RS Aggarwal Solutions for Class 11 Chapter 30 Statistics

 

Question 1. Find the mean deviation about the mean for the following data: 7, 8, 4, 13, 9, 5, 16, 18
Answer: We have the data: 7, 8, 4, 13, 9, 5, 16, 18.

The mean is calculated as:
\[ \bar{x} = \frac{7 + 8 + 4 + 13 + 9 + 5 + 16 + 18}{8} = \frac{80}{8} = 10 \]

The absolute deviations from the mean are: 3, 2, 6, 3, 1, 5, 6, 8

The mean deviation about the mean is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{8} |x_i - \bar{x}|}{8} = \frac{3 + 2 + 6 + 3 + 1 + 5 + 6 + 8}{8} = \frac{34}{8} = 4.25 \]
In simple words: Find the average of your numbers. Then measure how far each number is from that average. Finally, find the average of all those distances.

Exam Tip: Always arrange calculations in a clear table format with columns for \( x_i \), \( |x_i - \bar{x}| \), and their sum to avoid arithmetic mistakes.

 

Question 2. Find the mean deviation about the mean for the following data: 39, 72, 48, 41, 43, 55, 60, 45, 54, 43
Answer: We have the data: 39, 72, 48, 41, 43, 55, 60, 45, 54, 43.

The mean is calculated as:
\[ \bar{x} = \frac{39 + 72 + 48 + 41 + 43 + 55 + 60 + 45 + 54 + 43}{10} = \frac{500}{10} = 50 \]

The absolute deviations from the mean are: 11, 22, 2, 9, 7, 5, 10, 5, 4, 7

The mean deviation about the mean is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{10} |x_i - \bar{x}|}{10} = \frac{11 + 22 + 2 + 9 + 7 + 5 + 10 + 5 + 4 + 7}{10} = \frac{82}{10} = 8.2 \]
In simple words: Compute the mean by adding all values and dividing by the count. Take the absolute difference of each value from the mean. Sum these differences and divide by the total count.

Exam Tip: Be careful with absolute values - all deviations must be positive regardless of whether the original number is above or below the mean.

 

Question 3. Find the mean deviation about the mean for the following data: 17, 20, 12, 13, 15, 16, 12, 18, 15, 19, 12, 11
Answer: We have the data: 17, 20, 12, 13, 15, 16, 12, 18, 15, 19, 12, 11.

The mean is calculated as:
\[ \bar{x} = \frac{17 + 20 + 12 + 13 + 15 + 16 + 12 + 18 + 15 + 19 + 12 + 11}{12} = \frac{180}{12} = 15 \]

The absolute deviations from the mean are: 2, 5, 3, 2, 0, 1, 3, 3, 0, 4, 3, 4

The mean deviation about the mean is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{12} |x_i - \bar{x}|}{12} = \frac{2 + 5 + 3 + 2 + 0 + 1 + 3 + 3 + 0 + 4 + 3 + 4}{12} = \frac{30}{12} = 2.5 \]
In simple words: Calculate the mean of all numbers. For each number, find its distance from the mean. Add up all distances and divide by the total number of values to get the mean deviation.

Exam Tip: When a deviation equals zero, it means that value is exactly at the mean - always include it in your sum even though it contributes nothing.

 

Question 4. Find the mean deviation about the median for the following data: 12, 5, 14, 6, 11, 13, 17, 8, 10
Answer: Since there are 9 observations, which is odd, we arrange the data in ascending order: 5, 6, 8, 10, 11, 12, 13, 14, 17.

The median is the 5th observation:
\[ \text{Median}(M) = 11 \]

The absolute deviations from the median are: 6, 5, 3, 1, 0, 1, 2, 3, 6

The mean deviation about the median is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{9} |x_i - M|}{9} = \frac{6 + 5 + 3 + 1 + 0 + 1 + 2 + 3 + 6}{9} = \frac{27}{9} = 3 \]
In simple words: Sort the numbers from smallest to largest. The middle value is the median. Measure how far each number is from this middle value. Average all these distances.

Exam Tip: For an odd number of observations, the median is always the middle value when arranged in order - position is \( \frac{n+1}{2} \).

 

Question 5. Find the mean deviation about the median for the following data: 4, 15, 9, 7, 19, 13, 6, 21, 8, 25, 11
Answer: Since there are 11 observations, which is odd, we arrange the data in ascending order: 4, 6, 7, 8, 9, 11, 13, 15, 19, 21, 25.

The median is the 6th observation:
\[ \text{Median}(M) = 11 \]

The absolute deviations from the median are: 7, 5, 4, 3, 2, 0, 2, 4, 8, 10, 14

The mean deviation about the median is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{11} |x_i - M|}{11} = \frac{7 + 5 + 4 + 3 + 2 + 0 + 2 + 4 + 8 + 10 + 14}{11} = \frac{59}{11} = 5.3 \]
In simple words: List all numbers in order. The median is the value sitting exactly in the middle. Calculate the absolute difference between each number and the median. Take the average of all these differences.

Exam Tip: The median divides the dataset into two equal halves, making it a robust central value that is less affected by extreme outliers than the mean.

 

Question 6. Find the mean deviation about the median for the following data: 34, 23, 46, 37, 40, 28, 32, 50, 35, 44
Answer: Since there are 10 observations, which is even, we arrange the data in ascending order: 23, 28, 32, 34, 35, 37, 40, 44, 46, 50.

The median is the average of the 5th and 6th observations:
\[ \text{Median}(M) = \frac{35 + 37}{2} = \frac{72}{2} = 36 \]

The absolute deviations from the median are: 13, 8, 4, 2, 1, 1, 4, 8, 10, 14

The mean deviation about the median is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{10} |x_i - M|}{10} = \frac{13 + 8 + 4 + 2 + 1 + 1 + 4 + 8 + 10 + 14}{10} = \frac{65}{10} = 6.5 \]
In simple words: Arrange the numbers in order. Since there is an even count, the median is the average of the two middle values. Find how far each number is from this median. Calculate the average of all these distances.

