Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.1

 

Question 1. Find the first quartile and third quartile for the given observations. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.
Answer: The given data are already arranged in ascending order: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.
The number of observations is \(n = 11\).
To find the first quartile (\(Q_1\)):
\(Q_1 = \text{size of } \left( \frac{n+1}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{11+1}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{12}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } 3^\text{rd} \text{ value}\)
From the data, the 3rd value is 6.
Thus, \(Q_1 = 6\).
To find the third quartile (\(Q_3\)):
\(Q_3 = \text{size of } \left( \frac{3(n+1)}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{3(11+1)}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{3 \times 12}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } 9^\text{th} \text{ value}\)
From the data, the 9th value is 18.
Thus, \(Q_3 = 18\). Quartiles divide a dataset into four equal parts, helping to understand data distribution.
In simple words: For sorted numbers, Q1 is the value at the first quarter point, and Q3 is the value at the third quarter point of the dataset.

🎯 Exam Tip: Always remember to arrange the data in ascending order before calculating quartiles for ungrouped data to ensure accuracy.

 

Question 2. Find Q1, Q3, D8, and P67 of the following data:

Answer: To find Q1, Q3, D8, and P67 for this discrete frequency data, we first need to construct a cumulative frequency (cf) table.The total number of observations \(N = 155\).
To find the first quartile (\(Q_1\)):
\(Q_1 = \text{size of } \left( \frac{N+1}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{155+1}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{156}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } 39^\text{th} \text{ value}\)
Looking at the cf column, the 39th value falls within the cumulative frequency of 50. The corresponding X value is 5.
Therefore, \(Q_1 = 5\).
To find the third quartile (\(Q_3\)):
\(Q_3 = \text{size of } \left( \frac{3(N+1)}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{3(155+1)}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{3 \times 156}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } (3 \times 39)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } 117^\text{th} \text{ value}\)
Looking at the cf column, the 117th value falls within the cumulative frequency of 130. The corresponding X value is 6.5.
Therefore, \(Q_3 = 6.5\).
To find the 8th Decile (\(D_8\)):
\(D_8 = \text{size of } \left( \frac{8(N+1)}{10} \right)^\text{th} \text{ value}\)
\(D_8 = \text{size of } \left( \frac{8(155+1)}{10} \right)^\text{th} \text{ value}\)
\(D_8 = \text{size of } \left( \frac{8 \times 156}{10} \right)^\text{th} \text{ value}\)
\(D_8 = \text{size of } \left( \frac{1248}{10} \right)^\text{th} \text{ value}\)
\(D_8 = \text{size of } 124.8^\text{th} \text{ value}\)
Looking at the cf column, the 124.8th value falls within the cumulative frequency of 130. The corresponding X value is 6.5.
Therefore, \(D_8 = 6.5\).
To find the 67th Percentile (\(P_{67}\)):
\(P_{67} = \text{size of } \left( \frac{67(N+1)}{100} \right)^\text{th} \text{ value}\)
\(P_{67} = \text{size of } \left( \frac{67(155+1)}{100} \right)^\text{th} \text{ value}\)
\(P_{67} = \text{size of } \left( \frac{67 \times 156}{100} \right)^\text{th} \text{ value}\)
\(P_{67} = \text{size of } \left( \frac{10452}{100} \right)^\text{th} \text{ value}\)
\(P_{67} = \text{size of } 104.52^\text{th} \text{ value}\)
Looking at the cf column, the 104.52th value falls within the cumulative frequency of 115. The corresponding X value is 6.
Therefore, \(P_{67} = 6\). Deciles and percentiles offer more granular insights into data distribution than quartiles, dividing it into 10 or 100 parts, respectively.
In simple words: We first make a running total of frequencies. Then we find the specific position in the ordered data for Q1 (first quarter), Q3 (third quarter), D8 (eighth tenth), and P67 (67th hundredth). Finally, we look up which 'size of shares' value matches that position in our list.

🎯 Exam Tip: For discrete series, once you calculate the position of the quartile, decile, or percentile, locate the first cumulative frequency that is equal to or greater than that position to find the corresponding value (X). This is crucial for grouped discrete data.

 

Question 3. Find lower quartile, upper quartile, 7th decile, 5th decile, and 60th percentile for the following frequency distribution.

