GSEB Class 11 Statistics Solutions Chapter 3 Measures of Central Tendency Exercise 3.5

Exercise 3(E)

 

Question 1. The IQ levels of students in a class are given below. Find the modal value of IQ level of students. 146, 153, 138, 138, 146, 140, 135
Answer: Analyzing the provided dataset, the observation '138' appears three times, which is more frequent than any other value in the data. Therefore, the mode of the students' intelligence quotient (Mo) is 138.
In simple words: The mode is the number that appears most often in a dataset. Here, 138 shows up three times, making it the mode for the IQ levels.

🎯 Exam Tip: When finding the mode for ungrouped data, carefully count the frequency of each observation. The value with the highest frequency is the mode.

 

Question 2. The following table gives the number of cakes sold each day at a bakery. Find the mode for sale of cakes.

No. of cakes	10	12	13	16	17	18
No. of days	5	9	25	16	10	7

Answer: For the given frequency distribution of cake sales, the maximum frequency observed is 25, which corresponds to 13 cakes. Consequently, the mode for the sale of cakes (Mo) is 13 cakes.
In simple words: In this bakery data, 13 cakes were sold most frequently over a period of 25 days, making 13 the mode.

🎯 Exam Tip: For discrete frequency distributions, the mode is simply the observation that has the highest frequency. Identify the largest number in the frequency row and its corresponding data value.

 

Question 3. The distribution of ages of 48 persons in an old age home is given below. Which formula will be appropriate to find the mode? Why? Find the modal age of the persons in the old age home using the formula you have chosen.

Age (years)	50-60	60-65	65-70	70-85	85-100
No. of persons	6	10	19	9	4

Answer: Since the provided frequency distribution has class intervals of unequal lengths, the empirical formula \(M_0 = 3M - 2\bar{x}\) is the most suitable method for determining the mode. To apply this formula, we must first calculate the mean (\(\bar{x}\)) and the median (M) from the given data.

Age (year)	No. of persons \(f\)	Mid value \(x\)	\(f \cdot x\)	Cumulative frequency \(cf\)
50-60	6	55	330.0	6
60-65	10	62.5	625.0	16
65-70	19	67.5	1282.5	35
70-85	9	77.5	697.5	44
85-100	4	92.5	370.0	48

Total \(n = 48\), \( \Sigma fx = 3305\)
**Mean:** \( \bar{x} = \frac{\Sigma fx}{n} = \frac{3305}{48} = 68.85 \) years
**Median:** The median class is the one that contains the \( \left(\frac{n}{2}\right)^{th} \) observation. \( \left(\frac{48}{2}\right)^{th} = 24^{th} \) observation. Referring to the cumulative frequency column, the 24th observation falls within the class 65-70. So, the median class is 65-70.
Putting \(L = 65\), \( \left(\frac{n}{2}\right) = 24\), \(cf = 16\), \(f = 19\), and \(c = 5\) into the median formula: \( M = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
\( M = 65 + \frac{24-16}{19} \times 5 \)
\( M = 65 + \frac{8 \times 5}{19} \)
\( M = 65 + \frac{40}{19} \)
\( M = 65 + 2.11 \)
\( M = 67.11 \) years
**Mode:** Using the empirical formula for mode: \( M_0 = 3M - 2\bar{x} \) Substituting the calculated values of Median \( (M = 67.11) \) and Mean \( (\bar{x} = 68.85) \):
\( M_0 = 3(67.11) - 2(68.85) \)
\( M_0 = 201.33 - 137.7 \)
\( M_0 = 63.63 \) years Therefore, the modal age of individuals in the old age home is 63.63 years.
In simple words: Because the age groups are of different sizes, we first calculate the average (mean) and middle value (median) of the ages. Then, we use a special formula \( (3 \times \text{Median} - 2 \times \text{Mean}) \) to find the most common age, which turns out to be about 63.63 years.

🎯 Exam Tip: When dealing with grouped data with unequal class intervals, calculate the mean and median first. Remember the empirical formula \( M_0 = 3M - 2\bar{x} \) is essential for finding the mode in such scenarios and show all steps clearly.

 

Question 4. Comment on the mode for the following data showing the time taken (in seconds) by 8 competitors in a running race: 25.2, 26.5, 28.6, 32.1, 29.0, 29.3, 31.3, 27.8
Answer: The times taken by the 8 competitors in the running race are: 25.2, 26.5, 28.6, 32.1, 29.0, 29.3, 31.3, 27.8 seconds. Each of these observations appears only once in the dataset. By the standard definition of mode, which requires an observation to repeat, a mode cannot be directly determined for this data. However, if absolutely necessary, a mode could theoretically be estimated using an empirical formula if converted into a grouped frequency distribution, though it would not be a true mode of the raw data.
In simple words: Since every competitor has a unique race time and no time repeats, there isn't a "most common" time, so a mode cannot be found directly for this data.

🎯 Exam Tip: The mode exists only when at least one observation repeats in a dataset. If all observations are unique, the dataset is considered to have no mode. Clearly state this in your answer.

 

Question 5. The table below shows the data about weights of 86 apples from a garden. Find the mode for the weight of apples: Also find the mode of this distribution graphically.