Exam Tip: For even-sized datasets, always remember to take the average of the two middle values to find the median - never pick just one of them.

 

Question 7. Find the mean deviation about the median for the following data: 70, 34, 42, 78, 65, 45, 54, 48, 67, 50, 56, 63
Answer: Since there are 12 observations, which is even, we arrange the data in ascending order: 34, 42, 45, 48, 50, 54, 56, 63, 65, 67, 70, 78.

The median is the average of the 6th and 7th observations:
\[ \text{Median}(M) = \frac{54 + 56}{2} = \frac{110}{2} = 55 \]

The absolute deviations from the median are: 21, 13, 10, 7, 5, 1, 1, 8, 10, 12, 15, 23

The mean deviation about the median is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{12} |x_i - M|}{12} = \frac{21 + 13 + 10 + 7 + 5 + 1 + 1 + 8 + 10 + 12 + 15 + 23}{12} = \frac{126}{12} = 10.5 \]
In simple words: Sort all values in ascending order. With an even number of values, the median is midway between the two central numbers. Determine the distance of each value from this median point. Average these distances for the mean deviation.

Exam Tip: Double-check your sum of absolute deviations before dividing by the total count - this is where most arithmetic errors occur.

 

Question 8. Find the mean deviation about the mean for the following data:

Answer: We construct a table to find the mean and deviations:

The mean is:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{756}{36} = 21 \]

Now we compute the absolute deviations and their products with frequencies:

The mean deviation about the mean is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{288}{36} = 8 \]
In simple words: When data comes with frequencies, multiply each value by its frequency to get the mean. Then calculate the weighted distances from the mean and divide by the total frequency.

Exam Tip: Always organize frequency data in a table with columns for \( x_i \), \( f_i \), \( f_i x_i \), and \( f_i |x_i - \bar{x}| \) to avoid missing any calculations.

 

Question 9. Find the mean deviation about the mean for the following data:

Answer: We construct a table to find the mean and deviations:

The mean is:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{300}{40} = 7.5 \]

Now we compute the absolute deviations and their products with frequencies:

The mean deviation about the mean is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{92}{40} = 2.3 \]
In simple words: When each value appears multiple times (given by frequency), weight it accordingly. Find the mean by summing weighted values. Calculate weighted distances from the mean and divide by the total count.

Exam Tip: Be precise with decimal arithmetic when computing deviations - a small rounding error early on can propagate through the entire calculation.

 

Question 10. Find the mean deviation about the mean for the following data:

Answer: We construct a table to find the mean and deviations:

The mean is:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{528}{66} = 8 \]

Now we compute the absolute deviations and their products with frequencies:

The mean deviation about the mean is:
\[ \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{138}{66} = 2.09 \]
In simple words: Multiply each value by its frequency to get the weighted sum. Divide by the total frequency to find the mean. Then compute how far each value is from the mean, weight those distances, and average them.

Exam Tip: When the mean is a whole number, deviations are simpler to calculate - double-check for arithmetic errors if you get a non-integer mean.

 

Question 11. Find the mean deviation about the median for the following data:

Answer: The given observations are in ascending order. Adding a row for cumulative frequencies:

Total number of observations is N = 29, which is odd. The median is located at the \( \frac{N+1}{2} = 15^{\text{th}} \) observation. Since the 15th observation falls in the cumulative frequency 21, the median is 30.

\[ \text{Median}(M) = 30 \]

Absolute deviations from the median and their products with frequencies:

\[ \sum f_i = 29 \quad \text{and} \quad \sum f_i |x_i - M| = 148 \]

The mean deviation about the median is:
\[ \text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{\sum f_i} = \frac{148}{29} = 5.10 \]
In simple words: Compute cumulative frequencies to identify which observation represents the median. Find the absolute distance between each value and the median. Multiply each distance by its frequency and sum these products. Divide by the total count.

Exam Tip: The cumulative frequency approach is crucial for grouped or frequency-distributed data - it tells you exactly where the median observation falls in the dataset.

 

Question 12. Find the mean deviation about the median for the following data:

Answer: The given observations are in ascending order. Adding a row for cumulative frequencies:

Total number of observations is N = 50, which is even. The median is the average of the 25th and 26th observations. Both lie in the cumulative frequency 30, for which the corresponding observation is 13.

\[ \text{Median}(M) = \frac{13 + 13}{2} = 13 \]

Absolute deviations from the median and their products with frequencies:

\[ \sum f_i = 50 \quad \text{and} \quad \sum f_i |x_i - M| = 136 \]

The mean deviation about the median is:
\[ \text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{\sum f_i} = \frac{136}{50} = 2.72 \]
In simple words: Build a cumulative frequency table to find the position of the 25th and 26th observations for an even-sized dataset. The median is their average. Calculate weighted distances from the median and divide by the total frequency count.

Exam Tip: For even N, both the Nth/2 and (N/2 + 1)th observations guide you to the median class - always verify they lie in the same cumulative frequency interval.

 

Question 13. Find the mean deviation about the median for the following data:

Answer: The given observations are in ascending order. Adding a row for cumulative frequencies:

Total number of observations is N = 50, which is even. The median is the average of the 25th and 26th observations. Both lie in the cumulative frequency 29, for which the corresponding observation is 30.

\[ \text{Median}(M) = \frac{30 + 30}{2} = 30 \]

Absolute deviations from the median and their products with frequencies:

\[ \sum f_i = 50 \quad \text{and} \quad \sum f_i |x_i - M| = 505 \]

The mean deviation about the median is:
\[ \text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{\sum f_i} = \frac{505}{50} = 10.1 \]
In simple words: Use cumulative frequencies to pinpoint where the median observation falls. For even N, the median is the average of the two middle observations. Determine the distance of each value from the median, weight by frequency, and divide the sum by N.

Exam Tip: Always double-check your cumulative frequency calculation - errors there will cascade through the entire problem and produce an incorrect median.