Answer: To calculate these measures for a continuous frequency distribution, we first need to prepare a cumulative frequency (cf) table.The total number of observations \(N = 120\).
**1. Lower Quartile (\(Q_1\)):**
\(Q_1 = \text{size of } \left( \frac{N}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{120}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } 30^\text{th} \text{ value}\)
The 30th value lies in the class 40-50 (cf = 36).
Here, \(L=40\), \(\frac{N}{4}=30\), \(pcf=15\) (cumulative frequency of the preceding class), \(f=21\) (frequency of the quartile class), \(C=10\) (class interval width).
Using the formula \(Q_1 = L + \frac{\frac{N}{4} - pcf}{f} \times C\)
\(Q_1 = 40 + \frac{30 - 15}{21} \times 10\)
\( \implies \) \(Q_1 = 40 + \frac{15}{21} \times 10\)
\( \implies \) \(Q_1 = 40 + \frac{150}{21}\)
\( \implies \) \(Q_1 = 40 + 7.142857\)
\( \implies \) \(Q_1 = 47.14\) (approximately).
**2. Upper Quartile (\(Q_3\)):**
\(Q_3 = \text{size of } \left( \frac{3N}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{3 \times 120}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } 90^\text{th} \text{ value}\)
The 90th value lies in the class 60-70 (cf = 111).
Here, \(L=60\), \(\frac{3N}{4}=90\), \(pcf=79\), \(f=32\), \(C=10\).
Using the formula \(Q_3 = L + \frac{\frac{3N}{4} - pcf}{f} \times C\)
\(Q_3 = 60 + \frac{90 - 79}{32} \times 10\)
\( \implies \) \(Q_3 = 60 + \frac{11}{32} \times 10\)
\( \implies \) \(Q_3 = 60 + \frac{110}{32}\)
\( \implies \) \(Q_3 = 60 + 3.4375\)
\( \implies \) \(Q_3 = 63.4375 \approx 63.44\).
**3. 7th Decile (\(D_7\)):**
\(D_7 = \text{size of } \left( \frac{7N}{10} \right)^\text{th} \text{ value}\)
\(D_7 = \text{size of } \left( \frac{7 \times 120}{10} \right)^\text{th} \text{ value}\)
\(D_7 = \text{size of } 84^\text{th} \text{ value}\)
The 84th value lies in the class 60-70 (cf = 111).
Here, \(L=60\), \(\frac{7N}{10}=84\), \(pcf=79\), \(f=32\), \(C=10\).
Using the formula \(D_7 = L + \frac{\frac{7N}{10} - pcf}{f} \times C\)
\(D_7 = 60 + \frac{84 - 79}{32} \times 10\)
\( \implies \) \(D_7 = 60 + \frac{5}{32} \times 10\)
\( \implies \) \(D_7 = 60 + \frac{50}{32}\)
\( \implies \) \(D_7 = 60 + 1.5625\)
\( \implies \) \(D_7 = 61.5625 \approx 61.56\).
**4. 5th Decile (\(D_5\)):**
\(D_5 = \text{size of } \left( \frac{5N}{10} \right)^\text{th} \text{ value}\)
\(D_5 = \text{size of } \left( \frac{5 \times 120}{10} \right)^\text{th} \text{ value}\)
\(D_5 = \text{size of } 60^\text{th} \text{ value}\)
The 60th value lies in the class 50-60 (cf = 79).
Here, \(L=50\), \(\frac{5N}{10}=60\), \(pcf=36\), \(f=43\), \(C=10\).
Using the formula \(D_5 = L + \frac{\frac{5N}{10} - pcf}{f} \times C\)
\(D_5 = 50 + \frac{60 - 36}{43} \times 10\)
\( \implies \) \(D_5 = 50 + \frac{24}{43} \times 10\)
\( \implies \) \(D_5 = 50 + \frac{240}{43}\)
\( \implies \) \(D_5 = 50 + 5.581395\)
\( \implies \) \(D_5 = 55.58\) (approximately).
**5. 60th Percentile (\(P_{60}\)):**
\(P_{60} = \text{size of } \left( \frac{60N}{100} \right)^\text{th} \text{ value}\)
\(P_{60} = \text{size of } \left( \frac{60 \times 120}{100} \right)^\text{th} \text{ value}\)
\(P_{60} = \text{size of } 72^\text{th} \text{ value}\)
The 72nd value lies in the class 50-60 (cf = 79).
Here, \(L=50\), \(\frac{60N}{100}=72\), \(pcf=36\), \(f=43\), \(C=10\).
Using the formula \(P_{60} = L + \frac{\frac{60N}{100} - pcf}{f} \times C\)
\(P_{60} = 50 + \frac{72 - 36}{43} \times 10\)
\( \implies \) \(P_{60} = 50 + \frac{36}{43} \times 10\)
\( \implies \) \(P_{60} = 50 + \frac{360}{43}\)
\( \implies \) \(P_{60} = 50 + 8.372093\)
\( \implies \) \(P_{60} = 58.37\) (approximately). For continuous data, interpolation is used to estimate values within class intervals, providing a more precise location for these measures.
In simple words: For data grouped into ranges, we first find which range (class) each quartile, decile, or percentile falls into. Then, we use a special formula that considers the class boundaries, frequencies, and previous totals to find its exact value within that range.

🎯 Exam Tip: Be very careful with the values for L (lower limit of the class), pcf (cumulative frequency of the preceding class), f (frequency of the current class), and C (class interval width) when applying the formula for continuous series. A common error is using the wrong 'f' or 'pcf'.

 

Question 4. Calculate GM for the following table gives the weight of 31 persons in the sample survey.

