f_i y_i^2 - (\Sigma f_i y_i)^2] \) \[ = \frac{100}{22500} [150 \times 366 - 36] \] \[ = \frac{36}{225} [25 \times 61 - 1] \] \[ = \frac{4}{25} \times 1524 = \frac{6096}{25} = 243.84 \] \( \sigma = 15.62 \)
विचरण गुणांक, C.V. = \( \frac{\sigma}{\bar{x}} \times 100 \) \[ = \frac{15.62}{44.6} \times 100 = 35.02 \]
समूह B का विचरण गुणांक समूह A के विचरण गुणांक से अधिक है।
अतः समूह B में अंकों का बिखराव सूमह A के अंकों से अधिक है।
In simple words: To determine which group has more dispersion, calculate the Coefficient of Variation (CV) for both Group A and Group B. This involves first finding the mean and variance (and then standard deviation) for each group using the step-deviation method. The group with the higher CV has greater dispersion.

🎯 Exam Tip: The Coefficient of Variation (CV) is the best measure to compare the dispersion of two datasets, especially when their means are different. A higher CV indicates greater variability or dispersion. Be very careful with arithmetic operations, particularly when dealing with large numbers and fractions, to ensure accurate calculation of mean, variance, and finally, CV.

 

Question 2.

शेयरों X और Y के नीचे दिए गए मूल्यों से बताइए कि किसके मूल्यों में अधिक स्थिरता है ?
X 35 54 52 53 56 58 52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101

Answer:
हलः
माना शेयर X के आँकड़ों में कल्पित माध्य = 52
और शेयर Y के आँकड़ों में कल्पित माध्य = 105शेयर X के लिए :
\( \bar{x} = A + \frac{\Sigma y_i}{n} \) \[ = 52 + \frac{-10}{10} \] \[ = 52 - 1 = 51 \] \( \sigma^2 = \frac{1}{n^2} [n \Sigma y_i^2 - (\Sigma y_i)^2] \) \[ = \frac{1}{100} [10 \times 360 - (-10)^2] \] \[ = \frac{1}{100} [3600 - 100] \] \[ = \frac{3500}{100} = 35 \] \( \sigma = 5.916 \)
विचरण गुणांक, C.V. = \( \frac{\sigma}{\bar{x}} \times 100 \) \[ = \frac{5.916}{51} \times 100 = \frac{591.6}{51} = 11.6 \]
शेयर Y के लिए :
\( \bar{x} = A + \frac{\Sigma y_i}{n} \) \[ = 105 + \frac{0}{10} = 105 \] \( \sigma^2 = \frac{1}{n^2} [n \Sigma y_i^2 - (\Sigma y_i)^2] \) \[ = \frac{1}{100} [10 \times 40 - 0] \] \[ = \frac{400}{100} = 4 \] \( \sigma = 2 \)
विचरण गुणांक C.V. = \( \frac{\sigma}{\bar{x}} \times 100 \) \[ = \frac{2}{105} \times 100 = \frac{200}{105} = 1.9 \]
विचरण गुणांक Y शेयर में X शेयर की तुलना में कम है।
अतः शेयर Y में, शेयर X की तुलना में अधिक स्थिरता है।
In simple words: To compare stability, calculate the Coefficient of Variation (CV) for both Share X and Share Y. This involves finding the mean and standard deviation for each share, using the assumed mean method for easier calculations. A lower CV indicates more consistency or stability in the share prices.

🎯 Exam Tip: When asked to compare consistency or stability, the Coefficient of Variation (CV) is the most appropriate statistical measure. Ensure that you correctly apply the assumed mean method for both data sets. Accuracy in calculating \(\Sigma y_i\) and \(\Sigma y_i^2\) is paramount for getting the correct mean and standard deviation, which then feed into the CV calculation. Always interpret the result: lower CV means higher consistency.

 

Question 3.