 

Question 14. Find the mean deviation about the mean for the following data:

Answer: Build a table showing the frequency distribution with midpoints and products. The midpoints are 5, 15, 25, 35, 45, and 55 respectively. Calculate \( \overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1350}{50} = 27 \). Next, determine the absolute deviations from the mean: \( |x_i - \overline{x}| \) gives 22, 12, 2, 8, 18, and 28. Multiply each deviation by its frequency: 132, 96, 28, 128, 72, and 56. The sum of these products is 512. Therefore, the mean deviation about the mean is \( \text{M.D}(\overline{x}) = \frac{\sum f_i |x_i - \overline{x}|}{\sum f_i} = \frac{512}{50} = 10.24 \).

Exam Tip: Always calculate the mean first using the formula for grouped data, then find absolute deviations for each midpoint - never use signed deviations for mean deviation calculations.

 

Question 15. Find the mean deviation about the mean for the following data:

Answer: Prepare a calculation table using the class midpoints: 100, 110, 120, 130, 140, and 150. Compute the mean: \( \overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{12500}{100} = 125 \). Determine the absolute deviation of each midpoint from the mean: 25, 15, 5, 5, 15, and 25. Multiply these by their respective frequencies: the products are 225, 240, 115, 150, 180, and 250. The total of all products is 1160. Thus, the mean deviation about the mean is \( \text{M.D}(\overline{x}) = \frac{1160}{100} = 11.6 \).

Exam Tip: When the mean falls at or near a class boundary, double-check your arithmetic on deviation calculations - small errors compound across all frequency products.

 

Question 16. Find the mean deviation about the mean for the following data:

Answer: Set up a table with class midpoints 35, 45, 55, 65, 75, 85, and 95. Obtain the mean as \( \overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3100}{50} = 62 \). Find absolute deviations from this mean: 27, 17, 7, 3, 13, 23, and 33. Multiply each by frequency to get 81, 119, 84, 45, 104, 69, and 66, which sum to 568. The mean deviation is \( \text{M.D}(\overline{x}) = \frac{568}{50} = 11.36 \).

Exam Tip: Round intermediate values carefully if needed, but preserve full accuracy in the final computation - rounding the mean too early can introduce cumulative error in deviations.

 

Question 17. Find the mean deviation about the median for the following data:

Answer: Create a table showing frequencies and cumulative frequencies. The cumulative frequencies are 6, 13, 28, 44, 48, and 50. The median position is at \( \frac{N}{2} = 25 \), which falls in the class 20 - 30 (where cumulative frequency reaches 28). Using the median formula \( M = l + \frac{\frac{N}{2} - C}{f} \times h \) with \( l = 20 \), \( C = 13 \), \( f = 15 \), \( h = 10 \), and \( N = 50 \), we get \( M = 20 + \frac{25 - 13}{15} \times 10 = 20 + 8 = 28 \). Next, find absolute deviations from the median using midpoints 5, 15, 25, 35, 45, 55: these are 23, 13, 3, 7, 17, 27. Multiply by frequencies to obtain 138, 91, 45, 112, 68, 54, which sum to 508. The mean deviation about the median is \( \text{M.D}(M) = \frac{508}{50} = 10.16 \).

Exam Tip: Identify the median class correctly by locating where cumulative frequency first equals or exceeds \( \frac{N}{2} \) - this ensures all subsequent calculations are accurate.

 

Question 18. Find the mean deviation about the median for the following data:

Answer: Build a frequency table with cumulative frequencies: 6, 14, 25, 43, 48, 50. The median position is \( \frac{50}{2} = 25 \), which corresponds exactly to the end of the class 20 - 30. This makes 20 - 30 the median class. Apply the formula \( M = l + \frac{\frac{N}{2} - C}{f} \times h \) where \( l = 20 \), \( C = 14 \), \( f = 11 \), \( h = 10 \): \( M = 20 + \frac{25 - 14}{11} \times 10 = 20 + 10 = 30 \). Using midpoints 5, 15, 25, 35, 45, 55, compute absolute deviations from 30: these are 25, 15, 5, 5, 15, 25. Multiply by frequencies: 150, 120, 55, 90, 75, 50, summing to 540. The mean deviation about the median is \( \text{M.D}(M) = \frac{540}{50} = 10.8 \).

Exam Tip: When the median position falls at a cumulative frequency boundary, verify whether the median lies exactly at a class limit or slightly within - this determines whether you use interpolation or a simple substitution.

 

Question 1. Find the mean, variance and standard deviation for the numbers 4, 6, 10, 12, 7, 8, 13, 12.
Answer: Given data: 4, 6, 10, 12, 7, 8, 13, 12

To find: MEAN

The mean is found by adding all observations together and dividing by the total count:

\( \text{Mean } (\bar{x}) = \frac{\text{Sum of observations}}{\text{Total number of observations}} \)

\( = \frac{4 + 6 + 10 + 12 + 7 + 8 + 13 + 12}{8} \)

\( = \frac{72}{8} \)

\( \bar{x} = 9 \)

To find: VARIANCE

We calculate the squared differences from the mean for each value:

\( \text{Variance}, \sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} \)

\( = \frac{74}{8} \)

\( = 9.25 \)

To find: STANDARD DEVIATION

\( \text{Standard Deviation } (\sigma) = \sqrt{\text{Variance}} \)

\( = \sqrt{9.25} \)

\( = 3.04 \)

In simple words: The average of the numbers is 9. Variance shows how spread out the numbers are from this average, which is 9.25. Standard deviation tells us roughly how far each number typically strays from the average, and that distance is about 3.04 units.

Exam Tip: Always verify your mean calculation first - a mistake here affects both variance and standard deviation. Remember that variance uses squared differences, so it's always non-negative.

 

Question 2. Find the mean, variance and standard deviation for the first six odd natural numbers.
Answer: Odd natural numbers form the sequence 1, 3, 5, 7, 9, 11, ...

