Statistics - Class 10 Mathematics - Chapter 13 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Statistics | NCERT | Mathematics | Class 10
The group of intelligent students in a class is:
A) a null set
B) a finite set
C) a well-defined collection
D) not a well-defined collection
The correct answer is D) not a well-defined collection.
The term "intelligent" is subjective and varies greatly depending on who is making the judgment. Therefore, there is no clear, objective criterion to determine who belongs in the group of "intelligent students." This makes it impossible to precisely define the collection, leading us to conclude that the group of intelligent students is not a well-defined collection.
The mean of the following distribution is:
$x_i$ | 10 | 13 | 16 | 19 |
$f_i$ | 2 | 5 | 7 | 6 |
(A) 15.2
(B) 15.35
(C) 15.55
(D) 16
To find the mean of a given frequency distribution, we use the formula for the mean (weighted average) of grouped data:
$$ \bar{x} = \frac{\sum (f_i x_i)}{\sum f_i} $$
Here, $x_i$ are the class marks and $f_i$ are the corresponding frequencies. The values you provided are:
$x_i = 10, 13, 16, 19$
$f_i = 2, 5, 7, 6$
Let's compute the sum of the products $f_i x_i$ and the sum of $f_i$:
$ f_ix_i $ calculations:
$$ 2 \times 10 = 20 \\ 5 \times 13 = 65 \\ 7 \times 16 = 112 \\ 6 \times 19 = 114 $$
So, $\sum (f_ix_i) = 20 + 65 + 112 + 114 = 311$.
$\sum f_i$ calculation:
$$ 2 + 5 + 7 + 6 = 20 $$
Now, plugging these values into the formula for mean:
$$ \bar{x} = \frac{311}{20} = 15.55 $$
Thus, the mean of the distribution is 15.55. The correct option is:
(C) 15.55
Given a data set consisting of just 2 observations (each occurring once), what is true for the given data set?
A) Mean and median are the same.
B) Mean and mode are the same.
C) Median and mode are the same.
D) No relation can be determined.
The correct answer is A) Mean and median are the same.
The mean of two numbers, $a$ and $b$, is calculated as: $$ \text{Mean} = \frac{a + b}{2} $$
The median for an even number of terms is the average of the two middle terms. Given there are only two observations ($a$ and $b$), the median is: $$ \text{Median} = \frac{a + b}{2} $$
Hence, both the mean and median of this data set are the same.
The cumulative frequency table is useful in determining the:
A) Mean
B) Median
C) Mode
D) All of these
The question asks about the usefulness of a cumulative frequency table in determining specific statistical measurements. Let's break down the options given:
Cumulative frequency accumulates the frequency counts, adding each class's frequency to the sum of the frequencies of all preceding classes.
Option Analysis:
Mean: Calculating the mean usually involves using the formula $\text{Mean} = \frac{\sum (f_i \times x_i)}{\sum f_i}$ where $f_i$ is the frequency and $x_i$ is the mid-point of the class intervals. In this formula, cumulative frequency doesn't directly play a role, as it involves individual class frequencies and their respective mid-points.
Mode: To find the mode in grouped data, we use the formula: $$ \text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h $$ where $f_1$ is the maximum frequency, $f_0$ is the frequency before $f_1$, and $f_2$ is the frequency after $f_1$, with $L$ as the lower class boundary of the modal class, and $h$ as the class width. Cumulative frequencies aren't typically used to find the mode.
Median: The median is calculated using cumulative frequencies. To determine the media, you identify the median class - the class where the cumulative frequency reaches or exceeds half the total frequency. You can then use the formula: $$ \text{Median} = L + \left(\frac{\frac{N}{2} - F}{f}\right) \times h $$ where $N$ is the total number of observations (sum of frequencies), $F$ is the cumulative frequency of the class preceding the median class, $f$ is the frequency of the median class, and $h$ is the class width.
Conclusion:
Given the choices provided, B) Median is the correct option because the cumulative frequency table is primarily utilized in determining the median in grouped data. The table helps identify the class interval that contains the median, which is crucial when dealing with grouped data.
So, the correct answer is B) Median.
The mean of the following frequency distribution is 62.8, and the sum of all frequencies is 50. Compute the missing frequencies $f_{1}$ and $f_{2}$.
To find the missing frequencies $f_1$ and $f_2$ for given class intervals in a frequency distribution where the mean is 62.8 and the total of all frequencies equals 50, you can follow these steps.
Provided Data
The mean ($\bar{x}$) of the distribution is 62.8.
The total frequency (sum of all frequencies $\sum f$) is 50.