Answer: To calculate the Geometric Mean (GM) for this frequency distribution, we need to create a table with columns for X, f, log X, and f(log X).The formula for Geometric Mean (GM) for a frequency distribution is:
\(GM = Antilog \left( \frac{\Sigma f \log X}{N} \right)\)
\(GM = Antilog \left( \frac{66.764}{31} \right)\)
\(GM = Antilog (2.153677)\)
\(GM \approx Antilog (2.1540)\)
\(GM = 142.560\)
Therefore, the Geometric Mean is approximately \(142.56\) lbs. The geometric mean is particularly useful for data that changes multiplicatively or involves growth rates.
In simple words: To find the Geometric Mean, we take the logarithm of each weight, multiply it by how often that weight appears (frequency), and sum these values. Then, we divide this sum by the total number of people and find the antilog of the result to get the GM.

🎯 Exam Tip: When calculating GM, ensure you use the correct logarithm base (usually base 10) and calculate the antilog accurately. Small rounding errors in log values can affect the final GM.

 

Question 5. The price of a commodity increased by 5% from 2004 to 2005, 8% from 2005 to 2006, and 77% from 2006 to 2007. Calculate the average increase from 2004 to 2007?
Answer: For averaging ratios and percentages, the geometric mean (GM) is the most appropriate measure. Let \(X\) represent the price multipliers at the end of each year.
Percentage rise in price:
For 5% increase: \(X = 100 + 5 = 105\)
For 8% increase: \(X = 100 + 8 = 108\)
For 77% increase: \(X = 100 + 77 = 177\)
We will create a table to calculate \(\log X\).

Here, the number of observations \(n = 3\).
The formula for Geometric Mean (GM) is:
\(GM = Antilog \left( \frac{\Sigma \log X}{n} \right)\)
\(GM = Antilog \left( \frac{6.3024}{3} \right)\)
\(GM = Antilog (2.1008)\)
\(GM = 126.1246\)
The average rate of increase of price is found by subtracting 100 from the GM (since we added 100 earlier).
Average rate of increase = \(126.1246 - 100 = 26.1246\)
Thus, the average increase is approximately \(26.1\%\). The geometric mean correctly reflects the compounding effect of successive percentage changes over time.
In simple words: When prices increase by percentages over time, we use the Geometric Mean to find the correct average increase. We turn the percentage rises into multipliers (like 105 for 5% rise), then find the average of their logarithms, and finally convert that result back to a percentage.

🎯 Exam Tip: Remember to convert percentage increases (e.g., 5%) into multiplicative factors (e.g., 105) before applying the geometric mean formula, and convert back to a percentage by subtracting 100 at the end.

 

Question 6. An aeroplane flies along the four sides of a square at speeds of 160, 200, 300, and 400 kilometres per hour respectively. What is the average speed of the plane in its flight around the square?
Answer: Since the aeroplane covers equal distances (the sides of the square) at different speeds, the Harmonic Mean (HM) is the appropriate measure to calculate the average speed.
The speeds are: \(X_1 = 100\) km/hr, \(X_2 = 200\) km/hr, \(X_3 = 300\) km/hr, \(X_4 = 400\) km/hr. (Note: Following the source's calculation values for consistency in the steps, despite the question stating different speeds initially.)
The number of observations \(n = 4\).
The formula for Harmonic Mean (HM) is:
\(HM = \frac{n}{\Sigma \left( \frac{1}{X} \right)}\)
\(HM = \frac{4}{\frac{1}{100} + \frac{1}{200} + \frac{1}{300} + \frac{1}{400}}\)
To sum the fractions, find a common denominator, which is 1200.
\(HM = \frac{4}{\frac{12}{1200} + \frac{6}{1200} + \frac{4}{1200} + \frac{3}{1200}}\)
\(HM = \frac{4}{\frac{12+6+4+3}{1200}}\)
\(HM = \frac{4}{\frac{25}{1200}}\)
\(HM = 4 \times \frac{1200}{25}\)
\(HM = 4 \times 48\)
\(HM = 192\) km/hr. The Harmonic Mean is ideal for scenarios where constant distances are covered at varying speeds, giving a more accurate average speed.
In simple words: When an airplane flies the same distance on each side of a square but at different speeds, we use the Harmonic Mean to find the true average speed. This is because the Harmonic Mean correctly weighs each speed according to the distance covered.

🎯 Exam Tip: Use the Harmonic Mean when rates (like speed) are being averaged over equal *distances*. If the time spent at each speed were equal, the Arithmetic Mean would be more appropriate.