एक कारखाने की दो फर्मों A और B के कर्मचारियों को दिए मासिक वेतन के विश्लेषण का निम्नलिखित परिणाम है:

(i) A और B में से कौन सी फर्म अपने कर्मचारियों को वेतन के रूप में अधिक राशि देती है?
(ii) व्यक्तिगत वेतनों में किस फर्म A या B में अधिक विचरण है ?

Answer:
हलः
फर्म A के लिए :
वेतन पाने वाले कर्मचारियों की संख्या = 586
मासिक वेतन की माध्य = 5253 Rs.
फर्म A द्वारा दिया गया कुल वेतन = 5253 x 586 = 3078258 Rs.
वेतन बंटन का प्रसरण = 100
मानक विचलन = \( \sqrt{100} = 10 \)
विचरण गुणांक = \( \frac{10}{5253} \times 100 \) \[ = 0.19 \] फर्म B के लिए:
वेतन पाने वाले कर्मचारियों की संख्या = 648
मासिक वेतन का संख्या = 5253 Rs.
फर्म B द्वारा गया कुल वेतन = 5253 x 648 Rs. = 3403944 Rs.
वेतन बंटन का प्रसरण = 121
मानक विचलन = \( \sqrt{121} = 11 \)
विचरण गुणांक = \( \frac{11}{5253} \times 100 = 0.21 \)
(i) फर्म A द्वारा दिया गया कुल मासिक वेतन = 3078258 Rs.
फर्म B द्वारा दिया गया कुल मासिक वेतन = 3403944 Rs.
अतः फर्म B फर्म A की तुलना में अधिक मासिक वेतन देती है।
(ii) फर्म A के वेतन बंटन की विचरण गुणांक = 0.19 और फर्म A के वेतन बंटन का विचरण गुणांक = 0.21
अतः फर्म B के वेतन बंटन में अधिक बिखराव है।
In simple words: (i) To find which firm pays more, multiply the number of employees by the average monthly salary for each firm and compare the total amounts. (ii) To determine which firm has more variance in individual salaries, calculate the standard deviation (square root of variance) and then the Coefficient of Variation (CV) for each firm. The firm with the higher CV has greater dispersion or variability in salaries.

🎯 Exam Tip: When comparing total payments, multiply the number of employees by their average salary. To compare variability, use the Coefficient of Variation (CV), especially since the means are equal here. Remember, a higher CV indicates more dispersion. Accurately calculate standard deviation from variance as the first step for CV.

 

Question 4.

टीम A द्वारा एक सत्र में खेले गए फुटबॉल मैचों के आँकड़े नीचे दिए गए हैं:

टीम B द्वारा खेले गए मैचों में बनाए गए गोलोंका माध्य 2 प्रति मैच और गोलों का मानक विचलन 1.25 था।
किस टीम को अधिक संगत (consistent) समझा जाना चाहिए ?

Answer:
हल :टीम A के लिए :
किए गए गोलों का माध्य \( \bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{50}{25} = 2 \)
मानक विचलन = \( \frac{1}{N} \sqrt{N \Sigma f_i x_i^2 - (\Sigma f_i x_i)^2} \) \[ = \frac{1}{25} \sqrt{25 \times 130 - 50 \times 50} \] \[ = \frac{1}{25} \sqrt{3250 - 2500} \] \[ = \frac{1}{25} \sqrt{750} = \frac{1}{5} \sqrt{30} = 1.095 \] विचरण गुणांक = \( \frac{\sigma}{\bar{x}} \times 100 \) \[ = \frac{1.095}{2} \times 100 \] \[ = 54.75 \]
टीम B के लिए :
माध्य \( \bar{x} = 2 \)
मानक विचलन = 1.25
विचरण गुणांक = \( \frac{\sigma}{\bar{x}} \times 100 \) \[ = \frac{1.25}{2} \times 100 = 62.5 \]
टीम A का टीम B की तुलना में विचरण गुणांक कम है।
अतः टीम A में टीम B से अधिक स्थिरता है।
In simple words: To determine consistency, calculate the Coefficient of Variation (CV) for both Team A and Team B. For Team A, find the mean and standard deviation from the given frequency distribution. For Team B, the mean and standard deviation are provided. The team with a lower CV is considered more consistent.

