OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Exercise 28 (D)

Question 1. Find the mode of the given distribution by drawing a histogram.

Answer: First, we draw the rectangles for each class interval, making their heights match the given frequencies. The class with the highest frequency is 10-20, so this is our modal class. To find the mode graphically, we join the top-right corner of the rectangle just before the modal class (0-10) to the top-right corner of the modal class (10-20). Then, we join the top-left corner of the rectangle just after the modal class (20-30) to the top-left corner of the modal class. The point where these two lines cross is P. We then draw a straight line from P down to the x-axis. The value on the x-axis where this line touches (point A) is the mode. This method helps us visually locate the peak concentration of data. From the graph, the mode is clearly 15.15.

In simple words: First, draw a graph with bars representing how many items fall into each number range. The tallest bar shows the "modal class." To find the exact mode, draw two diagonal lines from the top corners of the tallest bar to the adjacent bars. Where these lines cross, draw a vertical line down to the bottom axis. This point on the bottom axis is the mode, which here is 15.15.

🎯 Exam Tip: When drawing a histogram for the mode, make sure your bars are correctly proportioned to their frequencies. The diagonal lines must connect the correct opposite corners of the modal rectangle with its neighboring rectangles to accurately find the mode.

Question 2.

Answer: First, we need to change this discontinuous distribution into a continuous one. We calculate an adjustment factor: \( \frac { 6-5 }{ 2 } = 0.5 \). We subtract 0.5 from each lower class limit and add 0.5 to each upper class limit to create continuous class intervals. This ensures there are no gaps between the classes, which is important for drawing a proper histogram. The new table is shown below:

Next, we draw rectangles for these continuous class intervals, with heights that match their frequencies. The modal class here is 15.5-20.5, because it has the highest frequency (32). To find the mode graphically, we connect the top-right corner of the bar before the modal class (10.5-15.5) to the top-right corner of the modal class (15.5-20.5). We also connect the top-left corner of the bar after the modal class (20.5-25.5) to the top-left corner of the modal class. These two lines will intersect at point P. From P, we draw a perpendicular line down to the x-axis, meeting it at point A. The x-coordinate of point A gives us the required mode. The required mode is 18.8.

In simple words: First, change the given data ranges so there are no gaps between them by adding and subtracting 0.5. Then, draw a bar graph. The tallest bar shows the group with the most items. To find the exact mode, draw lines from the top corners of the tallest bar to the corners of the bars next to it. The point where these lines cross, when dropped down to the bottom axis, gives the mode, which is 18.8.

🎯 Exam Tip: Always convert discontinuous class intervals to continuous ones before drawing a histogram. Missing this step will result in gaps between bars and an incorrect modal class identification.

Question 3.

Answer: We must first change the discontinuous distribution into a continuous one using an adjustment factor. The adjustment factor is calculated as \( \frac { 31-30 }{ 2 } = 0.5 \). We then subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class interval. This process ensures that the histogram bars touch each other, making the graphical mode calculation accurate.

Next, we draw rectangles for these continuous class intervals, with heights that match their frequencies. The modal class here is 25.5-30.5, as it has the highest frequency (18). To find the mode graphically, we join the top-right corner of the bar just before the modal class (assuming 20.5-25.5 with frequency 0) to the top-right corner of the modal class (25.5-30.5). Then, we join the top-left corner of the bar just after the modal class (30.5-35.5) to the top-left corner of the modal class. These lines intersect at point P. We draw a perpendicular line from P down to the x-axis, meeting it at point A. The x-coordinate of point A gives us the required mode. The mode is 28.96.

In simple words: First, adjust the number ranges to remove any gaps. Then, draw a bar graph. The tallest bar shows the mode group. To find the exact mode, draw diagonal lines connecting the top corners of the tallest bar to its neighbors. The point where these lines cross, when brought down to the bottom axis, gives the mode, which is 28.96.

🎯 Exam Tip: When the first or last class is the modal class, one of the diagonal lines for finding the mode will originate from the x-axis. Ensure you connect from the correct corner to get an accurate intersection point.

Question 4.

Answer: When given mid-values, we first need to find the class intervals. The common difference between mid-values is 4 (e.g., 64-60 = 4). So, the class width is 4. For each mid-value, the class interval will be \( \text{mid-value} - \frac{\text{class width}}{2} \) to \( \text{mid-value} + \frac{\text{class width}}{2} \). Thus, \( \frac{4}{2} = 2 \). We subtract 2 from the mid-value for the lower limit and add 2 for the upper limit. This forms the continuous frequency distribution table as shown:

Next, we draw rectangles for these class intervals, with heights corresponding to their frequencies. The modal class is 70-74, as it has the highest frequency (66). To find the mode graphically, we join the top-right corner of the bar before the modal class (66-70) to the top-right corner of the modal class (70-74). Then, we join the top-left corner of the bar after the modal class (74-78) to the top-left corner of the modal class. These lines intersect at point P. We draw a perpendicular line from P down to the x-axis, meeting it at point A. The x-coordinate of point A gives us the required mode. The mode is 71.90.