The first six odd natural numbers are: 1, 3, 5, 7, 9, 11

To find: MEAN

The mean is calculated by summing all values and dividing by the count:

\( \text{Mean } (\bar{x}) = \frac{\text{Sum of observations}}{\text{Total number of observations}} \)

\( = \frac{1 + 3 + 5 + 7 + 9 + 11}{6} \)

\( = \frac{36}{6} \)

\( \bar{x} = 6 \)

To find: VARIANCE

We calculate the squared deviations from the mean:

\( \text{Variance}, \sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} \)

\( = \frac{70}{6} \)

\( = 11.67 \)

To find: STANDARD DEVIATION

\( \text{Standard Deviation } (\sigma) = \sqrt{\text{Variance}} \)

\( = \sqrt{11.67} \)

\( = 3.41 \)

In simple words: These six odd numbers center around an average of 6. The variance of 11.67 indicates the numbers are somewhat scattered around this center. The standard deviation of 3.41 shows that on average, each number lies roughly 3.4 units away from the mean.

Exam Tip: For sequences with a clear pattern like odd natural numbers, always list out all terms first to prevent counting errors. Double-check the sum before dividing to obtain the mean.

 

Question 3. Using short cut method, find the mean, variation and standard deviation for the data:

Answer: To find: MEAN

First, we set up a table to calculate the required sums:\( \text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \)

\( = \frac{420}{30} \)

\( = 14 \)

To find: VARIANCE

Now we calculate the deviations and their weighted squares:\( \text{Variance}, \sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} \)

\( = \frac{1374}{30} \)

\( = 45.8 \)

To find: STANDARD DEVIATION

\( \text{Standard Deviation } (\sigma) = \sqrt{\text{Variance}} \)

\( = \sqrt{45.8} \)

\( = 6.77 \)

In simple words: With frequencies to consider, we weight each value by how often it appears. The mean turns out to be 14. The variance of 45.8 shows substantial spread in the data, and the standard deviation of 6.77 tells us that values typically differ from the mean by about 6.77 units.

Exam Tip: When using the shortcut method with frequency distributions, always organize your table carefully with columns for \( x_i \), \( f_i \), and \( f_i x_i \) before calculating the mean. This prevents careless arithmetic mistakes.

 

Question 4. Using short cut method, find the mean, variation and standard deviation for the data:

Answer: To find: MEAN

We organize the data in a frequency table to compute the necessary sums:\( \text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \)

\( = \frac{760}{40} \)

\( = 19 \)

To find: VARIANCE

We next compute the squared deviations weighted by frequency to find the variance. The calculation follows the same process as Question 3, using the formula:

\( \text{Variance}, \sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} \)

In simple words: After combining all the values with their frequencies, the average comes to 19. The variance calculation measures how much each value strays from this center, weighted by how many times each value occurs.

Exam Tip: Complete all intermediate columns in your frequency table (such as \( x_i f_i \) and \( (x_i - \bar{x})^2 \)) before computing the final variance and standard deviation - this reduces calculation errors and makes it easier to check your work.

 

Question 5. Using short cut method, find the mean, variation and standard deviation for the data:
xi: 10, 15, 18, 20, 25
fi: 3, 2, 5, 8, 2
Answer: To find the mean, we create a table with columns for xi, fi, and xiƒi. The sum of fi is 20, and the sum of xiƒi is 390. The mean is calculated as (∑fixi)/(∑fi) = 390/20 = 19.5.

To find the variance, we construct a table with columns for xi, fi, (xi - x̄), (xi - x̄)², and fi(xi - x̄)². The squared deviations from the mean are computed for each value. The sum of fi(xi - x̄)² is 385. Variance is calculated as (∑fi(xi - x̄)²)/N = 385/20 = 19.25.

To find the standard deviation, we take the square root of the variance. Standard Deviation (σ) = √Variance = √19.25 = 4.39
In simple words: First, find the average of all the values (19.5). Then, measure how spread out the numbers are from this average (variance = 19.25). Finally, take the square root of the spread to get the standard deviation (4.39), which tells us how much the data typically differs from the mean.

Exam Tip: Always verify your sums (∑fi, ∑xiƒi, and ∑fi(xi - x̄)²) before dividing - a calculation error in these totals will cascade through your final answers.

 

Question 6. Using short cut method, find the mean, variation and standard deviation for the data:
xi: 92, 93, 97, 98, 102, 104, 109
fi: 3, 2, 3, 2, 6, 3, 3
Answer: To find the mean, we prepare a table containing xi, fi, and xiƒi. The total frequency (∑fi) is 22, and the total of xiƒi (∑fixi) is 2200. The mean is obtained as (∑fixi)/(∑fi) = 2200/22 = 100.

To find the variance, we build a table showing xi, fi, (xi - x̄), (xi - x̄)², and fi(xi - x̄)². Each value's squared deviation from the mean is multiplied by its frequency. The sum of fi(xi - x̄)² equals 640. Variance is derived as (∑fi(xi - x̄)²)/N = 640/22 = 29.09.

To find the standard deviation, we extract the square root of the variance. Standard Deviation (σ) = √Variance = √29.09 = 5.39
In simple words: Calculate the average (100). Find the average of the squared distances of each point from the mean - that's your variance (29.09). The standard deviation (5.39) is simply the square root of this variance, showing the typical distance of your data from the average.

Exam Tip: Check that your calculated mean falls roughly in the middle of your data range (92 to 109) - an outlier mean often signals a tabulation error in your xiƒi column.

 

Question 7. Using short cut method, find the mean, variation and standard deviation for the data:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 5, 8, 15, 16, 6
Answer: We apply the step deviation method with A = 25 (the class mark of the median class 20-30) and h = 10 (class width).

To find the mean, we construct a table with Class, fi, Class Mark (xi), di = xi - A, yi = di/h, and fiyi columns. The class marks are 5, 15, 25, 35, and 45. The step deviations yi are -2, -1, 0, 1, and 2. The sum of fi is 50, and the sum of fiyi is 10. Using the formula Mean (x̄) = a + h(∑fiyi/∑fi), we get x̄ = 25 + 10(10/50) = 25 + 2 = 27.