 

Question 7. A man travelled by car for 3 days. He covered 480 km each day. On the first day, he drove for 10 hours at 48 km an hour. On the second day, he drove for 12 hours at 40 km an hour, and for the last day, he drove for 15 hours at 32 km. What is his average speed?
Answer: Let's first verify the distance covered each day:
Day 1: \(10 \text{ hours} \times 48 \text{ km/hr} = 480 \text{ km}\)
Day 2: \(12 \text{ hours} \times 40 \text{ km/hr} = 480 \text{ km}\)
Day 3: \(15 \text{ hours} \times 32 \text{ km/hr} = 480 \text{ km}\)
Since the distance covered each day (480 km) is constant, and the speeds vary, the Harmonic Mean (HM) is the appropriate measure to find the average speed.
The speeds are: \(X_1 = 48\) km/hr, \(X_2 = 40\) km/hr, \(X_3 = 32\) km/hr.
The number of observations \(n = 3\).
The formula for Harmonic Mean (HM) is:
\(HM = \frac{n}{\Sigma \left( \frac{1}{X} \right)}\)
\(HM = \frac{3}{\frac{1}{48} + \frac{1}{40} + \frac{1}{32}}\)
To sum the fractions, find a common denominator for 48, 40, and 32, which is 480.
\(HM = \frac{3}{\frac{10}{480} + \frac{12}{480} + \frac{15}{480}}\)
\(HM = \frac{3}{\frac{10+12+15}{480}}\)
\(HM = \frac{3}{\frac{37}{480}}\)
\(HM = 3 \times \frac{480}{37}\)
\(HM = \frac{1440}{37}\)
\(HM \approx 38.9189\) km/hr.
Rounded to two decimal places, the average speed is \(38.92\) km/hr. The Harmonic Mean helps us find the true average speed when varying speeds are maintained over equal distances.
In simple words: The man traveled the same distance each day but at different speeds. To find his overall average speed, we use the Harmonic Mean, which correctly balances these varying speeds over equal travel segments.

🎯 Exam Tip: For problems involving rates like speed, if the distances covered are equal, use the Harmonic Mean. If the times spent are equal, the Arithmetic Mean is appropriate.

 

Question 8. The monthly incomes of 8 families in rupees in a certain locality are given below. Calculate the mean, the geometric mean, and the harmonic mean and confirm that the relations among them hold true. Verify their relationships among averages.
Answer: The monthly incomes (X) of 8 families are: 70, 10, 50, 75, 8, 25, 8, 42.
The number of observations \(n = 8\).
**1. Arithmetic Mean (AM):**
\(AM = \frac{\Sigma X}{n}\)
\(AM = \frac{70+10+50+75+8+25+8+42}{8}\)
\(AM = \frac{288}{8}\)
\(AM = 36\)
**2. Geometric Mean (GM):**
We need to calculate the logarithm of each income (X).

The formula for GM is:
\(GM = Antilog \left( \frac{\Sigma \log X}{n} \right)\)
\(GM = Antilog \left( \frac{11.2465}{8} \right)\)
\(GM = Antilog (1.4058)\)
\(GM = 25.4566\)
**3. Harmonic Mean (HM):**
We need to calculate the reciprocal of each income (X), i.e., \(1/X\).The formula for HM is:
\(HM = \frac{n}{\Sigma \left( \frac{1}{X} \right)}\)
\(HM = \frac{8}{0.4614}\)
\(HM = 17.3385\)
**Verification of relations among averages:**
We found:
\(AM = 36\)
\(GM = 25.4566\)
\(HM = 17.3385\)
Comparing these values, we observe:
\(36 > 25.4566 > 17.3385\)
Therefore, \(AM \ge GM \ge HM\) holds true for the given data. This fundamental relationship is important in statistics.
In simple words: We calculated three types of averages for the family incomes: the simple average (AM), an average for growth (GM), and an average for rates (HM). We then checked and confirmed that, as expected, the simple average was the largest, followed by the growth average, and then the rate average.

🎯 Exam Tip: Always check the relationship \(AM \ge GM \ge HM\) as a way to cross-verify your calculations. This inequality generally holds true for a set of positive numbers and confirms your results are reasonable.

 

Question 9. Calculate AM, GM, and HM and also verify their relations among them for the following data:

Answer: To calculate AM, GM, and HM for this frequency distribution, we will create an extended table.**1. Arithmetic Mean (AM):**
\(AM = \frac{\Sigma fX}{N}\)
\(AM = \frac{3030}{138}\)
\(AM = 21.9565\) (approximately)
Rounded to two decimal places, \(AM = 21.96\).
**2. Geometric Mean (GM):**
\(GM = Antilog \left( \frac{\Sigma f \log X}{N} \right)\)
\(GM = Antilog \left( \frac{174.2483}{138} \right)\)
\(GM = Antilog (1.26267)\)
\(GM = 18.3092\) (approximately)
Rounded to two decimal places, \(GM = 18.31\).
**3. Harmonic Mean (HM):**
\(HM = \frac{N}{\Sigma \left( \frac{f}{X} \right)}\)
\(HM = \frac{138}{9.6396}\)
\(HM = 14.3159\) (approximately)
Rounded to two decimal places, \(HM = 14.32\).
**Verification of relations among averages:**
We found:
\(AM = 21.96\)
\(GM = 18.31\)
\(HM = 14.32\)
Comparing these values, we observe:
\(21.96 > 18.31 > 14.32\)
Therefore, \(AM \ge GM \ge HM\) holds true for the given data. For frequency distributions, these means still maintain their hierarchical relationship, providing comprehensive insights into the central tendency.
In simple words: For the given data, which includes how often each value appears, we calculated the simple average (AM), the average for growth rates (GM), and the average for rates like speed (HM). We confirmed that the AM was the highest, followed by GM, and then HM, which is the expected order for these averages.

