Statistics - Class 11 Mathematics - Chapter 13 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Statistics | NCERT | Mathematics | Class 11
Draw the frequency polygon representing the following frequency distribution.
Class interval | $30-34$ | $35-39$ | $40-44$ | $45-49$ | $50-54$ | $55-59$ |
Frequency | 12 | 16 | 20 | 8 | 10 | 4 |
To draw a frequency polygon for the given frequency distribution, we perform the following steps:
Calculate the Class Marks:
The class mark is the midpoint of each class interval and can be calculated using the formula: $$ \text{Class Mark} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} $$
Consider Imaginary Classes:
Add imaginary classes before the smallest class interval and after the largest class interval with frequency 0 to start and end the frequency polygon smoothly. Here, the imaginary classes are $25-29$ and $60-64$, each with zero frequency.
Construct an Extended Table:
Class interval
Class mark
Frequency
$25-29$
27
0
$30-34$
32
12
$35-39$
37
16
$40-44$
42
20
$45-49$
47
8
$50-54$
52
10
$55-59$
57
4
$60-64$
62
0
Plot Points on a Graph:
On the $x$-axis, plot the class marks.
On the $y$-axis, plot the frequencies associated with these class marks.
Plot points for each class mark and frequency pair, such as $A(27,0), B(32,12), C(37,16), D(42,20), E(47,8), F(52,10), G(57,4)$, and $H(62,0)$.
Connect the Points:
Join the plotted points with line segments in sequence from A to H. This connection of consecutive points forms the frequency polygon.
The result is a polygonal line graph which presents the frequency distribution visually, beginning at the first imaginary class mark and ending at the last, providing a complete view of data distribution.
Which symbols are used to mark the count in a frequency distribution table?
A) Tallies
B) Circle
C) Square
D) Triangle
The correct answer is A) Tallies. Tally marks are commonly used to mark the count in a frequency distribution table. Utilizing tally marks enables easy and accurate tracking of frequencies.
The number of times a particular observation occurs in a given data is called its
Frequency
The number of times an observation occurs in a given dataset is referred to as its frequency.
"Why are statistics used in economics?"
Statistics are utilized in economics for several pivotal reasons:
-
Interpretation of Data: Instead of relying solely on raw data, economists use statistical methods to interpret and analyze data, making it easier to understand complex economic phenomena.
-
Prediction and Forecasting: Statistics enable economists to make predictions and forecasts about economic trends by examining patterns and relationships within data sets.
-
Decision Making: The insights gained from statistical analyses aid policymakers and businesses in making informed decisions by providing a reliable basis for evaluating economic scenarios and potential outcomes.
Thus, statistics are indispensable in economics for transforming data into actionable intelligence that drives decision-making and future predictions.
Find the mean of the following data
1,7,9,3,4,5,6.
To calculate the mean of a given data set, you first need to use the formula:
$$ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Total number of observations}} $$
For the data set provided, which is {1, 7, 9, 3, 4, 5, 6}, we follow these steps:
Add up all the numbers:$$ 1 + 7 + 9 + 3 + 4 + 5 + 6 = 35 $$
Count the total number of observations:The data set contains 7 numbers.
Now substitute the values into the formula: $$ \text{Mean} = \frac{35}{7} = 5 $$
Therefore, the mean of the data set {1, 7, 9, 3, 4, 5, 6} is 5.
Find $\operatorname{cov}(X, Y)$ for the following data:
X | Y |
---|---|
4 | 5 |
6 | 9 |
8 | 12 |
10 | 14 |
12 | 15 |
A. 8
B. 10
C. 12
D. 15
Let's determine the covariance $\operatorname{cov}(X, Y)$ for the given dataset.