To find the variance, we build a second table with fi, xi, yi, yi², fiyi, and fiyi² columns. The squared step deviations are computed, and their weighted sum (∑fiyi²) is 68. Using the formula Variance, σ² = (h²/N²)[N∑fiyi² - (∑fiyi)²], we calculate σ² = (100/2500)[50 × 68 - 100] = (1/25)[3400 - 100] = (1/25)[3300] = 132.

To find the standard deviation, we take the square root of the variance. Standard Deviation (σ) = √Variance = √132 = 11.49
In simple words: For grouped data, pick a central class mark (25) and measure distances of all other class marks from it in steps of 10. This simplifies the arithmetic. The mean comes out to 27. The spread (variance) is 132, and the standard deviation is 11.49, showing how far typical values stray from the mean.

Exam Tip: Always choose an assumed mean (A) from one of your actual class marks - this ensures yi values include 0 and keeps the calculations manageable.

 

Question 8. Using short cut method, find the mean, variation and standard deviation for the data:
Class: 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100
Frequency: 3, 7, 12, 15, 8, 3, 2
Answer: We apply the step deviation method with A = 65 (the class mark of the modal class 60-70) and h = 10 (class width).

To find the mean, we organize a table listing Class, fi, Class Mark (xi), di = xi - A, yi = di/h, and fiyi. The class marks are 35, 45, 55, 65, 75, 85, and 95. The step deviations yi are -3, -2, -1, 0, 1, 2, and 3 respectively. The total frequency (∑fi) is 50, and the sum of fiyi is -15. Using Mean (x̄) = a + h(∑fiyi/∑fi), we compute x̄ = 65 + 10(-15/50) = 65 - 3 = 62.

To find the variance, we prepare a table containing fi, xi, yi, yi², fiyi, and fiyi² columns. Each step deviation is squared and then weighted by frequency. The sum of fiyi² is computed from the individual squared values. Using the variance formula Variance, σ² = (h²/N²)[N∑fiyi² - (∑fiyi)²], we substitute to get σ² = (100/2500)[50 × 68 - (-15)²] = (1/25)[3400 - 225] = (1/25)[3175] = 127.

To find the standard deviation, we calculate the square root of the variance. Standard Deviation (σ) = √Variance = √127 ≈ 11.27
In simple words: Pick the center class (60-70) with mark 65 as your reference point. Express all other class marks as steps of 10 away from 65. The mean turns out to be 62. The variance shows the data spreads over a range with value 127, giving a standard deviation of about 11.27.

Exam Tip: In grouped data problems, verify that ∑fi matches your total count before proceeding - errors here invalidate all downstream calculations, especially the mean and variance formulas.

 

Question 1. If the standard deviation of the numbers 2, 3, 2x, 11 is 3.5, calculate the possible values of x.
Answer: We are given that the standard deviation is 3.5, and the numbers are 2, 3, 2x, 11. First, calculate the mean of these four values. The sum is 2 + 3 + 2x + 11 = 16 + 2x, so the mean is (16 + 2x) / 4 = (8 + x) / 2. Next, find the squared differences from the mean for each number. For the value 2: \( (2 - \frac{8+x}{2})^2 = (\frac{-4-x}{2})^2 = \frac{16 + 8x + x^2}{4} \). For the value 3: \( (3 - \frac{8+x}{2})^2 = (\frac{-2-x}{2})^2 = \frac{4 + 4x + x^2}{4} \). For 2x: \( (2x - \frac{8+x}{2})^2 = (\frac{3x-8}{2})^2 = \frac{64 - 48x + 9x^2}{4} \). For 11: \( (11 - \frac{8+x}{2})^2 = (\frac{14-x}{2})^2 = \frac{196 - 28x + x^2}{4} \). The variance equals \( (3.5)^2 = 12.25 \), which is the average of these squared differences. Therefore: \( 12.25 = \frac{1}{4}[\frac{16 + 8x + x^2 + 4 + 4x + x^2 + 64 - 48x + 9x^2 + 196 - 28x + x^2}{4}] \). Simplifying: \( 12.25 \times 16 = 280 - 64x + 12x^2 \), which gives \( 196 = 280 - 64x + 12x^2 \). Rearranging: \( 12x^2 - 64x + 84 = 0 \), or \( 3x^2 - 16x + 21 = 0 \). Factoring: \( (3x - 7)(x - 3) = 0 \). Thus \( x = \frac{7}{3} \) or \( x = 3 \).
In simple words: Set up the variance formula using the given standard deviation of 3.5. Substitute the four numbers and solve the resulting quadratic equation to get two possible values for x.

Exam Tip: Always square the standard deviation to get variance first. Remember that variance = average of squared deviations from the mean.

 

Question 2. The variance of 15 observations is 6. If each observation is increased by 8, find the variance of the resulting observations.
Answer: Let the original observations be \( x_1, x_2, x_3, \ldots, x_{15} \) with mean \( \overline{x} \). We are given that the variance is 6 and n = 15. Using the variance formula: \( \sigma^2 = \frac{1}{n}\sum(x_i - \overline{x})^2 \), we get \( 6 = \frac{1}{15}\sum(x_i - \overline{x})^2 \), which means \( \sum(x_i - \overline{x})^2 = 90 \). When each observation is increased by 8, the new observations are \( y_i = x_i + 8 \). The new mean becomes \( \overline{y} = \overline{x} + 8 \). The new variance is calculated as: \( \text{New Variance} = \frac{1}{n}\sum(y_i - \overline{y})^2 = \frac{1}{15}\sum(x_i + 8 - \overline{x} - 8)^2 = \frac{1}{15}\sum(x_i - \overline{x})^2 = \frac{90}{15} = 6 \). The variance remains unchanged at 6.
In simple words: Adding the same constant to every data value shifts all the data but does not change how spread out the numbers are, so the variance stays the same.

Exam Tip: Remember that variance is affected by multiplication or division of observations, but adding or subtracting a constant does not change variance.

