OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Chapter Test

 

Question 1. The means of two sets of sizes 40 and 60 respectively are 15 and 16 and the standard deviations are 3 and 4. Obtain the mean and standard deviation of the composite set of 100 items when the two sets are pooled together.
Answer:
Given: \( n_1 = 40; n_2 = 60 \)
\( \overline{x}_1 = 15; \overline{x}_2 = 16 \)
\( \sigma_1 = 3 \) and \( \sigma_2 = 4 \)
First, we calculate the combined mean. This is found by summing the products of each set's size and mean, then dividing by the total number of items.
Combined Mean \( \overline{X}_{12} = \frac { n_1 \overline{x}_1 + n_2 \overline{x}_2 }{ n_1 + n_2 } \)
\( = \frac { 40 \times 15 + 60 \times 16 }{ 40 + 60 } \)
\( = \frac { 600 + 960 }{ 100 } \)
\( = \frac { 1560 }{ 100 } \)
\( = 15.6 \)
Next, we find the deviations for each mean from the combined mean:
\( d_1 = \overline{x}_1 - \overline{x}_{12} = 15 - 15.6 = -0.6 \)
\( d_2 = \overline{x}_2 - \overline{x}_{12} = 16 - 15.6 = 0.4 \)
Finally, we calculate the combined standard deviation. This formula involves the individual standard deviations and the deviations of their means from the combined mean.
Thus combined S.D \( = \sqrt { \frac { n_1 ( \sigma_1^2 + d_1^2 ) + n_2 ( \sigma_2^2 + d_2^2 ) }{ n_1 + n_2 } } \)
\( = \sqrt { \frac { 40[9+(0.6)^2] + 60[16 + (0.4)^2] }{ 40 + 60 } } \)
\( = \sqrt { \frac { 40[9+0.36] + 60[16+0.16] }{ 100 } } \)
\( = \sqrt { \frac { 40 \times 9.36 + 60 \times 16.16 }{ 100 } } \)
\( = \sqrt { \frac { 374.4 + 969.6 }{ 100 } } \)
\( = \sqrt { \frac { 1344 }{ 100 } } \)
\( = \sqrt { 13.44 } \)
\( \approx 3.67 \)
So, the mean of the composite set is 15.6 and the standard deviation is approximately 3.67.
In simple words: We first find the average of all 100 items together. Then we calculate how much each original group's average is different from this new overall average. Using these differences, we find the spread (standard deviation) of all 100 items combined.

🎯 Exam Tip: Remember to calculate both the combined mean and the combined standard deviation using the correct formulas. Pay close attention to squaring the deviations \( d_1 \) and \( d_2 \) correctly.

 

Question 2. Find the median of the following values : 7 cm, 9 cm, 10 cm, 12 cm, 15 cm, 18 cm, 20 cm.
Answer:
Given data is already in ascending order: 7 cm, 9 cm, 10 cm, 12 cm, 15 cm, 18 cm, 20 cm.
Here, the number of observations \( n = 7 \) (which is an odd number).
For an odd number of observations, the median is the middle value. We can find its position using the formula \( \frac { n+1 }{ 2 } \)th item.
\( \implies \) Median \( = \text{size of } \left( \frac { n+1 }{ 2 } \right)\text{th item} \)
\( = \text{size of } \left( \frac { 7+1 }{ 2 } \right)\text{th item} \)
\( = \text{size of } \left( \frac { 8 }{ 2 } \right)\text{th item} \)
\( = \text{size of } 4\text{th item} \)
Looking at the ordered list, the 4th item is 12.
Thus, the median value is 12 cm.
In simple words: First, put all the numbers in order from smallest to largest. Since there is an odd number of values, the median is simply the number that sits exactly in the middle of the list.

🎯 Exam Tip: Always arrange the data in ascending or descending order before finding the median. If the number of observations (n) is odd, the median is the \( \left( \frac { n+1 }{ 2 } \right) \)th item.