First, we will use the following table structure to calculate the necessary values:
X | Y | $\mathrm{X}-\overline{\mathrm{X}}$ | $\mathrm{Y}-\overline{\mathrm{Y}}$ | $(\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})$ |
---|---|---|---|---|
4 | 5 | -4 | -6 | 24 |
6 | 9 | -2 | -2 | 4 |
8 | 12 | 0 | 1 | 0 |
10 | 14 | 2 | 3 | 6 |
12 | 15 | 4 | 4 | 16 |
Now, compute the mean values ($ \overline{\mathrm{X}} $ and $ \overline{\mathrm{Y}} $):
$$ \begin{array}{l} \bar{X} = \frac{\Sigma X}{N} = \frac{4 + 6 + 8 + 10 + 12}{5} = \frac{40}{5} = 8 \ \bar{Y} = \frac{\Sigma Y}{N} = \frac{5 + 9 + 12 + 14 + 15}{5} = \frac{55}{5} = 11 \ \end{array} $$
Next, calculate each $ (\mathrm{X} - \overline{\mathrm{X}}) $ and $ (\mathrm{Y} - \overline{\mathrm{Y}}) $:
$$ \begin{array}{l} (\mathrm{X}-\overline{\mathrm{X}})_1 = 4 - 8 = -4 \ (\mathrm{X}-\overline{\mathrm{X}})_2 = 6 - 8 = -2 \ (\mathrm{X}-\overline{\mathrm{X}})_3 = 8 - 8 = 0 \ (\mathrm{X}-\overline{\mathrm{X}})_4 = 10 - 8 = 2 \ (\mathrm{X}-\overline{\mathrm{X}})_5 = 12 - 8 = 4 \ \end{array} $$
$$ \begin{array}{l} (\mathrm{Y}-\overline{\mathrm{Y}})_1 = 5 - 11 = -6 \ (\mathrm{Y}-\overline{\mathrm{Y}})_2 = 9 - 11 = -2 \ (\mathrm{Y}-\overline{\mathrm{Y}})_3 = 12 - 11 = 1 \ (\mathrm{Y}-\overline{\mathrm{Y}})_4 = 14 - 11 = 3 \ (\mathrm{Y}-\overline{\mathrm{Y}})_5 = 15 - 11 = 4 \ \end{array} $$
Then, compute the product $ (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}}) $ for each data point:
$$ \begin{array}{l} (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})_1 = (-4)(-6) = 24 \ (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})_2 = (-2)(-2) = 4 \ (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})_3 = (0)(1) = 0 \ (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})_4 = (2)(3) = 6 \ (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})_5 = (4)(4) = 16 \ \end{array} $$
Sum these products:
$$ \sum (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}}) = 24 + 4 + 0 + 6 + 16 = 50 $$
Finally, the covariance $\operatorname{cov}(X, Y)$ is calculated as follows:
$$ \operatorname{cov}(X, Y) = \frac{\sum (\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})}{N} = \frac{50}{5} = 10 $$
Therefore, the covariance $\operatorname{cov}(X, Y)$ is 10.
The given data shows the population of different cities. What is the median of this data?
Cities | Population |
---|---|
Delhi | 55000 |
Mumbai | 60000 |
Kolkatta | 44000 |
Ranchi | 39500 |
Chennai | 48500 |
Bengaluru | 57000 |
Hyderabad | 51000 |
Chandigarh | 43000 |
Option 1) 50500
Option 2) 48750
Option 3) 46900
Option 4) 49750
The correct option is Option 4: 49750
To determine the median, we first need to arrange the population data in ascending order: $$ 39500, 43000, 44000, 48500, 51000, 55000, 57000, 60000 $$
Given that there are 8 observations, the median will be the average of the 4th and 5th terms.
4th term: 48500
5th term: 51000
Thus, the median is calculated as: $$ \text{Median} = \frac{48500 + 51000}{2} = 49750 $$
Find the M.D. from mean of the following data:
Class Interval | 2-4 | 4-6 | 6-8 | 8-10 |
---|---|---|---|---|
Frequency | 3 | 4 | 2 | 1 |
Find the M.D. from median of the following data:
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
---|---|---|---|---|---|
Frequency | 5 | 7 | 4 | 8 | 10 |
To solve the problems, we need to find:
The Mean Deviation (M.D.) from the mean of the given data.
The Mean Deviation (M.D.) from the median of the given data.
Let's solve each part step by step.