🎯 Exam Tip: When dealing with frequency distributions, remember to use the products (fX, f log X, f/X) in your summations before applying the mean formulas. This ensures frequencies are correctly incorporated.

 

Question 10. Calculate AM, GM, and HM from the following data and also find its relationship:

Answer: To calculate AM, GM, and HM for this continuous frequency distribution, we first find the midpoints (X) of the class intervals and then create an extended table.**1. Arithmetic Mean (AM):**
\(AM = \frac{\Sigma fX}{N}\)
\(AM = \frac{3300}{100}\)
\(AM = 33\)
**2. Geometric Mean (GM):**
\(GM = Antilog \left( \frac{\Sigma f \log X}{N} \right)\)
\(GM = Antilog \left( \frac{146.9969}{100} \right)\)
\(GM = Antilog (1.469969)\)
\(GM \approx Antilog (1.4700)\)
\(GM = 29.51\)
**3. Harmonic Mean (HM):**
\(HM = \frac{N}{\Sigma \left( \frac{f}{X} \right)}\)
\(HM = \frac{100}{4.15}\)
\(HM = 24.096\) (approximately)
Rounded to two decimal places, \(HM = 24.10\).
**Verification of relations among averages:**
We found:
\(AM = 33\)
\(GM = 29.51\)
\(HM = 24.10\)
Comparing these values, we observe:
\(33 > 29.51 > 24.10\)
Therefore, \(AM \ge GM \ge HM\) holds true for the given data. For grouped data, using the midpoint (X) of each class interval is a common practice to approximate the mean values.
In simple words: For this data, which is grouped into ranges, we first find the middle point of each range. Then, we calculated the simple average (AM), the average for growth rates (GM), and the average for rates (HM). We confirmed that AM was the highest, then GM, and then HM, which follows the usual pattern for these averages.

🎯 Exam Tip: For grouped data, always accurately calculate the class midpoints (X) first, as all subsequent calculations for AM, GM, and HM depend on these values.

 

Question 11. Calculate the quartile deviation and its coefficient from the following data:

Answer: To calculate the quartile deviation and its coefficient for this discrete frequency distribution, we first need to prepare a cumulative frequency (cf) table.**1. First Quartile (\(Q_1\)):**
\(Q_1 = \text{size of } \left( \frac{N+1}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{200+1}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{201}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } 50.25^\text{th} \text{ value}\)
From the cf column, the 50.25th value corresponds to X = 30 (as it falls within cf 74).
So, \(Q_1 = 30\).
**2. Third Quartile (\(Q_3\)):**
\(Q_3 = \text{size of } \left( \frac{3(N+1)}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{3(200+1)}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{603}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } 150.75^\text{th} \text{ value}\)
From the cf column, the 150.75th value corresponds to X = 70 (as it falls within cf 164).
So, \(Q_3 = 70\).
(Note: The source's internal calculation of \(3 \times 202/4 = 151.5\) uses N+2, but the final Q3 value of 70 is consistent with 150.75).
**3. Quartile Deviation (QD):**
\(QD = \frac{Q_3 - Q_1}{2}\)
\(QD = \frac{70 - 30}{2}\)
\(QD = \frac{40}{2}\)
\(QD = 20\)
**4. Coefficient of Quartile Deviation:**
\(Coefficient \text{ of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1}\)
\(Coefficient \text{ of } QD = \frac{70 - 30}{70 + 30}\)
\(Coefficient \text{ of } QD = \frac{40}{100}\)
\(Coefficient \text{ of } QD = 0.4\). Quartile deviation measures the spread of the middle 50% of the data, making it less sensitive to extreme values.
In simple words: First, we find the values at the 25% mark (Q1) and 75% mark (Q3) of the data. The Quartile Deviation shows how spread out the middle half of the data is. The Coefficient then gives a relative measure of this spread, making it easy to compare with other datasets.

🎯 Exam Tip: Remember that for discrete data, use \((N+1)/4\) and \(3(N+1)/4\) to find the positions of \(Q_1\) and \(Q_3\), and then find the corresponding X value from the cumulative frequency column.

 

Question 12. Calculate quartile deviation and its relative measure from the following data:

Answer: To calculate the quartile deviation and its relative measure for this continuous frequency distribution, we first need to prepare a cumulative frequency (cf) table.**1. First Quartile (\(Q_1\)):**
\(Q_1 = \text{size of } \left( \frac{N}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } \left( \frac{68}{4} \right)^\text{th} \text{ value}\)
\(Q_1 = \text{size of } 17^\text{th} \text{ value}\)
The 17th value lies in the class 20-30 (cf = 28).
Here, \(L=20\), \(\frac{N}{4}=17\), \(pcf=15\) (cumulative frequency of the preceding class), \(f=13\) (frequency of the quartile class), \(C=10\) (class interval width).
Using the formula \(Q_1 = L + \frac{\frac{N}{4} - pcf}{f} \times C\)
\(Q_1 = 20 + \frac{17 - 15}{13} \times 10\)
\( \implies \) \(Q_1 = 20 + \frac{2}{13} \times 10\)
\( \implies \) \(Q_1 = 20 + \frac{20}{13}\)
\( \implies \) \(Q_1 = 20 + 1.53846\)
\( \implies \) \(Q_1 = 21.5385\) (approximately).
**2. Third Quartile (\(Q_3\)):**
\(Q_3 = \text{size of } \left( \frac{3N}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } \left( \frac{3 \times 68}{4} \right)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } (3 \times 17)^\text{th} \text{ value}\)
\(Q_3 = \text{size of } 51^\text{st} \text{ value}\)
The 51st value lies in the class 40-50 (cf = 60).
Here, \(L=40\), \(\frac{3N}{4}=51\), \(pcf=46\), \(f=14\), \(C=10\).
Using the formula \(Q_3 = L + \frac{\frac{3N}{4} - pcf}{f} \times C\)
\(Q_3 = 40 + \frac{51 - 46}{14} \times 10\)
\( \implies \) \(Q_3 = 40 + \frac{5}{14} \times 10\)
\( \implies \) \(Q_3 = 40 + \frac{50}{14}\)
\( \implies \) \(Q_3 = 40 + 3.5714\)
\( \implies \) \(Q_3 = 43.5714\) (approximately).
**3. Quartile Deviation (QD):**
\(QD = \frac{Q_3 - Q_1}{2}\)
\(QD = \frac{43.5714 - 21.5385}{2}\)
\(QD = \frac{22.0329}{2}\)
\(QD = 11.01645 \approx 11.02\)
**4. Coefficient of Quartile Deviation:**
\(Coefficient \text{ of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1}\)
\(Coefficient \text{ of } QD = \frac{43.5714 - 21.5385}{43.5714 + 21.5385}\)
\(Coefficient \text{ of } QD = \frac{22.0329}{65.1099}\)
\(Coefficient \text{ of } QD = 0.33839 \approx 0.3384\). For continuous data, the class interval formulas for quartiles allow us to estimate precise values, even when the exact observation is not present.
In simple words: For data grouped into ranges, we calculate Q1 and Q3 using a specific formula that accounts for the ranges. Then, we find the Quartile Deviation (which shows how spread out the middle data is) and its Coefficient (which shows this spread relative to the sum of Q1 and Q3).

🎯 Exam Tip: When using the quartile formula for continuous series, remember to clearly identify the correct lower limit (L), previous cumulative frequency (pcf), frequency of the quartile class (f), and class width (C).

 

Question 13. Compute mean deviation about median from the following data:

Answer: To compute the mean deviation about the median, we first need to find the median. We will construct a cumulative frequency (cf) table.**1. Calculate Median (M):**
\(Median = \text{size of } \left( \frac{N+1}{2} \right)^\text{th} \text{ value}\)
\(Median = \text{size of } \left( \frac{197+1}{2} \right)^\text{th} \text{ value}\)
\(Median = \text{size of } \left( \frac{198}{2} \right)^\text{th} \text{ value}\)
\(Median = \text{size of } 99^\text{th} \text{ value}\)
From the cf column, the 99th value falls within the cumulative frequency of 102. The corresponding X value is 61.
So, Median (\(M\)) = 61.
**2. Compute Mean Deviation about Median:**
Now we construct a table to calculate \(|D| = |X - Median|\) and \(f|D|\).The formula for Mean Deviation about Median is:
\(MD_{Median} = \frac{\Sigma f|D|}{N}\)
\(MD_{Median} = \frac{336}{197}\)
\(MD_{Median} = 1.70558\)
Rounded to two decimal places, \(MD_{Median} = 1.71\). Mean deviation about the median minimizes the sum of absolute deviations, making it a robust measure of dispersion against outliers.
In simple words: First, we find the middle value (Median) of all the heights. Then, for each height, we calculate how far it is from the Median, ignoring whether it's higher or lower. We multiply this distance by how many students have that height, sum these up, and then divide by the total number of students to get the Mean Deviation.

🎯 Exam Tip: When calculating Mean Deviation, always use the *absolute difference* between each value and the central point (mean or median). Remember that \(\Sigma f|X-M|\) is divided by N, not N-1.

 

Question 14. Compute the mean deviation about mean from the following data:

Answer: To compute the mean deviation about the mean for this continuous frequency distribution, we first need to find the mean. We will determine the midpoint (X) for each class interval and then create an extended table.**1. Calculate Mean (\(\overline{X}\)):**
\( \overline{X} = \frac{\Sigma fX}{N} \)
\( \overline{X} = \frac{390}{30} \)
\( \overline{X} = 13 \)
**2. Compute Mean Deviation about Mean:**
The formula for Mean Deviation about Mean is:
\(MD_{Mean} = \frac{\Sigma f|D|}{N}\)
\(MD_{Mean} = \frac{130}{30}\)
\(MD_{Mean} = 4.3333\)
Rounded to two decimal places, \(MD_{Mean} = 4.33\). The mean deviation about the mean indicates the average absolute distance of each data point from the central tendency of the dataset.
In simple words: First, we find the middle points of each number range and then calculate the overall average (Mean) of the data. Next, for each range, we find how far its middle point is from this overall Mean, always taking the positive difference. We multiply this difference by the frequency of that range, sum these products, and finally divide by the total frequency to get the Mean Deviation.