 

Question 3. The marks obtained by 12 students out of 50 are as under : 25, 24, 23, 32, 40, 27, 30, 25, 20, 15, 16, 45 Find the median marks.
Answer:
First, we need to arrange the given data in ascending order:
15, 16, 20, 23, 24, 25, 25, 27, 30, 32, 40, 45.
Here, the number of observations \( n = 12 \) (which is an even number).
For an even number of observations, the median is the average of the two middle values. These values are the \( \left( \frac { n }{ 2 } \right) \)th item and the \( \left( \frac { n }{ 2 } + 1 \right) \)th item.
\( \implies \) Median \( = \text{size of } \left( \frac { \left( \frac { n }{ 2 } \right)\text{th item} + \left( \frac { n }{ 2 } + 1 \right)\text{th item} }{ 2 } \right) \)
\( = \text{size of } \left( \frac { \left( \frac { 12 }{ 2 } \right)\text{th item} + \left( \frac { 12 }{ 2 } + 1 \right)\text{th item} }{ 2 } \right) \)
\( = \text{size of } \left( \frac { \text{6th item} + \text{7th item} }{ 2 } \right) \)
From the ordered list: The 6th item is 25, and the 7th item is 25.
\( = \frac { 25 + 25 }{ 2 } \)
\( = \frac { 50 }{ 2 } \)
\( = 25 \)
Thus, the median marks are 25.
In simple words: Arrange the marks from lowest to highest. Since there's an even count of students, find the two marks that are in the exact middle. The median is the average of these two middle marks.

🎯 Exam Tip: When n is even, it's crucial to identify the two middle values correctly and then take their average. Don't just pick one of them.

 

Question 4. Compute Q3, D6 and P70 for the following data : 28, 17, 12, 25, 26, 19, 13, 27, 21, 16
Answer:
First, arrange the given data in ascending order:
12, 13, 16, 17, 19, 21, 25, 26, 27, 28
Here, the number of observations \( n = 10 \) (even).

**To compute Q3 (Third Quartile):**
Q3 is the \( 3 \left( \frac { n+1 }{ 4 } \right) \)th value.
\( \implies \) Q3 \( = 3 \left( \frac { 10+1 }{ 4 } \right)\text{th value} \)
\( = 3 \left( \frac { 11 }{ 4 } \right)\text{th value} \)
\( = 3 \times 2.75\text{th value} \)
\( = 8.25\text{th value} \)
This means Q3 is between the 8th and 9th values. We interpolate using the formula:
Q3 \( = \text{8th value} + 0.25 (\text{9th value} - \text{8th value}) \)
From the ordered data, 8th value = 26, 9th value = 27.
Q3 \( = 26 + 0.25 (27 - 26) \)
\( = 26 + 0.25 (1) \)
\( = 26 + 0.25 \)
\( = 26.25 \)

**To compute D6 (Sixth Decile):**
D6 is the \( 6 \left( \frac { n+1 }{ 10 } \right) \)th value.
\( \implies \) D6 \( = 6 \left( \frac { 10+1 }{ 10 } \right)\text{th value} \)
\( = 6 \left( \frac { 11 }{ 10 } \right)\text{th value} \)
\( = 6 \times 1.1\text{th value} \)
\( = 6.6\text{th value} \)
This means D6 is between the 6th and 7th values.
D6 \( = \text{6th value} + 0.6 (\text{7th value} - \text{6th value}) \)
From the ordered data, 6th value = 21, 7th value = 25.
D6 \( = 21 + 0.6 (25 - 21) \)
\( = 21 + 0.6 \times 4 \)
\( = 21 + 2.4 \)
\( = 23.4 \)

**To compute P70 (70th Percentile):**
P70 is the \( 70 \left( \frac { n+1 }{ 100 } \right) \)th value.
\( \implies \) P70 \( = 70 \left( \frac { 10+1 }{ 100 } \right)\text{th value} \)
\( = 70 \left( \frac { 11 }{ 100 } \right)\text{th value} \)
\( = 70 \times 0.11\text{th value} \)
\( = 7.7\text{th value} \)
This means P70 is between the 7th and 8th values.
P70 \( = \text{7th value} + 0.7 (\text{8th value} - \text{7th value}) \)
From the ordered data, 7th value = 25, 8th value = 26.
P70 \( = 25 + 0.7 (26 - 25) \)
\( = 25 + 0.7 \times 1 \)
\( = 25 + 0.7 \)
\( = 25.7 \)
In simple words: We first put the numbers in order. Then, to find Q3, D6, and P70, we use specific formulas that tell us where these values are located in the sorted list. If the position is not a whole number, we find the value by looking at the numbers on either side and calculating a weighted average.