(a) Mean Deviation (M.D.) from the Mean
Given Data:
Class Intervals: 2-4, 4-6, 6-8, 8-10
Frequencies: 3, 4, 2, 1
Step 1: Calculate Class Midpoints ($ x $)
$$ x = \frac{(\text{Lower Limit} + \text{Upper Limit})}{2} $$
$$ \begin{aligned} &\text{For 2-4}: \quad x = \frac{(2+4)}{2} = 3 \ &\text{For 4-6}: \quad x = \frac{(4+6)}{2} = 5 \ &\text{For 6-8}: \quad x = \frac{(6+8)}{2} = 7 \ &\text{For 8-10}: \quad x = \frac{(8+10)}{2} = 9 \ \end{aligned} $$
Step 2: Calculate the Mean ($ \bar{x} $)
$$ \bar{x} = \frac{\sum{f_i x_i}}{\sum{f_i}} = \frac{(3 \times 3) + (4 \times 5) + (2 \times 7) + (1 \times 9)}{3 + 4 + 2 + 1} = \frac{9 + 20 + 14 + 9}{10} = \frac{52}{10} = 5.2 $$
Step 3: Calculate $ |x_i - \bar{x}| $ and then find $\sum{f_i |x_i - \bar{x}|}$
$$ \begin{aligned} &\text{For } x = 3: \quad |3 - 5.2| = 2.2 \ &\text{For } x = 5: \quad |5 - 5.2| = 0.2 \ &\text{For } x = 7: \quad |7 - 5.2| = 1.8 \ &\text{For } x = 9: \quad |9 - 5.2| = 3.8 \ \end{aligned} $$
$$ \sum{f_i |x_i - \bar{x}|} = (3 \times 2.2) + (4 \times 0.2) + (2 \times 1.8) + (1 \times 3.8) = 6.6 + 0.8 + 3.6 + 3.8 = 14.8 $$
Step 4: Calculate M.D. from Mean
$$ \text{M.D. from Mean} = \frac{\sum{f_i |x_i - \bar{x}|}}{\sum{f_i}} = \frac{14.8}{10} = 1.48 $$
(b) Mean Deviation (M.D.) from the Median
Given Data:
Marks: 0-10, 10-20, 20-30, 30-40, 40-50
Frequencies: 5, 7, 4, 8, 10
Step 1: Calculate Cumulative Frequencies
$$ \begin{aligned} &\text{Cumulative Frequencies:} \ &5, 12, 16, 24, 34 \end{aligned} $$
Step 2: Find the Median Class
$$ \text{Total Frequency (N)} = 34 \quad \Rightarrow \quad \frac{N}{2} = 17 $$
The cumulative frequency just greater than 17 is 24, which corresponds to the class interval 30-40. So, the median class is 30-40.
Step 3: Calculate the Median
$$ \text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times h $$
Where:
$ L $ = lower boundary of median class = 30
$ N $ = total frequency = 34
$ CF $ = cumulative frequency of the class preceding the median class = 16
$ f $ = frequency of the median class = 8
$ h $ = class width = 10
$$ \text{Median} = 30 + \left( \frac{17 - 16}{8} \right) \times 10 = 30 + \left( \frac{1}{8} \right) \times 10 = 30 + 1.25 = 31.25 $$
Step 4: Calculate $ |x_i - \text{Median}| $ and then find $\sum{f_i |x_i - \text{Median}}|$
Midpoints:
$$
\begin{aligned}
&0-10: \quad x = 5 \
&10-20: \quad x = 15 \
&20-30: \quad x = 25 \
&30-40: \quad x = 35 \
&40-50: \quad x = 45 \
\end{aligned}
$$
$$ \begin{aligned} &|5 - 31.25| = 26.25 \ &|15 - 31.25| = 16.25 \ &|25 - 31.25| = 6.25 \ &|35 - 31.25| = 3.75 \ &|45 - 31.25| = 13.75 \ \end{aligned} $$
$$ \sum{f_i |x_i - \text{Median}|} = (5 \times 26.25) + (7 \times 16.25) + (4 \times 6.25) + (8 \times 3.75) + (10 \times 13.75) $$
$$ = 131.25 + 113.75 + 25 + 30 + 137.5 = 437.5 $$
Step 5: Calculate M.D. from Median
$$ \text{M.D. from Median} = \frac{\sum{f_i |x_i - \text{Median}|}}{\sum{f_i}} = \frac{437.5}{34} = 12.87 $$
Final Answers:
(a) M.D. from Mean: 1.48
(b) M.D. from Median: 12.87
The standard deviation for the set of numbers 1, 4, 5, 7, 8 is 2.45. If 10 is added to each number, then the new standard deviation is:
A. 2.45
B. 24.5
C. 0.245
D. 12.45
To determine the new standard deviation after adding a constant to each number in a dataset, let's review the properties of standard deviation.