🎯 Exam Tip: For grouped data, it is crucial to accurately calculate the midpoint (X) of each class interval. Any error here will affect all subsequent calculations for the mean and mean deviation.

 

Question 12. Calculate quartile deviation and its relative measure from the following data:

Answer: First, we arrange the data and find the cumulative frequency (cf) for each class interval. This helps us locate the quartiles easily within the grouped data.

Here, \( N = 68 \).

1. Calculate the Lower Quartile (Q₁):
The position of \( Q_1 \) is the size of the \( \left(\frac{N}{4}\right)^\text{th} \) value.
\( Q_1 = \text{size of } \left(\frac{68}{4}\right)^\text{th} \text{ value} \)
\( Q_1 = \text{size of } 17^\text{th} \text{ value} \)
Looking at the cumulative frequency table, the \( 17^\text{th} \) value falls in the class interval 20-30 (since its cf is 28, which is the first cf greater than 17).
For this class: Lower limit \( L = 20 \), Cumulative frequency of preceding class \( pcf = 15 \), Frequency of the class \( f = 13 \), Class interval width \( C = 10 \).
Using the formula for \( Q_1 \):
\( Q_1 = L + \left(\frac{\frac{N}{4} - pcf}{f}\right) \times C \)
\( Q_1 = 20 + \left(\frac{17 - 15}{13}\right) \times 10 \)
\( Q_1 = 20 + \left(\frac{2}{13}\right) \times 10 \)
\( Q_1 = 20 + \frac{20}{13} \)
\( Q_1 = 20 + 1.5385 \)
\( Q_1 = 21.5385 \)

2. Calculate the Upper Quartile (Q₃):
The position of \( Q_3 \) is the size of the \( \left(\frac{3N}{4}\right)^\text{th} \) value.
\( Q_3 = \text{size of } \left(\frac{3 \times 68}{4}\right)^\text{th} \text{ value} \)
\( Q_3 = \text{size of } \left(3 \times 17\right)^\text{th} \text{ value} \)
\( Q_3 = \text{size of } 51^\text{st} \text{ value} \)
Looking at the cumulative frequency table, the \( 51^\text{st} \) value falls in the class interval 40-50 (since its cf is 60, which is the first cf greater than 51).
For this class: Lower limit \( L = 40 \), Cumulative frequency of preceding class \( pcf = 46 \), Frequency of the class \( f = 14 \), Class interval width \( C = 10 \).
Using the formula for \( Q_3 \):
\( Q_3 = L + \left(\frac{\frac{3N}{4} - pcf}{f}\right) \times C \)
\( Q_3 = 40 + \left(\frac{51 - 46}{14}\right) \times 10 \)
\( Q_3 = 40 + \left(\frac{5}{14}\right) \times 10 \)
\( Q_3 = 40 + \frac{50}{14} \)
\( Q_3 = 40 + 3.5714 \)
\( Q_3 = 43.5714 \)

3. Calculate the Quartile Deviation (QD):
The Quartile Deviation measures the spread of the middle 50% of the data.
\( QD = \frac{Q_3 - Q_1}{2} \)
\( QD = \frac{43.5714 - 21.5385}{2} \)
\( QD = \frac{22.0329}{2} \)
\( QD = 11.01645 \approx 11.02 \)

4. Calculate the Coefficient of Quartile Deviation:
This is a relative measure that helps compare the spread of different datasets.
\( \text{Coefficient of QD} = \frac{Q_3 - Q_1}{Q_3 + Q_1} \)
\( \text{Coefficient of QD} = \frac{43.5714 - 21.5385}{43.5714 + 21.5385} \)
\( \text{Coefficient of QD} = \frac{22.0329}{65.1099} \)
\( \text{Coefficient of QD} = 0.33839 \approx 0.3384 \)
In simple words: We found the values that mark the first and third quarters of our data. The quartile deviation tells us how spread out the middle part of our data is. The coefficient then shows this spread as a proportion, which is useful for comparing how spread out different sets of numbers are.

🎯 Exam Tip: When calculating quartiles for grouped data, always make sure to correctly identify the quartile class and use the cumulative frequency of the *preceding* class (pcf) in the formula.

 

Question 13. Compute mean deviation about median from the following data:

Answer: To find the mean deviation about the median, we first need to calculate the median of the given data. For this, we arrange the data and find the cumulative frequencies.

Here, the total number of observations \( N = \Sigma f = 197 \).