In simple words: When you are given mid-points, first figure out the full class ranges. Then, draw a bar graph. The tallest bar shows the group with the most data, which is the modal class. To find the exact mode, draw diagonal lines from the top corners of the tallest bar to the adjacent bars. The point where these lines cross, when dropped straight down to the bottom axis, gives the mode, which is 71.90.

🎯 Exam Tip: Remember to calculate the class intervals correctly from mid-values before proceeding. The class width is usually the difference between consecutive mid-values.

Question 5.

Answer: We begin by drawing rectangles that represent each class interval, with their heights matching the given frequencies. From the histogram, we can see that the modal class, which is the class with the highest frequency, is 30-40. To find the mode graphically, we connect the top-right corner of the rectangle before the modal class (20-30) to the top-right corner of the modal class (30-40). Then, we connect the top-left corner of the rectangle after the modal class (40-50) to the top-left corner of the modal class. These two lines intersect at a point, let's call it P. From point P, we draw a perpendicular line down to the x-axis, meeting it at point A. The x-coordinate of point A will be our required mode. This process visually pinpoints the most frequent value. The required mode is 32.8.

In simple words: Draw a bar graph where the height of each bar shows how often something occurs in that range. Find the tallest bar, which is the "modal class." Then, draw two diagonal lines: one from the top-right of the bar before the modal class to the top-right of the modal class, and another from the top-left of the modal class to the top-left of the bar after it. Where these two lines cross, drop a straight line down to the bottom axis. This point on the bottom axis is the mode, which is 32.8.

🎯 Exam Tip: Always clearly label your axes and ensure the bars in the histogram touch each other if the data is continuous. This provides a clear visual representation for identifying the modal class and intersecting lines.

Question 6. The daily profits in rupees of 100 shops are distributed as under :

Answer: First, we draw a histogram where the height of each bar shows the number of shops (frequency) for each profit range. By simply looking at the bars, we can see that the modal class, which is the class with the highest frequency, is 200-300. This is because it has 27 shops, more than any other range. To find the mode graphically, we connect the top-right corner of the bar for 100-200 to the top-right corner of the modal class bar (200-300). Next, we connect the top-left corner of the bar for 300-400 to the top-left corner of the modal class bar (200-300). The point where these two lines cross is P. We then draw a line straight down from P to the x-axis, meeting it at point A. The x-coordinate of point A gives us the required mode. This graphic method helps visualize the peak of the distribution.

For verification by direct calculation using the formula:
\( \text{Mode} = l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i \)
Here, from inspection, the modal class is 200-300.
So, \( l = 200 \) (lower limit of modal class)
\( f_m = 27 \) (frequency of modal class)
\( f_{m-1} = 18 \) (frequency of preceding class)
\( f_{m+1} = 20 \) (frequency of succeeding class)
\( i = 100 \) (class width)

\( \implies \) \( \text{Mode} = 200+\frac{27-18}{2(27)-18-20} \times 100 \)
\( \implies \) \( \text{Mode} = 200+\frac{9}{54-18-20} \times 100 \)
\( \implies \) \( \text{Mode} = 200+\frac{9}{16} \times 100 \)
\( \implies \) \( \text{Mode} = 200+0.5625 \times 100 \)
\( \implies \) \( \text{Mode} = 200+56.25 \)
\( \implies \) \( \text{Mode} = 256.25 \)
Thus, the mode found graphically and by calculation is 256.25.

In simple words: First, draw a bar graph with profit ranges on the bottom and number of shops on the side. The tallest bar tells you the most common profit range. To find the exact mode, draw two diagonal lines at the top of this tallest bar, connecting its corners to the corners of the bars next to it. Where these lines cross, drop a line down to the profit axis. This point, which is 256.25, is the most frequent profit amount. The formula method gives the same answer, confirming the graph.

🎯 Exam Tip: Always verify your graphical mode with the formula \( \text{Mode} = l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i \). This ensures accuracy and allows for cross-checking your answer.

Question 7. Determine the value of mode of the following distribution graphically and verify the result.