Given the original set of numbers: (1, 4, 5, 7, 8), we know that the standard deviation (SD) is 2.45.
If 10 is added to each number, the new set of numbers will be: (11, 14, 15, 17, 18).
Key Concept:
Adding a constant to every number in a dataset does not change the standard deviation. This is because standard deviation measures the spread of numbers in the dataset relative to their mean. When we add a constant to each number, it shifts the entire dataset without affecting the relative distances between the numbers.
Mathematical Explanation:
Standard Deviation Formula:
$$ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2} $$
Here, $ \sigma $ is the standard deviation, ( N ) is the number of observations, ($ x_i $ is each individual observation, and $ \bar{x} $ is the mean of the observations.
Let's denote the original numbers by $ x_i $ and the new numbers by $ x_i' = x_i + 10 $.
The mean of the original numbers $( \bar{x} )$ is:
$$ \bar{x} = \frac{1 + 4 + 5 + 7 + 8}{5} = 5 $$
The mean of the new numbers $( \bar{x}' )$ will be:
$$ \bar{x}' = \frac{11 + 14 + 15 + 17 + 18}{5} = \bar{x} + 10 = 15 $$
The standard deviation formula for the new numbers can be written as:
$$ \sigma' = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i' - \bar{x}')^2} $$
Substituting $ x_i' = x_i + 10 $ and $ \bar{x}' = \bar{x} + 10 $:
$$ x_i' - \bar{x}' = (x_i + 10) - (\bar{x} + 10) = x_i - \bar{x} $$
Thus, the expression for the standard deviation remains unchanged:
$$ \sigma' = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2} = \sigma $$
Therefore, the new standard deviation is the same as the old standard deviation, which means:
New Standard Deviation = 2.45
Final Answer:
A. 2.45
Identify the correct formula for calculating $r$ using the step deviation method, where $U$ and $V$ are the step deviations of $X$ and $Y$:
A. $r=\frac{N \Sigma\left(U_{i} V_{i}\right)-N\left(\Sigma U_{i}\right)\left(\Sigma V_{i}\right)}{\sqrt{N \Sigma U_{i}^{2}-\left(\Sigma U_{i}\right)^{2}} \sqrt{N \Sigma V_{i}^{2}-\left(\Sigma V_{i}\right)^{2}}}$
B. $r=\frac{N \Sigma\left(U_{i} V_{i}\right)-\left(\Sigma U_{i}\right)\left(\Sigma V_{i}\right)}{\sqrt{N \Sigma U_{i}^{2}-\left(\Sigma U_{i}\right)^{2}} \sqrt{N \Sigma V_{i}^{2}-\left(\Sigma V_{i}\right)^{2}}}$
C. $r=\frac{N \Sigma\left(U_{i} V_{i}\right)-\left(\Sigma U_{i}\right)\left(\Sigma V_{i}\right)}{\left(N \Sigma U_{i}^{2}-\left(\Sigma U_{i}\right)^{2}\right)\left(N \Sigma V_{i}^{2}-\left(\Sigma V_{i}\right)^{2}\right)}$
D. $r=\frac{N \Sigma\left(U_{i} V_{i}\right)-N\left(\Sigma U_{i}\right)\left(\Sigma V_{i}\right)}{\left(N \Sigma U_{i}^{2}-\left(\Sigma U_{i}\right)^{2}\right)\left(N \Sigma V_{i}^{2}-\left(\Sigma V_{i}\right)^{2}\right)}$
The correct option is B:
$$ r=\frac{N \sum\left(U_{i} V_{i}\right)-\left(\sum U_{i}\right)\left(\sum V_{i}\right)}{\sqrt{N \sum U_{i}^{2}-\left(\sum U_{i}\right)^{2}} \sqrt{N \sum V_{i}^{2}-\left(\sum V_{i}\right)^{2}}} $$
The shortcut formula for calculating $r$ using the step deviation method is:
$$ r = \frac{N \sum(U_i V_i) - (\sum U_i)(\sum V_i)}{\sqrt{N \sum U_i^2 - (\sum U_i)^2} \cdot \sqrt{N \sum V_i^2 - (\sum V_i)^2}} $$
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