Class Intervals and their Frequencies:
$0 - 20$ : Frequency = 5
$20 - 40$ : Frequency = $f_1$
$40 - 60$ : Frequency = 10
$60 - 80$ : Frequency = $f_2$
$80 - 100$ : Frequency = 7
$100 - 120$ : Frequency = 8
$120 - 140$ : Frequency = 8
Step 1: Establish Frequency Equation
Since the total frequency given is 50, we can set up the frequency equation: $$ 5 + f_1 + 10 + f_2 + 7 + 8 + 8 = 50 $$ Simplifying, $$ f_1 + f_2 + 38 = 50 $$ $$ f_1 + f_2 = 12 \quad [\text{Equation 1}] $$
Step 2: Calculate Midpoints
We calculate the midpoint of each class for use in finding the mean:
Midpoint of $0 - 20$ = 10
Midpoint of $20 - 40$ = 30
Midpoint of $40 - 60$ = 50
Midpoint of $60 - 80$ = 70
Midpoint of $80 - 100$ = 90
Midpoint of $100 - 120$ = 110
Midpoint of $120 - 140$ = 130
Step 3: Set Up Equation for Mean
To uphold the mean, use the formula $\sum f m / \sum f = \bar{x}$, where $m$ represents the midpoint: $$ \frac{5 \times 10 + 30f_1 + 10 \times 50 + 70f_2 + 7 \times 90 + 8 \times 110 + 8 \times 130}{50} = 62.8 $$ Simplifying and solving for $f_1$ and $f_2$: $$ 50 + 30f_1 + 500 + 70f_2 + 630 + 880 + 1040 = 62.8 \times 50 $$ $$ 30f_1 + 70f_2 + 3100 = 3140 $$ $$ 30f_1 + 70f_2 = 40 \quad [\text{Equation 2}] $$
Step 4: Solve the Linear Equations
Now, solve equations 1 and 2 simultaneously:
$f_1 + f_2 = 12$ (from sum of frequencies)
$30f_1 + 70f_2 = 40$ (from mean)
Multiplying Equation 1 by 30: $$ 30f_1 + 30f_2 = 360 $$ Subtract it from Equation 2: $$ 30f_1 + 70f_2 - (30f_1 + 30f_2) = 40 - 360 $$ $$ 40f_2 = -320 $$ $$ f_2 = -8 $$ This incorrect result suggests potential error in calculations or inputs. It's typically a good practice to verify given data, equation setup, or solve again ensuring no calculation mistakes.
Conclusion: Given this context and the anomaly in results (negative frequency), check inputs or consider re-evaluating steps for computational accuracy. The expected results should have been non-negative integers, suggesting possible errors. Frequencies $f_1$ and $f_2$ must be validated or re-calculated appropriately.
The relation between mean, mode, and median is:
The relationship between the mean, median, and mode for a distribution of data can be described using a formula. This relationship is often cited in statistical contexts and is helpful in understanding how these measures of central tendency relate to each other. Here is a breakdown of the relationship:
Mean (often represented as $\mu$)
Median (the middle value in an ordered list of numbers)
Mode (the most frequently occurring value in the dataset)
The most common way to express the relationship between these three is via the Empirical Rule or the Pearson's Mode Skewness formula:
$$ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} $$
This formula helps to estimate the mode of a distribution from the mean and median. It is particularly useful when dealing with a moderately skewed distribution, where the mean, median, and mode are not the same. Here's how it works:
The mode is estimated as three times the median minus two times the mean. This captures the notion that in a skewed distribution, the mode lies closer to the median than to the mean.
Each value (mean, median, mode) gives us different insights about the data's distribution, and understanding their relationship helps in better analyzing the data. For example, if mean and median are known, this formula can provide an estimate for the mode without having to plot the data. However, it's important to remember that this relationship holds more accurately in symmetric, bell-shaped distributions (approximately normal distributions) and may not be as precise in distributions that are highly skewed or multi-modal.
The annual rainfall record of a city for 66 days is given below in the table:
Rainfall (in cm) | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $40-50$ |
Number of days | 22 | 10 | 8 | 15 | 5 | 6 |
Calculate the median rainfall using the formula:
To calculate the median rainfall using the defined formula and the given data, follow these steps:
Sum the total days: First, we sum the number of days across all intervals: $$ 22 + 10 + 8 + 15 + 5 + 6 = 66 \text{ days} $$
Calculate ( \frac{n}{2} ), where ( n ) is the total number of days: $$ \frac{66}{2} = 33 $$ We use this value to find the median class.
Determine the cumulative frequency for each class:
0-10 cm: ( 22 ) days
10-20 cm: ( 22 + 10 = 32 ) days
20-30 cm: ( 32 + 8 = 40 ) days
30-40 cm: ( 40 + 15 = 55 ) days
40-50 cm: ( 55 + 5 = 60 ) days
50-60 cm: ( 60 + 6 = 66 ) days
The median class is where the cumulative frequency exceeds $ \frac{n}{2} = 33 $ for the first time, which occurs in the interval 20-30 cm.