🎯 Exam Tip: Always sort the data first. Remember the different formulas for quartiles (divided by 4), deciles (divided by 10), and percentiles (divided by 100). Interpolate correctly when the position is not an integer.

 

Question 5. Obtain the median for the following data and describe the information conveyed by it.

Answer:
To find the median for this grouped data, we first need to construct a cumulative frequency (c.f) table. This helps us locate the middle value.

Here, the total number of observations \( N = 147 \).
The median position is \( \left( \frac { N+1 }{ 2 } \right) \)th item.
\( \implies \) Median \( = \left( \frac { 147+1 }{ 2 } \right)\text{th item} \)
\( = \left( \frac { 148 }{ 2 } \right)\text{th item} \)
\( = 74\text{th item} \)
Now, we look at the cumulative frequency column. The 74th item falls in the row where c.f. is 140 (which corresponds to x = 12), because 74 is greater than 70 and less than or equal to 140.
Therefore, the median is 12.

**Information conveyed:**
The median value of 12 students absent tells us that on about half of the observed days, 12 or fewer students were absent. Conversely, on the remaining half of the days, 12 or more students were absent. This gives us a central point of student absenteeism.
In simple words: We add up the "number of days" to get a running total. Then we find the middle position in this total. The "number of students absent" that matches this middle position is our median. This means half the time, fewer than 12 students were absent, and half the time, more than 12 students were absent.

🎯 Exam Tip: For discrete frequency distributions, calculate the cumulative frequency first. The median is the value of 'x' corresponding to the cumulative frequency just greater than or equal to \( \frac { N+1 }{ 2 } \).

 

Question 6. Calculate the median for the following distribution:

Answer:
To calculate the median, we first need to arrange the 'x' values in ascending order and then create a cumulative frequency table.

Here, the total number of observations \( N = 96 \) (which is an even number).
The median position is the average of the \( \left( \frac { N }{ 2 } \right) \)th value and the \( \left( \frac { N }{ 2 } + 1 \right) \)th value.
\( \frac { N }{ 2 } = \frac { 96 }{ 2 } = 48 \)
So, we need the 48th value and the 49th value.
\( \implies \) Median \( = \left( \frac { \left( \frac { N }{ 2 } \right)\text{th value} + \left( \frac { N }{ 2 } + 1 \right)\text{th value} }{ 2 } \right) \)
\( = \left( \frac { 48\text{th value} + 49\text{th value} }{ 2 } \right) \)
From the cumulative frequency table: The 48th item corresponds to \( x=5 \) (since c.f. 48 includes values up to 5). The 49th item corresponds to \( x=6 \) (since c.f. 57 includes values up to 6).
\( = \frac { 5 + 6 }{ 2 } \)
\( = \frac { 11 }{ 2 } \)
\( = 5.5 \)
Thus, the median for this distribution is 5.5.
In simple words: We arrange the 'x' values in order and create a running total of frequencies. Since the total count is even, we find the two middle positions. The median is the average of the 'x' values that correspond to these two middle positions in the running total.

🎯 Exam Tip: For discrete frequency data with an even total frequency (N), remember to average the \( (\frac{N}{2}) \)th and \( (\frac{N}{2}+1) \)th items. Ensure your 'x' values are sorted before calculating cumulative frequencies.

 

Question 7. Calculate the median, first quartile and third decile for the following data:

Answer:
To find the median, first quartile (Q1), and third decile (D3), we first create a cumulative frequency (c.f) table.

Here, the total number of workers \( N = 50 \).