🎯 Exam Tip: Consistency is best measured by the Coefficient of Variation (CV). Remember the formula for standard deviation from grouped data, \(\sigma = \frac{1}{N} \sqrt{N \Sigma f_i x_i^2 - (\Sigma f_i x_i)^2}\). Always compare the CVs; the lower the CV, the more consistent the data set. Be meticulous in calculations, especially when dealing with square roots.

 

Question 5. पचास वनस्पति उत्पादों की लंबाई x (सेमी में) और भार y (ग्राम में) के योग और वर्गों के योग नीचे दिए गए हैं।

\[ \sum_{i=1}^{50} x_i = 212, \sum_{i=1}^{50} x_i^2 = 902.8, \sum_{i=1}^{50} y_i = 261, \sum_{i=1}^{50} y_i^2 = 1457.6 \] लंबाई या भार में किसमें अधिक विचरण है ?
Answer:
हलः
लंबाई के लिए:
\[ n = 50, \sum_{i=1}^{50} x_i = 212 \] माध्य \( \bar{x} = \frac{\sum x_i}{n} = \frac{212}{50} = 4.24 \)
मानक विचलन \( \sigma = \frac{1}{n} \sqrt{n \sum x_i^2 - (\sum x_i)^2} \)
\[ = \frac{1}{50} \sqrt{50 \times 902.8 - (212)^2} \] \[ = \frac{1}{50} \sqrt{45140 - 44944} \] \[ = \frac{\sqrt{196}}{50} = \frac{14}{50} = 0.28 \] विचरण गुणांक, \( C.V. = \frac{\sigma}{\bar{x}} \times 100 \)
\[ = \frac{0.28}{4.24} \times 100 = 6.60 \]
भार के लिए:
\[ n = 50, \sum_{i=1}^{50} y_i = 261, \sum_{i=1}^{50} y_i^2 = 1457.6 \] माध्य \( \bar{x} = \frac{\sum y_i}{n} = \frac{261}{50} = 5.22 \)
मानक विचलन \( \sigma = \frac{1}{n} \sqrt{n \sum y_i^2 - (\sum y_i)^2} \)
\[ = \frac{1}{50} \sqrt{50 \times 1457.6 - (261)^2} \] \[ = \frac{1}{50} \sqrt{72880 - 68121} \] \[ = \frac{\sqrt{4759}}{50} = \frac{68.9855}{50} = 1.38 \] विचरण गुणांक, \( C.V. = \frac{\sigma}{\bar{x}} \times 100 \)
\[ = \frac{1.38}{5.22} \times 100 = 26.44 \] भार का विचरण गुणांक, लंबाई के विचरण गुणांक से अधिक है।
अतः भार के बंटन में अधिक विचरण है।
In simple words: We calculated the coefficient of variation (C.V.) for both length and weight. Since the C.V. for weight (26.44) is higher than that for length (6.60), it indicates that there is more variation in the distribution of weight.

🎯 Exam Tip: Remember that a higher coefficient of variation indicates greater variability relative to the mean, which helps compare the dispersion of two datasets with different units or means.

अध्याय 15 पर विविध प्रश्नावली

 