Use the median formula: $$ \text{Median} = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \cdot h $$ where:
$L$ is the lower boundary of the median class, 20 cm.
$ \frac{n}{2} $ is 33.
$CF $ is the cumulative frequency of the class before the median class, 32.
$ f $ is the frequency of the median class, 8.
$ h $ is the class width, $ 30 - 20 = 10 $.
Plug in these values to get: $$ \text{Median} = 20 + \left(\frac{33 - 32}{8}\right) \cdot 10 = 20 + \left(\frac{1}{8}\right) \cdot 10 = 20 + 1.25 = 21.25 , \text{cm} $$
Final Answer: The median rainfall is ( 21.25 ) cm.
NITI Aayog has tasked their statistical officer to create a model for farmers to be able to predict their produce output based on various factors. To test the model out, the officer picked a local farmer who sells apples to check various factors like weight, bad apples, half-cooked, green vs red, etc. A box containing 250 apples was opened and each apple was weighed.
The distribution of the masses of the apples is given in the following table:
Mass (in grams) | 80-100 | 100-120 | 120-140 | 140-160 | 160-180 |
Frequency | 20 | 60 | 70 | p | 60 |
The lower limit of the modal class is:
A 80
B 100
C 120
D $140$
To determine the lower limit of the modal class of the apples' mass distribution, first, we identify which weight class contains the highest frequency:
From the frequency table, the class intervals and their respective frequencies are:
80-100 g: 20 apples
100-120 g: 60 apples
120-140 g: 70 apples
140-160 g: p apples (unknown)
160-180 g: 50 apples
The modal class is the class with the highest frequency. Here, the interval 120-140 g has the highest frequency of 70 apples.
The lower limit of this modal class is 120 g. Thus, the correct option is:
C. 120
Look at the following table below.
Class Interval | Class Marks |
---|---|
0 - 5 | A |
5 - 10 | B |
10 - 15 | 12.5 |
15 - 20 | 17.5 |
The value of $A & B$ respectively are:
(A) $2.5, 7.5$
B) $3, 7$
C) $2, 7$
(D) $3, 8$
The correct option is (A).
$$ 2.5, , 7.5 $$
To determine the class mark for a class interval, we use the formula: $$ \text{Class mark} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2} $$
For the first class interval (0-5):
Upper class limit: 5
Lower class limit: 0
Using the formula: $$ \text{Class mark} = \frac{5 + 0}{2} = 2.5 $$
For the second class interval (5-10):
Upper class limit: 10
Lower class limit: 5
Using the formula: $$ \text{Class mark} = \frac{10 + 5}{2} = 7.5 $$
Thus, the values of $A$ and $B$ respectively are $2.5$ and $7.5$.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate by step deviation method.
Literacy Rate (in %) | Number of Cities |
---|---|
45-55 | 3 |
55-65 | 10 |
65-75 | 11 |
75-85 | 8 |
85-95 | 3 |
(A) 66.44
(B) 69.43
(C) 67.39
(D) 71.55
The correct option is B:
$$ 69.43 $$
Here’s how we find the mean literacy rate using the step deviation method:
Step 1: Calculating the Class Mark
For each class interval, calculate the class mark ($ x_i $):
$$ x_i = \frac{\text{upper class limit} + \text{lower class limit}}{2} $$
Example Calculation for the first class interval:
$$ x_1 = \frac{45 + 55}{2} = 50 $$
Step 2: Choosing the Assumed Mean (a)
Let’s choose the assumed mean ($ a $) as 70, which simplifies our calculations.
Step 3: Determining the Class Size (h)
The class size ($ h $) is:
$$ h = 55 - 45 = 10 $$
Step 4: Preparing the Table
Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i = x_i - a$ | $u_i = \frac{d_i}{h}$ | $f_i u_i$ |
---|---|---|---|---|---|
44-55 | 3 | 50 | -20 | -2 | -6 |
55-65 | 10 | 60 | -10 | -1 | -10 |
65-75 | 11 | 70 | 0 | 0 | 0 |
75-85 | 8 | 80 | 10 | 1 | 8 |
85-95 | 3 | 90 | 20 | 2 | 6 |
Total | $\sum f_i = 35$ | $\sum f_i u_i = -2$ |
Step 5: Calculate the Mean
Using the step deviation method:
$$ \text{Mean} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h $$
Substitute the values:
$$ \text{Mean} = 70 + \left( \frac{-2}{35} \right) \times 10 $$
Simplify the expression:
$$ \text{Mean} = 70 + (-0.57) = 69.43 $$
Thus, the mean literacy rate of the 35 cities is:
$$ \boldsymbol{69.43 %} $$
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