**1. Calculate the Median (Md):**
Median position \( = \left( \frac { N+1 }{ 2 } \right)\text{th value} \)
\( = \left( \frac { 50+1 }{ 2 } \right)\text{th value} \)
\( = \frac { 51 }{ 2 }\text{th value} \)
\( = 25.5\text{th value} \)
This means the median is between the 25th and 26th values. We interpolate.
Md \( = 25\text{th value} + 0.5 (26\text{th value} - 25\text{th value}) \)
From the c.f. table: The 25th value is 61 (c.f. 26), and the 26th value is also 61 (it's the last value within the c.f. group of 26 before it jumps to 36). Actually, looking at the cumulative frequency, 25th value is within the c.f. of 26, meaning it is 61. The 26th value is also within the c.f. of 26, meaning it is also 61. This indicates that for all values from 12th to 26th (inclusive), the income is 61. So, both the 25th and 26th values are 61.
Md \( = 61 + 0.5 (61 - 61) \)
\( = 61 + 0.5 (0) \)
\( = 61 \)

**2. Calculate the First Quartile (Q1):**
Q1 position \( = \left( \frac { N+1 }{ 4 } \right)\text{th value} \)
\( = \left( \frac { 50+1 }{ 4 } \right)\text{th value} \)
\( = \frac { 51 }{ 4 }\text{th value} \)
\( = 12.75\text{th value} \)
This means Q1 is between the 12th and 13th values. We interpolate.
Q1 \( = 12\text{th value} + 0.75 (13\text{th value} - 12\text{th value}) \)
From the c.f. table: The 12th value is 61 (c.f. 26). The 13th value is also 61.
Q1 \( = 61 + 0.75 (61 - 61) \)
\( = 61 + 0.75 (0) \)
\( = 61 \)
(Since 12th to 26th values are the same, Q1 is simply 61).

**3. Calculate the Third Decile (D3):**
D3 position \( = 3 \left( \frac { N+1 }{ 10 } \right)\text{th value} \)
\( = 3 \left( \frac { 50+1 }{ 10 } \right)\text{th value} \)
\( = 3 \left( \frac { 51 }{ 10 } \right)\text{th value} \)
\( = 3 \times 5.1\text{th value} \)
\( = 15.3\text{th value} \)
This means D3 is between the 15th and 16th values. We interpolate.
D3 \( = 15\text{th value} + 0.3 (16\text{th value} - 15\text{th value}) \)
From the c.f. table: The 15th value is 61 (c.f. 26). The 16th value is also 61.
D3 \( = 61 + 0.3 (61 - 61) \)
\( = 61 + 0.3 (0) \)
\( = 61 \)
In simple words: First, we make a running total of the workers for each income. Then, for the median, Q1, and D3, we find their positions using special formulas. We look at the running total to find which income value matches each position. If a position falls between two whole numbers, we check the income values at those positions.

🎯 Exam Tip: When using cumulative frequency for percentiles, deciles, and quartiles, if the calculated position (e.g., 25.5th) falls within a range where 'x' values are constant (e.g., all 61), then the quartile/decile is that constant 'x' value. No interpolation is needed if the values are identical.

 

Question 8. Draw an ogive for the following distribution :

Hence, find Q1, Q3, D9, P4 and P40 from it.
Answer:
To draw an ogive (cumulative frequency curve), we first need to prepare cumulative frequency distributions of both 'less than' and 'more than' types.

**Less than type cumulative frequency distribution:**

The points to plot for the 'less than' ogive are (30,25), (60,100), (90,150), (120,175), (150,325), (180,375), (210,400) and (0,0). Joining these points by a free hand curve gives the required less than type ogive.

**More than type cumulative frequency distribution:**

The points to plot for the 'more than' ogive are (0,400), (30,375), (60,300), (90,250), (120,225), (150,75), (180,25), (210,0). Joining these points by a free hand curve gives the required more than type ogive.

The median is found where both ogives intersect. Drawing a perpendicular from this intersection point (P) to the x-axis (at M) gives the required median. The graph shows the median is 125.
Thus, the required median = 125.

**Finding Q1 (First Quartile) from the ogive:**
Q1 position \( = \left( \frac { N }{ 4 } \right)\text{th value} \)
Here, \( N = 400 \). So, Q1 position \( = \left( \frac { 400 }{ 4 } \right)\text{th value} = 100\text{th item} \).
From the ogive, the x-value corresponding to the 100th observation (on the y-axis of the 'less than' ogive) is 60.
So, Q1 = 60.

**Finding Q3 (Third Quartile) from the ogive:**
Q3 position \( = \left( \frac { 3N }{ 4 } \right)\text{th item} \)
\( = \left( \frac { 3 \times 400 }{ 4 } \right)\text{th item} = 300\text{th item} \).
From the ogive, the x-value corresponding to the 300th observation is 150.
So, Q3 = 150.