Question 1. आठ प्रेक्षणों का माध्य तथा प्रसरण क्रमशः 9 और 9.25 है। यदि इनमें से छः प्रेक्षण 6, 7, 10, 12, 12, और 13 हैं, तो शेष दो प्रेक्षण ज्ञात कीजिए।
Answer: हल : मान लीजिए वे दो संख्याएँ x और y हैं।
तब, माध्य \( \bar{x} = 9 \)
\[ 9 = \frac{6+7+10+12+12+13+x+y}{8} \] \[ 72 = 60 + x + y \]
\( \implies x + y = 12 \) ...(1)
प्रसरण \( \sigma^2 = \frac{1}{n^2} [n \sum x_i^2 - (\sum x_i)^2] \)
\[ \bar{x} = \frac{\sum x_i}{n} \implies \sum x_i = n \bar{x} = 8 \times 9 = 72 \] \[ 9.25 = \frac{1}{64} [8 \times \sum x_i^2 - (72)^2] \]
\( \implies 8 \times \sum x_i^2 = 9.25 \times 64 + 72 \times 72 \)
\[ = 592 + 5184 = 5776 \] \[ \sum x_i^2 = \frac{5776}{8} = 722 \] \[ 722 = 6^2 + 7^2 + 10^2 + 12^2 + 12^2 + 13^2 + x^2 + y^2 \] \[ 722 = 36 + 49 + 100 + 144 + 144 + 169 + x^2 + y^2 \]
\( \implies 722 = 642 + x^2 + y^2 \)
\[ x^2 + y^2 = 722 - 642 = 80 \] ...(2)
समीकरण (1) और (2) से
\( x^2 + (12-x)^2 = 80 \)
\( x^2 + 144 - 24x + x^2 = 80 \)
\( 2x^2 - 24x + 144 - 80 = 0 \)
\( 2x^2 - 24x + 64 = 0 \)
\( x^2 - 12x + 32 = 0 \)
\( (x-4)(x-8) = 0 \)

\( \implies x = 4 \) या \( x = 8 \)
यदि \( x = 4 \), तो समीकरण (1) से \( 4+y = 12 \implies y = 8 \)
यदि \( x = 8 \), तो समीकरण (1) से \( 8+y = 12 \implies y = 4 \)
अतः वे दो संख्याएँ 4 और 8 हैं।
In simple words: We used the given mean and variance formulas to create two equations involving the two unknown observations. Solving these simultaneous equations yielded the values 4 and 8 for the missing observations.

🎯 Exam Tip: For problems involving missing data and statistical measures, setting up equations from the definition of mean and variance is a standard approach. Be careful with algebraic manipulations.

 

Question 2. सात प्रेक्षणों का माध्य तथा प्रसरण क्रमशः 8 और 16 हैं। यदि इनमें से पाँच प्रेक्षण 2, 4, 10, 12, 14 हैं तो शेष दो प्रेक्षण ज्ञात कीजिए ।
Answer: हल : माना कि वे दो संख्याएँ x और y हैं।
माध्य \( \bar{x} = 8 \)
\[ 8 = \frac{2+4+10+12+14+x+y}{7} \] \[ 56 = 42 + x + y \]
\( \implies x + y = 56 - 42 = 14 \) ...(1)
प्रसरण \( \sigma^2 = \frac{1}{n^2} [n \sum x_i^2 - (\sum x_i)^2] \)
यहाँ \( \sum x_i = n \bar{x} = 7 \times 8 = 56 \)
\[ 16 = \frac{1}{49} [7 \sum x_i^2 - (56)^2] \]
\( \implies 7 \sum x_i^2 = 49 \times 16 + 56 \times 56 \)
\[ \sum x_i^2 = 7 \times 16 + 8 \times 56 = 112 + 448 = 560 \] \[ 560 = 2^2 + 4^2 + 10^2 + 12^2 + 14^2 + x^2 + y^2 \] \[ 560 = 4 + 16 + 100 + 144 + 196 + x^2 + y^2 \] \[ 560 = 460 + x^2 + y^2 \]
\( \implies x^2 + y^2 = 560 - 460 = 100 \) ...(2)
समीकरण (1) और (2) से
\( x^2 + (14-x)^2 = 100 \)
\( x^2 + 196 - 28x + x^2 = 100 \)
\( 2x^2 - 28x + 196 - 100 = 0 \)
\( 2x^2 - 28x + 96 = 0 \)
\( x^2 - 14x + 48 = 0 \)
\( (x-6)(x-8) = 0 \)

\( \implies x = 6 \) या \( x = 8 \)
यदि \( x = 6 \), तो समीकरण (1) से \( 6+y = 14 \implies y = 8 \)
यदि \( x = 8 \), तो समीकरण (1) से \( 8+y = 14 \implies y = 6 \)
अतः वे दो संख्याएँ 6 और 8 हैं।
In simple words: Similar to the previous problem, we used the definitions of mean and variance for seven observations to set up two algebraic equations. Solving these equations for the two unknown observations gave us the values 6 and 8.