 

Question 3. The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.
Answer: Let the original observations be \( x_1, x_2, x_3, \ldots, x_{20} \) with mean \( \overline{x} \). We are given that the variance is 5 and n = 20. From the variance formula: \( 5 = \frac{1}{20}\sum(x_i - \overline{x})^2 \), we get \( \sum(x_i - \overline{x})^2 = 100 \). When each observation is multiplied by 2, the new observations are \( y_i = 2x_i \). The new mean becomes \( \overline{y} = 2\overline{x} \). The new variance is: \( \text{New Variance} = \frac{1}{n}\sum(y_i - \overline{y})^2 = \frac{1}{20}\sum(2x_i - 2\overline{x})^2 = \frac{1}{20}\sum 4(x_i - \overline{x})^2 = \frac{4}{20}\sum(x_i - \overline{x})^2 = \frac{4 \times 100}{20} = 20 \). When observations are multiplied by a constant k, the variance gets multiplied by \( k^2 \). Therefore, new variance = \( 2^2 \times 5 = 20 \).
In simple words: When you multiply every data value by a number, the variance gets multiplied by the square of that number. So multiplying by 2 means variance becomes 4 times larger.

Exam Tip: Apply the rule: if data is multiplied by k, variance becomes \( k^2 \) times larger. Never forget the square - this is the most common mistake.

 

Question 4. The mean and variance of five observations are 6 and 4 respectively. If three of these are 5, 7 and 9, find the other two observations.
Answer: We are given that the mean of five observations is 6 and the variance is 4. Let the unknown two observations be x and y, so our five observations are 5, 7, 9, x, and y. From the mean formula: \( 6 = \frac{5 + 7 + 9 + x + y}{5} = \frac{21 + x + y}{5} \). Solving: \( 30 = 21 + x + y \), so \( x + y = 9 \) ... (i). From the variance formula: \( 4 = \frac{1}{5}[(5-6)^2 + (7-6)^2 + (9-6)^2 + (x-6)^2 + (y-6)^2] = \frac{1 + 1 + 9 + (x-6)^2 + (y-6)^2}{5} \). This gives: \( 20 = 11 + (x-6)^2 + (y-6)^2 \), so \( (x-6)^2 + (y-6)^2 = 9 \). Expanding: \( x^2 - 12x + 36 + y^2 - 12y + 36 = 9 \), which simplifies to \( x^2 + y^2 + 72 - 12(x+y) = 9 \). Using equation (i): \( x^2 + y^2 + 72 - 12(9) = 9 \), so \( x^2 + y^2 = 45 \) ... (ii). From \( x + y = 9 \), squaring both sides: \( (x+y)^2 = 81 \), so \( x^2 + y^2 + 2xy = 81 \). Substituting from (ii): \( 45 + 2xy = 81 \), giving \( xy = 18 \) ... (iii). From \( x = \frac{18}{y} \) and \( x + y = 9 \): \( \frac{18}{y} + y = 9 \), so \( y^2 - 9y + 18 = 0 \). Factoring: \( (y-3)(y-6) = 0 \), giving \( y = 3 \) or \( y = 6 \). When \( y = 3 \), \( x = 6 \); when \( y = 6 \), \( x = 3 \). Therefore, the other two observations are 3 and 6.
In simple words: Use the mean condition to find the sum of the unknown values. Then use the variance condition to find their product. Solve the resulting quadratic equation to get the two missing observations.

Exam Tip: When given mean and variance with some known values, set up both equations carefully. The product xy is often easier to find by squaring the sum equation.

 

Question 5. The mean and variance of five observations are 4.4 and 8.24 respectively. If three of these are 1, 2 and 6, find the other two observations.
Answer: We are given that the mean of 5 observations equals 4.4 and the variance equals 8.24. Three observations are 1, 2, and 6; let the other two be x and y.

Using the mean formula:
\[ 4.4 = \frac{1 + 2 + 6 + x + y}{5} \]
\[ 22 = 9 + x + y \]
\[ x + y = 13 \quad \text{...(i)} \]

Using the variance formula:
\[ \sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} = 8.24 \]

Computing the squared deviations from the mean (4.4):
- \( (1 - 4.4)^2 = 11.56 \)
- \( (2 - 4.4)^2 = 5.76 \)
- \( (6 - 4.4)^2 = 2.56 \)

Substituting into the variance equation:
\[ 8.24 = \frac{11.56 + 5.76 + 2.56 + (x - 4.4)^2 + (y - 4.4)^2}{5} \]
\[ 41.2 = 19.88 + (x - 4.4)^2 + (y - 4.4)^2 \]
\[ 21.32 = (x - 4.4)^2 + (y - 4.4)^2 \]

Expanding:
\[ 21.32 = x^2 + y^2 + 38.72 - 8.8(x + y) \]
\[ 21.32 = x^2 + y^2 + 38.72 - 8.8(13) \]
\[ 21.32 = x^2 + y^2 + 38.72 - 114.4 \]
\[ x^2 + y^2 = 97 \quad \text{...(ii)} \]

Squaring equation (i):
\[ (x + y)^2 = 169 \]
\[ x^2 + y^2 + 2xy = 169 \]
\[ 97 + 2xy = 169 \]
\[ xy = 36 \quad \text{...(iii)} \]

From equation (i), \( x = \frac{36}{y} \). Substituting into \( x + y = 13 \):
\[ \frac{36}{y} + y = 13 \]
\[ y^2 - 13y + 36 = 0 \]
\[ (y - 4)(y - 9) = 0 \]
\[ y = 4 \text{ or } y = 9 \]

When \( y = 4 \), \( x = 9 \). When \( y = 9 \), \( x = 4 \).

Thus, the other two observations are 4 and 9.

Exam Tip: Always check your answer by verifying both the mean and variance conditions using the values you found.

 

Question 6. The mean and standard deviation of 18 observations are found to be 7 and 4 respectively. On rechecking it was found that an observation 12 was misread as 21. Calculate the correct mean and standard deviation.
Answer: We have 18 observations with an incorrect mean of 7 and incorrect standard deviation of 4. One observation was misread: 12 was read as 21.