**Finding D9 (Ninth Decile) from the ogive:**
D9 position \( = \left( \frac { 9 \times N }{ 10 } \right)\text{th item} \)
\( = \left( \frac { 9 \times 400 }{ 10 } \right)\text{th item} = 360\text{th item} \).
From the ogive, the x-value corresponding to the 360th item is 171 sec.
So, D9 = 171 sec.

**Finding P4 (Fourth Percentile) from the ogive:**
P4 position \( = \left( \frac { 4 \times N }{ 100 } \right)\text{th item} \)
\( = \left( \frac { 4 \times 400 }{ 100 } \right)\text{th item} = 16\text{th item} \).
From the ogive, the x-value corresponding to the 16th item is 19 sec.
So, P4 = 19 sec.

**Finding P40 (40th Percentile) from the ogive:**
P40 position \( = \left( \frac { 40 \times N }{ 100 } \right)\text{th item} \)
\( = \left( \frac { 40 \times 400 }{ 100 } \right)\text{th item} = 160\text{th item} \).
From the ogive, the x-value corresponding to the 160th item is 102 sec.
So, P40 = 102 sec.
In simple words: First, we make two tables: one for "less than" cumulative frequencies and one for "more than" cumulative frequencies. We plot these points on a graph to draw two special curves called ogives. The point where these curves cross gives us the median. We then use these curves to find the values for Q1, Q3, D9, P4, and P40 by locating their respective positions on the vertical axis and reading the corresponding value on the horizontal axis.

🎯 Exam Tip: When drawing ogives, plot 'less than' cumulative frequencies against the upper class limits and 'more than' cumulative frequencies against the lower class limits. Ensure clear labelling of axes and points for accurate reading of quartiles, deciles, and percentiles.

 

Question 9. Calculate the median, Q3, D7 and P70 for the following distribution :

Answer:
To calculate the median, Q3, D7, and P70 for this grouped frequency distribution, we first need to construct a cumulative frequency table.

Here, the total frequency \( N = 50 \).

**1. Calculate Q3 (Third Quartile):**
Position of Q3 \( = \frac { 3N }{ 4 } = \frac { 3 \times 50 }{ 4 } = \frac { 150 }{ 4 } = 37.5 \)
This value (37.5) lies in the class 40-50 (since c.f. 37 is for 30-40, and c.f. 43 is for 40-50).
So, the Q3 class is 40-50.
Here, \( l = 40 \) (lower limit of Q3 class)
\( f = 6 \) (frequency of Q3 class)
\( c = 37 \) (cumulative frequency of class preceding Q3 class)
\( i = 10 \) (class interval size)
Formula for Q3 \( = l + \left( \frac { \frac { 3N }{ 4 } - c }{ f } \right) \times i \)
\( \implies \) Q3 \( = 40 + \left( \frac { 37.5 - 37 }{ 6 } \right) \times 10 \)
\( = 40 + \left( \frac { 0.5 }{ 6 } \right) \times 10 \)
\( = 40 + \frac { 5 }{ 6 } \)
\( = 40 + 0.833... \)
\( = 40.83 \)

**2. Calculate Median (Md):**
Position of Median \( = \frac { N }{ 2 } = \frac { 50 }{ 2 } = 25 \)
This value (25) lies in the class 20-30 (since c.f. 13 is for 10-20, and c.f. 30 is for 20-30).
So, the Median class is 20-30.
Here, \( l = 20 \) (lower limit of Median class)
\( f = 17 \) (frequency of Median class)
\( c = 13 \) (cumulative frequency of class preceding Median class)
\( i = 10 \) (class interval size)
Formula for Median \( = l + \left( \frac { \frac { N }{ 2 } - c }{ f } \right) \times i \)
\( \implies \) Median \( = 20 + \left( \frac { 25 - 13 }{ 17 } \right) \times 10 \)
\( = 20 + \left( \frac { 12 }{ 17 } \right) \times 10 \)
\( = 20 + \frac { 120 }{ 17 } \)
\( = 20 + 7.058... \)
\( = 27.06 \)