🎯 Exam Tip: It's crucial to correctly calculate \( \sum x_i \) and \( \sum x_i^2 \) from the given data and then solve the system of equations. Double-check your arithmetic to avoid errors.

 

Question 3. छः प्रेक्षणों को माध्य तथा मानक विचलन क्रमशः 8 तथा 4 हैं। यदि प्रत्येक प्रेक्षण को 3 से गुणा कर दिया जाए तो परिणामी प्रेक्षणों का माध्य व मानक विचलन ज्ञात कीजिए।
Answer: हल :
माना दिए गए छः प्रेक्षण \( x_1, x_2, x_3, x_4, x_5, x_6 \) हैं।
माध्य \( \bar{x} = \frac{\sum x_i}{n} = 8 \)
मानक विचलन \( \sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = 4 \)
यदि प्रत्येक प्रेक्षण को 3 से गुणा किया जाए, तो नए प्रेक्षण \( 3x_1, 3x_2, ..., 3x_6 \) होंगे।
नया माध्य \( \bar{x}' = \frac{\sum (3x_i)}{n} = \frac{3 \sum x_i}{n} = 3 \times \frac{\sum x_i}{n} = 3 \bar{x} = 3 \times 8 = 24 \)
नया मानक विचलन \( \sigma' = \sqrt{\frac{\sum (3x_i - 3\bar{x})^2}{n}} \)
\[ = \sqrt{\frac{\sum 3^2 (x_i - \bar{x})^2}{n}} \] \[ = \sqrt{\frac{9 \sum (x_i - \bar{x})^2}{n}} \] \[ = 3 \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = 3 \sigma \] \[ = 3 \times 4 = 12 \] अतः नया माध्य 24 तथा नया मानक विचलन 12 होगा।
In simple words: When each observation in a dataset is multiplied by a constant (in this case, 3), the new mean is the original mean multiplied by that constant, and the new standard deviation is also the original standard deviation multiplied by that constant.

🎯 Exam Tip: Remember these properties for scaling data: multiplying each observation by 'a' scales the mean by 'a' and the standard deviation by `|a|`. Adding a constant to each observation shifts the mean but does not change the standard deviation.

 

Question 4. यदि \( n \) प्रेक्षणों का माध्य \( \bar{x} \) तथा प्रसरण \( \sigma^2 \) है तो सिद्ध कीजिए कि प्रेक्षणों \( ax_1, ax_2, ax_3, ......, ax_n \) का माध्य और प्रसरण क्रमशः \( a\bar{x} \) तथा \( a^2\sigma^2 \) (जहाँ \( a \neq 0 \)) है।
Answer: हल : (i) माध्य
दिए गए प्रेक्षणों का माध्य \( \bar{x} = \frac{\sum x_i}{n} \)
नए प्रेक्षणों \( ax_1, ax_2, ..., ax_n \) का नया माध्य:
नया माध्य \( = \frac{\sum (ax_i)}{n} = \frac{a \sum x_i}{n} = a \bar{x} \)
(ii) प्रसरण
दिए गए प्रेक्षणों का प्रसरण \( \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} \)
नए प्रेक्षणों \( ax_1, ax_2, ..., ax_n \) का नया माध्य \( a\bar{x} \)
नया प्रसरण \( = \frac{\sum (ax_i - a\bar{x})^2}{n} \)
\[ = \frac{\sum a^2 (x_i - \bar{x})^2}{n} \] \[ = a^2 \frac{\sum (x_i - \bar{x})^2}{n} = a^2 \sigma^2 \] इति सिद्धम् ।
In simple words: This proof shows that if you multiply every data point by a constant 'a', the mean also gets multiplied by 'a', and the variance gets multiplied by \( a^2 \). This is a fundamental property of how scaling affects central tendency and dispersion.