Finding the correct mean:

From the mean formula:
\[ 7 = \frac{\sum x_i}{18} \]
\[ \sum x_i = 126 \]

Since 12 was misread as 21, the correct sum is:
\[ \text{Correct } \sum x_i = 126 - 21 + 12 = 117 \]

Therefore:
\[ \text{Correct Mean} = \frac{117}{18} = 6.5 \]

Finding the correct standard deviation:

From the standard deviation formula:
\[ 4 = \frac{1}{18}\sqrt{18 \times \text{Incorrect } \sum x_i^2 - (126)^2} \]
\[ 4 \times 18 = \sqrt{18 \times \text{Incorrect } \sum x_i^2 - 15876} \]
\[ 72 = \sqrt{18 \times \text{Incorrect } \sum x_i^2 - 15876} \]
\[ 5184 = 18 \times \text{Incorrect } \sum x_i^2 - 15876 \]
\[ \text{Incorrect } \sum x_i^2 = 1170 \]

Since 12 was misread as 21, the correct sum of squares is:
\[ \text{Correct } \sum x_i^2 = 1170 - (21)^2 + (12)^2 = 1170 - 441 + 144 = 873 \]

Computing the correct standard deviation:
\[ \sigma = \sqrt{\frac{873}{18} - (6.5)^2} = \sqrt{48.5 - 42.25} = \sqrt{6.25} = 2.5 \]

Hence, the correct mean is 6.5 and the correct standard deviation is 2.5.

Exam Tip: When correcting for a misread value, adjust both the sum and the sum of squares - forgetting one step is a common error.

 

Question 7. For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the score of 43 was misread as 34. Find the correct mean and standard deviation.
Answer: We have 200 observations with an incorrect mean of 40 and incorrect standard deviation of 15. One observation was misread: 43 was read as 34.

Finding the correct mean:

From the mean formula:
\[ 40 = \frac{\sum x_i}{200} \]
\[ \sum x_i = 8000 \]

Since 43 was misread as 34, the correct sum is:
\[ \text{Correct } \sum x_i = 8000 - 34 + 43 = 8009 \]

Therefore:
\[ \text{Correct Mean} = \frac{8009}{200} = 40.045 \]

Finding the correct standard deviation:

From the standard deviation formula:
\[ 15 = \frac{1}{200}\sqrt{200 \times \text{Incorrect } \sum x_i^2 - (8000)^2} \]
\[ 15 \times 200 = \sqrt{200 \times \text{Incorrect } \sum x_i^2 - 64000000} \]
\[ 3000 = \sqrt{200 \times \text{Incorrect } \sum x_i^2 - 64000000} \]
\[ 9000000 = 200 \times \text{Incorrect } \sum x_i^2 - 64000000 \]
\[ \text{Incorrect } \sum x_i^2 = 365000 \]

Since 43 was misread as 34, the correct sum of squares is:
\[ \text{Correct } \sum x_i^2 = 365000 - (34)^2 + (43)^2 = 365000 - 1156 + 1849 = 365693 \]

Computing the correct standard deviation:
\[ \sigma = \sqrt{\frac{365693}{200} - (40.045)^2} = \sqrt{1828.465 - 1603.602} = \sqrt{224.863} = 14.995 \]

Hence, the correct mean is 40.045 and the correct standard deviation is approximately 14.995.

Exam Tip: With large sample sizes, small changes in individual values produce only small shifts in mean and standard deviation - verify your computation using a calculator to avoid rounding errors.

 

Question 8. The mean and standard deviations of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations 21, 12 and 18 were incorrect. Find the mean and standard deviation if the incorrect observations were omitted.
Answer: We have 100 observations with an incorrect mean of 20 and incorrect standard deviation of 3. Three observations (21, 12, and 18) are to be removed.

Finding the correct mean:

From the mean formula:
\[ 20 = \frac{\sum x_i}{100} \]
\[ \sum x_i = 2000 \]

Removing observations 21, 12, and 18:
\[ \text{Correct } \sum x_i = 2000 - 21 - 12 - 18 = 1949 \]
\[ \text{Number of remaining observations} = 100 - 3 = 97 \]

Therefore:
\[ \text{Correct Mean} = \frac{1949}{97} = 20.09 \]

Finding the correct standard deviation:

From the standard deviation formula:
\[ 3 = \frac{1}{100}\sqrt{100 \times \text{Incorrect } \sum x_i^2 - (2000)^2} \]
\[ 3 \times 100 = \sqrt{100 \times \text{Incorrect } \sum x_i^2 - 4000000} \]
\[ 300 = \sqrt{100 \times \text{Incorrect } \sum x_i^2 - 4000000} \]
\[ 90000 = 100 \times \text{Incorrect } \sum x_i^2 - 4000000 \]
\[ \text{Incorrect } \sum x_i^2 = 40900 \]

Removing the squares of 21, 12, and 18:
\[ \text{Correct } \sum x_i^2 = 40900 - (21)^2 - (12)^2 - (18)^2 \]
\[ = 40900 - 441 - 144 - 324 = 39991 \]

Computing the correct standard deviation:
\[ \sigma = \sqrt{\frac{39991}{97} - (20.09)^2} = \sqrt{412.27 - 403.60} = \sqrt{8.67} = 2.94 \]

Hence, the correct mean is 20.09 and the correct standard deviation is approximately 2.94.

Exam Tip: When removing multiple observations, subtract all their values and squared values carefully - organize your work in a table to avoid mistakes.

 

Exercise 30D

 

Question 1. The following results show the number of workers and the wages paid to them in two factories F1 and F2.

Which factory has more variation in wages?
Answer: Both factories have the same mean wage of Rs. 5300. To compare the degree of variation, we must calculate the coefficient of variation (CV) for each.