Weight of apple (gram)	120-130	130-140	140-150	150-160	160-170	170-180	180-190
No. of apples	8	13	19	23	10	8	5

Answer: To find the mode by formula and graphically for the apple weight distribution:
**Mode by formula:** First, identify the maximum frequency in the distribution. Here, the maximum frequency is 23, which corresponds to the class interval 150-160. This is our modal class.
From the table, we have:
\( L = 150 \) (Lower limit of the modal class)
\( f_m = 23 \) (Frequency of the modal class)
\( f_1 = 19 \) (Frequency of the class preceding the modal class)
\( f_2 = 10 \) (Frequency of the class succeeding the modal class)
\( c = 10 \) (Class length of the modal class)
Now, applying the mode formula: \( M_0 = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times c \)
\( M_0 = 150 + \frac{23 - 19}{2(23) - 19 - 10} \times 10 \)
\( M_0 = 150 + \frac{4}{46 - 19 - 10} \times 10 \)
\( M_0 = 150 + \frac{4}{17} \times 10 \)
\( M_0 = 150 + \frac{40}{17} \)
\( M_0 = 150 + 2.35 \)
\( M_0 = 152.35 \) grams.
**Mode by graphical method:**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह हिस्टोग्राम सेबों के वजन की आवृत्ति वितरण को दर्शाता है। X-अक्ष पर 'Weight of apple (gram)' को 1 cm = 5 ग्राम के पैमाने पर दिखाया गया है, जबकि Y-अक्ष पर 'No. of apples (Frequency f)' को 2 cm = 5 सेब के पैमाने पर दर्शाया गया है। प्रत्येक वर्ग अंतराल के लिए आयत बनाए गए हैं, और उच्चतम आयत के शीर्ष को आसन्न आयतों के शीर्षों से तिरछी रेखाओं (A से B, C से D) से जोड़कर मोड बिंदु (P) को इंगित किया गया है। मोड उस बिंदु पर स्थित होता है जहाँ ये रेखाएँ X-अक्ष पर प्रतिच्छेद करती हैं, और इस आरेख से मोड का मान लगभग 152.5 ग्राम प्राप्त होता है।
From the histogram, by drawing lines from the top corners of the highest bar (modal class 150-160) to the adjacent bars' top corners, their intersection point (P) is found. Dropping a perpendicular from P to the X-axis (labeled Q in the original diagram) gives the mode. Graphically, the mode is approximately 152.5 grams.
In simple words: We find the mode using two methods. First, using a formula, we identify the most frequent weight range (modal class) and calculate the exact mode, which is about 152.35 grams. Second, we draw a bar chart (histogram) and find the tallest bar; by drawing specific lines from its corners, we pinpoint the mode graphically, which is also around 152.5 grams.

🎯 Exam Tip: For continuous data, be prepared to calculate the mode using both the formula (\( M_0 = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times c \)) and by constructing a histogram and identifying the peak. Ensure your graphical representation is accurate with labeled axes and scales.

 

Question 6. The data about the monthly house rent paid by 50 families is given in the following table: Find the mode for house rent using the graphical method.

House rent (thousand Rs.)	0-5	5-10	10-20	20-30	30-50
No. of families	1	7	14	16	12

Answer: The given frequency distribution for house rent has unequal class lengths. To accurately draw a histogram and find the mode graphically, we first need to calculate the equi-proportional frequency for each class relative to the least class length, which is 5.
**Calculation of Equi-Proportional Frequencies:**

House rent (thousand Rs.)	No. of families \(f\)	Class length \(c\)	Equi-proportional frequency \( = \frac{\text{frequency of class}}{\text{class length}} \times \text{least class length}\)
0-5	1	5	\( \frac{1}{5} \times 5 = 1 \)
5-10	7	5	\( \frac{7}{5} \times 5 = 7 \)
10-20	14	10	\( \frac{14}{10} \times 5 = 7 \)
20-30	16	10	\( \frac{16}{10} \times 5 = 8 \)
30-50	12	20	\( \frac{12}{20} \times 5 = 3 \)

**Mode by graphical method:**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह हिस्टोग्राम मासिक घर के किराए की आवृत्ति वितरण को दर्शाता है, जहाँ वर्ग अंतराल की लंबाई असमान है। X-अक्ष पर 'House rent (Rs.)' को 1 cm = 5 हजार के पैमाने पर दिखाया गया है, जबकि Y-अक्ष पर 'No. of families (Frequency f)' को 1 cm = 1 परिवार के पैमाने पर दर्शाया गया है। असमान वर्ग अंतरालों के कारण, आयत की ऊंचाई समायोजित (equi-proportional frequency) की गई है। मोड ज्ञात करने के लिए, सबसे ऊँचे आयत (20-30 हजार Rs. वर्ग) के शीर्ष को आसन्न आयतों (10-20 हजार Rs. और 30-50 हजार Rs. वर्ग) के शीर्षों से तिरछी रेखाओं (C से D और E से F) से जोड़ा गया है। इन रेखाओं का प्रतिच्छेदन बिंदु (P) X-अक्ष पर एक लंबवत रेखा (OQ) गिराने पर मोड का मान देता है।
From the histogram, by drawing lines from the upper corners of the highest bar (modal class 20-30 thousand Rs.) to the adjacent bars' upper corners, their intersection point (P) is found. Dropping a perpendicular from P to the X-axis (labeled Q in the original diagram) gives the graphical mode. The graph indicates that mode \( = OQ = 22 \). Hence, the mode of house rent is Rs. 22 thousand.
In simple words: Since the rent categories have different ranges, we adjust the frequencies to make them comparable. Then, we draw a special bar chart (histogram). By drawing lines from the tallest bar to its neighbors, we find where they cross, and that point on the rent axis gives us the most common house rent, which is Rs. 22 thousand.

🎯 Exam Tip: For grouped data with unequal class intervals, always calculate equi-proportional frequencies before constructing a histogram to find the mode graphically. Clearly label your axes and mark the steps to find the mode on the graph.