**3. Calculate D7 (Seventh Decile):**
Position of D7 \( = \frac { 7N }{ 10 } = \frac { 7 \times 50 }{ 10 } = \frac { 350 }{ 10 } = 35 \)
This value (35) lies in the class 30-40 (since c.f. 30 is for 20-30, and c.f. 37 is for 30-40).
So, the D7 class is 30-40.
Here, \( l = 30 \) (lower limit of D7 class)
\( f = 7 \) (frequency of D7 class)
\( c = 30 \) (cumulative frequency of class preceding D7 class)
\( i = 10 \) (class interval size)
Formula for D7 \( = l + \left( \frac { \frac { 7N }{ 10 } - c }{ f } \right) \times i \)
\( \implies \) D7 \( = 30 + \left( \frac { 35 - 30 }{ 7 } \right) \times 10 \)
\( = 30 + \left( \frac { 5 }{ 7 } \right) \times 10 \)
\( = 30 + \frac { 50 }{ 7 } \)
\( = 30 + 7.142... \)
\( = 37.14 \)

**4. Calculate P70 (70th Percentile):**
Position of P70 \( = \frac { 70N }{ 100 } = \frac { 70 \times 50 }{ 100 } = \frac { 3500 }{ 100 } = 35 \)
This value (35) also lies in the class 30-40 (same as D7).
So, the P70 class is 30-40.
Here, \( l = 30 \) (lower limit of P70 class)
\( f = 7 \) (frequency of P70 class)
\( c = 30 \) (cumulative frequency of class preceding P70 class)
\( i = 10 \) (class interval size)
Formula for P70 \( = l + \left( \frac { \frac { 70N }{ 100 } - c }{ f } \right) \times i \)
\( \implies \) P70 \( = 30 + \left( \frac { 35 - 30 }{ 7 } \right) \times 10 \)
\( = 30 + \left( \frac { 5 }{ 7 } \right) \times 10 \)
\( = 30 + \frac { 50 }{ 7 } \)
\( = 30 + 7.142... \)
\( = 37.14 \)
In simple words: For grouped data, we first make a running total of students. Then, for median, Q3, D7, and P70, we find which group (class interval) they fall into. Once we know the group, we use a special formula that considers the lower limit of that group, its frequency, the running total before it, and the size of the group to calculate the exact value.

🎯 Exam Tip: For continuous grouped data, the formula \( l + \left( \frac { \frac { KN }{ X } - c }{ f } \right) \times i \) is essential. Identify the correct class interval, lower limit (l), frequency (f), preceding cumulative frequency (c), and class interval size (i) carefully for each calculation.

 

Question 10. Graphically find the mode of the following distribution :

Answer:
To find the mode graphically, we first need to draw a histogram for the given frequency distribution. A histogram uses bars to show the frequency of data within different ranges. The highest bar in the histogram shows the modal class.

Steps to find mode graphically:
1. Draw a histogram by plotting rectangles for each class interval, with heights corresponding to their frequencies (number of students).
2. Identify the 'modal class', which is the class interval with the highest frequency. In this case, the highest frequency is 11, which corresponds to the class interval 60-70. So, 60-70 is the modal class.
3. Draw two diagonal lines: one from the top-right corner of the rectangle of the pre-modal class (50-60) to the top-right corner of the modal class (60-70). The other diagonal line goes from the top-left corner of the post-modal class (70-80) to the top-left corner of the modal class (60-70).
4. These two diagonal lines will intersect at a point, let's call it P. From point P, draw a perpendicular line down to the x-axis. The point where this perpendicular line meets the x-axis (let's call it A) gives the required mode.

Based on such a graphical construction, the required mode is 64.
In simple words: We draw a bar graph (histogram) with the heights of the bars showing how many students are in each height group. The tallest bar is the "mode" group. We then draw diagonal lines from the top corners of the mode group's bar to the top corners of the bars next to it. Where these lines cross, we drop a straight line down to the bottom axis. The number on the bottom axis where this line lands is the mode.

🎯 Exam Tip: For graphical mode, drawing an accurate histogram is crucial. Ensure you correctly identify the modal class and draw the diagonal lines to the adjacent class bars to pinpoint the intersection for the mode value.