🎯 Exam Tip: Understanding how linear transformations affect statistical measures like mean, variance, and standard deviation is crucial for data analysis. This identity is frequently tested in exams.

 

Question 5. बीस प्रेक्षणों का माध्य तथा मानक विचलन क्रमशः 10 तथा 2 हैं। जांच करने पर यह पाया गया कि प्रेक्षण 8 गलत है। निम्न में से प्रत्येक का सही मध्य तथा मानक विचलन ज्ञात कीजिए यदि
(i) गलत प्रेक्षण हटा दिया जाए।
(ii) उसे 12 से बदल दिया जाए।
Answer: हल :
माध्य \( \bar{x} = \frac{\sum x_i}{n} \)
दिया है, \( n = 20, \bar{x} = 10, \sigma = 2 \)
\[ 10 = \frac{\sum x_i}{20} \implies \sum x_i = 10 \times 20 = 200 \] मानक विचलन \( \sigma = \frac{1}{n} \sqrt{n \sum x_i^2 - (\sum x_i)^2} \)
\[ 2 = \frac{1}{20} \sqrt{20 \sum x_i^2 - (200)^2} \]
\( \implies 4 = \frac{1}{400} [20 \sum x_i^2 - 40000] \)

\( \implies 1600 = 20 \sum x_i^2 - 40000 \)

\( \implies 20 \sum x_i^2 = 41600 \)

\( \implies \sum x_i^2 = \frac{41600}{20} = 2080 \)
(i) गलत प्रेक्षण (8) हटा दिया जाए।
प्रेक्षणों की संख्या \( n' = 20 - 1 = 19 \)
सही योग \( \sum x_i' = 200 - 8 = 192 \)
नया माध्य \( \bar{x}' = \frac{192}{19} = 10.11 \)
सही \( \sum x_i^2' = 2080 - 8^2 = 2080 - 64 = 2016 \)
नया मानक विचलन \( \sigma' = \frac{1}{n'} \sqrt{n' \sum x_i^2' - (\sum x_i')^2} \)
\[ = \frac{1}{19} \sqrt{19 \times 2016 - (192)^2} \] \[ = \frac{1}{19} \sqrt{38304 - 36864} \] \[ = \frac{1}{19} \sqrt{1440} = \frac{37.947}{19} = 1.997 \] (ii) गलत प्रेक्षण (8) को 12 से बदल दिया जाए।
प्रेक्षणों की संख्या \( n'' = 20 \) (कोई बदलाव नहीं)
सही योग \( \sum x_i'' = 200 - 8 + 12 = 204 \)
नया माध्य \( \bar{x}'' = \frac{204}{20} = 10.2 \)
सही \( \sum x_i^2'' = 2080 - 8^2 + 12^2 \)
\[ = 2080 - 64 + 144 = 2160 \] नया मानक विचलन \( \sigma'' = \frac{1}{n''} \sqrt{n'' \sum x_i^2'' - (\sum x_i'')^2} \)
\[ = \frac{1}{20} \sqrt{20 \times 2160 - (204)^2} \] \[ = \frac{1}{20} \sqrt{43200 - 41616} \] \[ = \frac{1}{20} \sqrt{1584} = \frac{39.799}{20} = 1.99 \]
In simple words: We first calculated the sum of observations and the sum of their squares from the initial mean and standard deviation. Then, for each scenario (removing the wrong observation or replacing it), we adjusted these sums and re-calculated the new mean and standard deviation using the corrected values.

🎯 Exam Tip: When correcting statistical data, remember to adjust both the sum of observations \( (\sum x_i) \) and the sum of squares \( (\sum x_i^2) \). A common mistake is to only correct the sum of observations.