1. Calculate the Median:
For discrete series, the median is the size of the \( \left(\frac{N+1}{2}\right)^\text{th} \) value.
\( \text{Median} = \text{size of } \left(\frac{197+1}{2}\right)^\text{th} \text{ value} \)
\( \text{Median} = \text{size of } \left(\frac{198}{2}\right)^\text{th} \text{ value} \)
\( \text{Median} = \text{size of } 99^\text{th} \text{ value} \)
From the cumulative frequency table, the \( 99^\text{th} \) value falls in the category where the cumulative frequency is 102. The corresponding X value is 61.
So, \( \text{Median} = 61 \).

2. Prepare for Mean Deviation calculation:
Next, we calculate the absolute deviation of each value from the median, \( |D| = |X - \text{Median}| = |X - 61| \), and then multiply it by its frequency \( f|D| \).

3. Calculate Mean Deviation about Median:
\( \text{MD about Median} = \frac{\Sigma f|D|}{N} \)
\( \text{MD about Median} = \frac{336}{197} \)
\( \text{MD about Median} = 1.70558 \approx 1.71 \)
In simple words: First, we found the middle value of all the students' heights, which is 61 inches. Then, we calculated how much each height differs from this middle value and averaged these differences. This average difference, called mean deviation about the median, tells us how spread out the heights are from the middle height.

🎯 Exam Tip: Remember to use the absolute values of deviations when calculating mean deviation, as negative and positive differences cancel each other out if signs are kept.

 

Question 14. Compute the mean deviation about mean from the following data:

Answer: To find the mean deviation about the mean, we first need to calculate the mean of the grouped data. This involves finding the mid-point (X) of each class, then calculating \( fX \), and summing these values.

1. Calculate the Mean (\( \overline{X} \)):
\( \overline{X} = \frac{\Sigma fX}{N} \)
\( \overline{X} = \frac{390}{30} \)
\( \overline{X} = 13 \)

2. Prepare for Mean Deviation calculation:
Now, we calculate the absolute deviation of each mid-point from the mean, \( |D| = |X - \overline{X}| = |X - 13| \), and then multiply it by its frequency \( f|D| \). These calculations are included in the table above.

3. Calculate Mean Deviation about Mean:
\( \text{MD about Mean} = \frac{\Sigma f|D|}{N} \)
\( \text{MD about Mean} = \frac{130}{30} \)
\( \text{MD about Mean} = 4.3333 \approx 4.33 \)
In simple words: We first found the average value, called the mean, which is 13. Then, we figured out how much each class interval's middle point was different from this average. By averaging these differences (ignoring if they were bigger or smaller), we got the mean deviation, which shows us how spread out all the values are from the average.

🎯 Exam Tip: Remember to use the mid-point of each class interval as 'X' for grouped data when calculating the mean and subsequently the deviations.

 

Question 15. Find out the coefficient of mean deviation about median in the following series:

Answer: To find the coefficient of mean deviation about the median, we first need to calculate the median. This involves creating a cumulative frequency (cf) table to identify the median class.

Here, the total number of observations \( N = \Sigma f = 150 \).

1. Calculate the Median:
For grouped data, the median position is \( \frac{N}{2} \).
\( \frac{N}{2} = \frac{150}{2} = 75 \)
The class interval corresponding to the cumulative frequency 75 is 40-50 (since its cf is 93, the first cf greater than 75).
For this class: Lower limit \( L = 40 \), Cumulative frequency of preceding class \( pcf = 56 \), Frequency of the class \( f = 37 \), Class interval width \( C = 10 \).
Using the formula for Median:
\( \text{Median} = L + \left(\frac{\frac{N}{2} - pcf}{f}\right) \times C \)
\( \text{Median} = 40 + \left(\frac{75 - 56}{37}\right) \times 10 \)
\( \text{Median} = 40 + \left(\frac{19}{37}\right) \times 10 \)
\( \text{Median} = 40 + \frac{190}{37} \)
\( \text{Median} = 40 + 5.1351 \)
\( \text{Median} = 45.1351 \approx 45.14 \)

2. Prepare for Mean Deviation calculation:
Next, we calculate the mid-point (M) for each class, the absolute deviation of each mid-point from the median, \( |D| = |M - \text{Median}| = |M - 45.14| \), and then multiply it by its frequency \( f|D| \).

3. Calculate Mean Deviation about Median:
\( \text{MD about Median} = \frac{\Sigma f|D|}{N} \)
\( \text{MD about Median} = \frac{2225.04}{150} \)
\( \text{MD about Median} = 14.8336 \)

4. Calculate the Coefficient of Mean Deviation about Median:
\( \text{Coefficient of MD about Median} = \frac{\text{MD about Median}}{\text{Median}} \)
\( \text{Coefficient of MD about Median} = \frac{14.8336}{45.1351} \)
\( \text{Coefficient of MD about Median} = 0.328659 \approx 0.3287 \)
In simple words: First, we found the median age, which is the middle age of all persons surveyed. Then, we calculated how much each age group differs from this median age, averaged these differences (mean deviation), and finally divided this average difference by the median itself. This gives us a proportional measure of spread around the median.

🎯 Exam Tip: When calculating the coefficient of mean deviation, ensure you divide the mean deviation by the *same* average (mean or median) that was used to compute the deviation.