Answer: We will draw rectangles for each class interval with heights proportional to their frequencies. By observing the histogram, the modal class (the class with the highest frequency) is 20-30, having 14 students. To find the mode graphically, we join the ends of the opposite rectangles that surround the modal class. Specifically, we draw a line from the top-right corner of the 10-20 bar to the top-right corner of the 20-30 bar. We also draw a line from the top-left corner of the 30-40 bar to the top-left corner of the 20-30 bar. These two lines will intersect at a point, let's call it P. From point P, we draw a perpendicular line down to the x-axis, meeting it at point A. The x-coordinate of point A gives us the required mode. This graphical method provides a visual estimate of the mode.

For verification by direct calculation using the formula:
\( \text{Mode} = l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i \)
Here, from inspection, the modal class is 20-30.
So, \( l = 20 \) (lower limit of modal class)
\( f_m = 14 \) (frequency of modal class)
\( f_{m-1} = 12 \) (frequency of preceding class)
\( f_{m+1} = 10 \) (frequency of succeeding class)
\( i = 10 \) (class width)

\( \implies \) \( \text{Mode} = 20+\frac{14-12}{2(14)-12-10} \times 10 \)
\( \implies \) \( \text{Mode} = 20+\frac{2}{28-12-10} \times 10 \)
\( \implies \) \( \text{Mode} = 20+\frac{2}{6} \times 10 \)
\( \implies \) \( \text{Mode} = 20+\frac{1}{3} \times 10 \)
\( \implies \) \( \text{Mode} = 20+3.33 \)
\( \implies \) \( \text{Mode} = 23.33 \)
Thus, the mode found graphically and by calculation is 23.3 (or 23.33).

In simple words: Draw a bar graph where the heights of the bars show how many students got marks in each range. Find the tallest bar; that's the mode group. To get the exact mode, draw diagonal lines from the top corners of the tallest bar to the corners of the bars next to it. Where these lines cross, drop a line down to the bottom axis. This point is the mode, which is 23.3. When you calculate it with the formula, you get the same answer.

🎯 Exam Tip: Always state the formula for calculating mode in grouped data and clearly show each step of the calculation to earn full marks, in addition to the graphical representation.

Question 8.
(i) In an asymmetrical distribution mean is 58 and median is 61 . Calculate mode.
(ii) If mode in a tolerably asymmetrical distribution is 12 and median is 16, what would be the most probable mean?
(iii) Find the median if mean is 40 and mode is 36.
Answer:
(i) We know the empirical relationship between mean, median, and mode for a moderately asymmetrical distribution:
\( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \)
Given: Mean \( = 58 \), Median \( = 61 \)
\( \implies \) \( \text{Mode} = 3 \times 61 - 2 \times 58 \)
\( \implies \) \( \text{Mode} = 183 - 116 \)
\( \implies \) \( \text{Mode} = 67 \)
When data is not perfectly symmetrical, this formula provides a good estimate for the mode.
(ii) We use the same empirical formula:
\( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \)
Given: Mode \( = 12 \), Median \( = 16 \)
\( \implies \) \( 12 = 3 \times 16 - 2 \times \text{Mean} \)
\( \implies \) \( 12 = 48 - 2 \times \text{Mean} \)
\( \implies \) \( 2 \times \text{Mean} = 48 - 12 \)
\( \implies \) \( 2 \times \text{Mean} = 36 \)
\( \implies \) \( \text{Mean} = 18 \)
This calculation shows how to find the mean when the mode and median are known.
(iii) Again, we use the empirical formula:
\( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \)
Given: Mean \( = 40 \), Mode \( = 36 \)
\( \implies \) \( 36 = 3 \times \text{Median} - 2 \times 40 \)
\( \implies \) \( 36 = 3 \times \text{Median} - 80 \)
\( \implies \) \( 3 \times \text{Median} = 36 + 80 \)
\( \implies \) \( 3 \times \text{Median} = 116 \)
\( \implies \) \( \text{Median} = \frac { 116 }{ 3 } \)
\( \implies \) \( \text{Median} = 38.67 \)
The formula is very useful for finding any one of these central tendencies if the other two are known.
In simple words: For data that is not perfectly balanced (asymmetrical), there's a simple rule: the mode is roughly three times the median minus two times the mean. You can use this rule to find any one of these three values if you know the other two. For example, if the average (mean) is 58 and the middle value (median) is 61, then the most common value (mode) is 67. If mode is 12 and median is 16, mean is 18. If mean is 40 and mode is 36, median is 38.67.

🎯 Exam Tip: Memorize the empirical formula for asymmetrical distribution (Mode = 3 Median - 2 Mean). This formula is frequently tested and can be directly applied to solve various problems quickly.