 

Question 6. एक कक्षा के पचास छात्रों द्वारा तीन विषयों गणित, भौतिक शास्त्र व रसायन शास्त्र में प्राप्तांकों का माध्य व मानक विचलन नीचे दिए गए हैं:

किस विषय में सबसे अधिक विचलन है तथा किसमें सबसे कम विचलन है?
Answer: हल :
विचरण गुणांक \( (C.V.) = \frac{\sigma}{\bar{x}} \times 100 \)
गणित विषय में विचरण गुणांक \( = \frac{12}{42} \times 100 = 28.57 \)
भौतिक विषय में विचरण गुणांक \( = \frac{15}{32} \times 100 = 46.875 \)
रसायन विषय में विचरण गुणांक \( = \frac{20}{40.9} \times 100 = 48.9 \)
अतः रसायन विषय में सबसे अधिक विचलन है तथा गणित में सबसे कम विचलन है।
In simple words: To compare the variability of scores across different subjects with different means, we use the Coefficient of Variation (C.V.). The subject with the highest C.V. (Chemistry) has the most variation, and the subject with the lowest C.V. (Mathematics) has the least variation.

🎯 Exam Tip: The Coefficient of Variation is a key tool for comparing the relative variability of two or more data sets, especially when they have different units or widely different means. A higher C.V. implies greater relative dispersion.

 

Question 7. 100 प्रेक्षणों का माध्य और मानक विचलन क्रमशः 20 और 3 हैं। बाद में यह पाया गया कि तीन प्रेक्षण 21, 21 तथा 18 गलत थे। यदि गलत प्रेक्षणों को हटा दिया जाए तो माध्य व मानक विचलन ज्ञात कीजिए।
Answer: हल :
माध्य \( \bar{x} = \frac{\sum x_i}{n} \)
दिया है, \( n = 100, \bar{x} = 20, \sigma = 3 \)
\[ 20 = \frac{\sum x_i}{100} \implies \sum x_i = 100 \times 20 = 2000 \] मानक विचलन \( \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} \)
हमें पता है कि \( \sigma^2 = \frac{n \sum x_i^2 - (\sum x_i)^2}{n^2} \)

\( \implies n^2 \sigma^2 = n \sum x_i^2 - (\sum x_i)^2 \)

\( \implies n \sum x_i^2 = n^2 \sigma^2 + (\sum x_i)^2 \)

\( \implies 100 \sum x_i^2 = (100)^2 \times 3^2 + (2000)^2 \)
\[ = 10000 \times 9 + 4000000 \] \[ = 90000 + 4000000 = 4090000 \]
\( \implies \sum x_i^2 = \frac{4090000}{100} = 40900 \)
गलत प्रेक्षणों को हटा दिया जाए:
प्रेक्षणों की संख्या \( n' = 100 - 3 = 97 \)
सही योग \( \sum x_i' = 2000 - (21+21+18) = 2000 - 60 = 1940 \)
नया माध्य \( \bar{x}' = \frac{1940}{97} = 20 \)
सही \( \sum x_i^2' = 40900 - (21^2 + 21^2 + 18^2) \)
\[ = 40900 - (441 + 441 + 324) \] \[ = 40900 - 1206 = 39694 \] नया मानक विचलन \( \sigma' = \frac{1}{n'} \sqrt{n' \sum x_i^2' - (\sum x_i')^2} \)
\[ = \frac{1}{97} \sqrt{97 \times 39694 - (1940)^2} \] \[ = \frac{1}{97} \sqrt{3850318 - 3763600} \] \[ = \frac{1}{97} \sqrt{86718} = \frac{294.479}{97} = 3.036 \]
In simple words: We calculated the initial sum of observations and sum of squares using the given mean and standard deviation. Then, we removed the incorrect observations by subtracting their values and squares from the sums. Finally, we recalculated the new mean and standard deviation with the corrected number of observations and sums.

🎯 Exam Tip: This problem highlights the impact of erroneous data on statistical measures. Always adjust both the sum of observations and the sum of their squares when correcting data, as both are crucial for calculating variance and standard deviation correctly.