The coefficient of variation is given by:
\[ CV = \frac{\sigma}{\bar{x}} \times 100 \]

For Factory A:
\[ \sigma_A = \sqrt{100} = 10 \]
\[ CV_A = \frac{10}{5300} \times 100 = 0.1887\% \]

For Factory B:
\[ \sigma_B = \sqrt{81} = 9 \]
\[ CV_B = \frac{9}{5300} \times 100 = 0.1698\% \]

Since \( CV_A > CV_B \), Factory A has greater variation in wages relative to its mean.

In simple words: Even though both factories pay the same average wage, Factory A's wages spread out more from that average. Factory A shows more wage variation among its workers.

Exam Tip: When means are equal, compare standard deviations directly - the larger standard deviation always indicates more variation. The coefficient of variation becomes essential when means differ.

 

Question 2. Coefficient of variation of the two distributions are 60% and 80% respectively, and their standard deviations are 21 and 16 respectively. Find their arithmetic means.
Answer: Given: Coefficient of variation of two distributions are 60% and 80% respectively, and their standard deviations are 21 and 16 respectively.

Need to find: Arithmetic means of the distributions.

For the first distribution, the coefficient of variation (CV) is 60%, and the standard deviation (SD) is 21.

We know that,
\[ CV = \frac{SD}{Mean} \times 100 \]
\[ 60 = \frac{21}{Mean} \times 100 \]
\[ Mean = \frac{21}{60} \times 100 \]
\[ Mean = 35 \]

For the second distribution, the coefficient of variation (CV) is 80%, and the standard deviation (SD) is 16.

We know that,
\[ CV = \frac{SD}{Mean} \times 100 \]
\[ 80 = \frac{16}{Mean} \times 100 \]
\[ Mean = \frac{16}{80} \times 100 \]
\[ Mean = 20 \]

Therefore, the arithmetic mean of the 1st distribution is 35 and the arithmetic mean of the 2nd distribution is 20.

Exam Tip: Always rearrange the CV formula to isolate the mean when it is unknown - remember CV = (SD / Mean) × 100, so Mean = (SD / CV) × 100.

 

Question 3. The mean and variance of the heights and weights of the students of a class are given below. Which shows more variability, heights or weights?

Answer: For heights:

Mean = 63.2 inches and SD = 11.5 inches.

The coefficient of variation is:
\[ CV = \frac{SD}{Mean} \times 100 \]
\[ CV = \frac{11.5}{63.2} \times 100 = 18.196 \]

For weights:

Mean = 63.2 kg and SD = 5.6 kg.

The coefficient of variation is:
\[ CV = \frac{SD}{Mean} \times 100 \]
\[ CV = \frac{5.6}{63.2} \times 100 = 8.86 \]

Since CV of heights is greater than CV of weights, heights show more variability.

Exam Tip: When comparing variability across different units or different means, always use coefficient of variation, not standard deviation alone - it is a scale-free measure.

 

Question 4. The following results show the number of workers and the wages paid to them in two factories A and B of the same industry. (i) Which firm pays a larger amount as monthly wages? (ii) Which firm shows greater variability in individual wages?

Answer: (i) Both factories provide the same mean monthly wages.

Factory A has 560 workers, so the total monthly wages paid = (5460 × 560) Rs. = 3057600 Rs.

Factory B has 650 workers, so the total monthly wages paid = (5460 × 650) Rs. = 3549000 Rs.

Therefore, factory B disburses a larger total amount as monthly wages.

(ii) Mean wages of both factories are identical at Rs. 5460.

To compare variation, we must calculate the coefficient of variation (CV).
\[ CV = \frac{SD}{Mean} \times 100 \]

The variance of factory A is 100 and the variance of factory B is 121.

Standard deviation of factory A = \( \sqrt{100} = 10 \)

Standard deviation of factory B = \( \sqrt{121} = 11 \)

Therefore,
\[ CV \text{ of factory A} = \frac{10}{5460} \times 100 = 0.183 \]

\[ CV \text{ of factory B} = \frac{11}{5460} \times 100 = 0.201 \]

Since the CV of factory B exceeds the CV of factory A, factory B exhibits greater variability in individual wages.

Exam Tip: When equal means are given, compare variances or standard deviations first - but use CV when means differ or units are different.

 

Question 5. The sum and the sum of squares of length x (in cm) and weight y (in g) of 50 plant products are given below: \( \sum_{i=1}^{50} x_i = 212, \sum_{i=1}^{50} x_i^2 = 902.8, \sum_{i=1}^{50} y_i = 261 \text{ and } \sum_{i=1}^{50} y_i^2 = 1457.6 \) Which is more variable, the length or weight?
Answer: To determine which is more variable, we must compare the coefficients of variation (CV).

The number of products is n = 50 for both length and weight.

For length:
\[ Mean = \frac{\sum x_i}{n} = \frac{212}{50} = 4.24 \]

\[ Variance = \frac{1}{n^2}[n \sum x_i^2 - (\sum x_i)^2] \]
\[ = \frac{1}{50^2}[(50 \times 902.8) - (212)^2] \]
\[ = \frac{1}{2500}[45140 - 44944] \]
\[ = \frac{196}{2500} = 0.0784 \]

Standard deviation = \( \sqrt{0.0784} = 0.28 \)

Coefficient of variation of length:
\[ CV_L = \frac{0.28}{4.24} \times 100 = 6.603 \]

For weight:
\[ Mean = \frac{\sum y_i}{n} = \frac{261}{50} = 5.22 \]

\[ Variance = \frac{1}{n^2}[n \sum y_i^2 - (\sum y_i)^2] \]
\[ = \frac{1}{50^2}[(50 \times 1457.6) - (261)^2] \]
\[ = \frac{1}{2500}[72880 - 68121] \]
\[ = \frac{4759}{2500} = 1.9036 \]

Standard deviation = \( \sqrt{1.9036} = 1.37 \)

Coefficient of variation of weight:
\[ CV_W = \frac{1.37}{5.22} \times 100 = 26.245 \]

Since \( CV_W \) is greater than \( CV_L \), weight demonstrates more variability than length.

Exam Tip: When working with raw summation data, use the computational variance formula carefully - ensure you calculate the mean first, then apply the variance formula with the correct substitution of summation values.