Statistics - Class 9 Mathematics - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Statistics | NCERT | Mathematics | Class 9
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:
Number of letters | Number of surnames |
---|---|
4-6 | 30 |
6-8 | 44 |
8-12 | 16 |
12-20 | 4 |
(ii) Write the class interval in which the maximum number of surnames lie.
The class interval in which the maximum number of surnames lie is $6-8$. This interval contains 44 surnames, the highest compared to the other intervals given in the frequency distribution. This indicates it is the most common surname length in the sample from the local telephone directory.
In the given frequency distribution, what is the third value in cumulative frequency of lesser than type?
Age in years | Number of people |
---|---|
15 | 3 |
16 | 8 |
17 | 10 |
18 | 10 |
19 | 5 |
20 | 4 |
A) 31
B) 21
C) 40
D) 36
To determine the third value in the cumulative frequency of a lesser than type from the provided age distribution table, you simply add up the number of people cumulatively until the third age group:
Age in years | Number of people | Cumulative Frequency |
---|---|---|
15 | 3 | 3 |
16 | 8 | $3+8=11$ |
17 | 10 | $11+10=21$ |
18 | 10 | $21+10=31$ |
19 | 5 | $31+5=36$ |
20 | 4 | $36+4=40$ |
The third value, corresponding to age 17, is 21.
Therefore, the correct answer is: A) 31.
Variable | 1 | 2 | x | 4 | 5 |
Frequency | 2 | 3 | 4 | 5 | 6 |
The mean of the above frequency distribution is 3.5. Then the value of $x$ is:
A) 2
B) 3
C) 4
D) 5
The correct answer is option B) 3.
The problem states that the mean of the frequency distribution is $3.5$. We calculate the mean by using the formula:
$$ \text{Mean} = \frac{\sum_{i} (f_i x_i)}{\sum_{i} f_i} $$
Where $x_i$ represents the variables and $f_i$ their corresponding frequencies. Substituting the given values into the formula, we have:
$$ 3.5 = \frac{1 \cdot 2 + 2 \cdot 3 + x \cdot 4 + 4 \cdot 5 + 5 \cdot 6}{2+3+4+5+6} $$
Calculating the values in the numerator:
$1 \cdot 2 = 2$
$2 \cdot 3 = 6$
$x \cdot 4 = 4x$
$4 \cdot 5 = 20$
$5 \cdot 6 = 30$
Thus,
$$ 3.5 = \frac{2 + 6 + 4x + 20 + 30}{20} $$
which simplifies to
$$ 3.5 = \frac{58 + 4x}{20} $$
To find $x$, solve the equation:
$$ 70 = 58 + 4x $$
Resulting in
$$ 4x = 12 $$
Hence,
$$ x = 3 $$
This means the value of x in the distribution must be 3 to achieve the given mean of $3.5$.
Concentration of $\mathrm{SO}_2$ (in ppm) | Number of days (frequency) |
---|---|
0.00 − 0.04 | 4 |
0.04-0.08 | 9 |
0.08-0.12 | 9 |
0.12-0.16 | 2 |
0.16 - 0.20 | 4 |
0.20-0.24 | 2 |
The above frequency distribution table represents the concentration of Sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of Sulphur dioxide in the interval 0.12-0.16 on any of these days.
To determine the probability that the concentration of Sulphur dioxide (SO₂) falls within the interval $0.12$ to $0.16$ ppm, we use the provided frequency data:
Total number of days provided is $30$
Number of days the concentration was between $0.12$ and $0.16$ is $2$
The formula for probability is: $$ P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total outcomes}} $$
Substitute the relevant values into the formula: $$ P(\text{SO}_2 \text{ concentration in } 0.12-0.16) = \frac{2}{30} $$
Simplify this fraction to get the probability: $$ P = \frac{1}{15} $$
Therefore, the probability of the concentration of SO₂ being within the range $0.12$ to $0.16$ ppm on any given day is $\frac{1}{15}$.
"The red line indicates the girls' height distribution. The blue line indicates the boys' height distribution. Find:
i) Maximum number of girls belong to which height group?
ii) The difference in the number of boys to the number of girls in the class.
A) $140-150, 1$
B) $150-160, 2$
C) $160-170, 0$
D) $170-180, \text{Not enough data}$
The correct answer is Option C: $$ 160-170, 0 $$
Analysis:
i) Maximum number of girls' height group: Observing the graph, the $160-170$ cm group contains 7 girls, which is the highest count among the groups.
ii) Difference in the number of boys to the number of girls:
Total number of boys is calculated as follows: $$ 4 + 3 + 2 + 4 + 3 = 16 $$
Total number of girls is: $$ 1 + 7 + 6 + 2 = 16 $$
Difference in number: $$ 16 - 16 = 0 $$
Thus, the difference between the number of boys and girls in the class is 0.
If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?
The mean of a set of observations is calculated by the formula:
$$ \text{Mean} = \frac{\text{Sum of observations}}{\text{Total number of observations}} $$
Given that the mean of 25 observations is 27, we can find the sum of all observations by rearranging the formula:
$$ 27 = \frac{\text{Sum of observations}}{25} $$
Solving for the sum of the observations:
$$ \text{Sum of observations} = 27 \times 25 = 675 $$
If each observation is decreased by 7, the total decrease across all observations will be:
$$ 25 \times 7 = 175 $$
Thus, the new sum of the observations is:
$$ 675 - 175 = 500 $$
Finally, the new mean after decreasing each observation by 7 can be calculated as:
$$ \text{New mean} = \frac{500}{25} = 20 $$
Therefore, the new mean of the observations after the decrease is 20.
What does the $\mathrm{I}$ in $mode=I+\left(\frac{\mathrm{f}{1}-f{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h$ stand for?
Concept Understanding: 1 Mark
In the provided formula for the mode: $$ \text{Mode} = I + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h $$ $I$ represents the lower limit of the modal class.
Find the mean of $56, 34, 89, 13$.
A. 18
B. 58
C. $\mathbf{28}$
D. 48
To find the mean of the numbers $56$, $34$, $89$, and $13$, you apply the formula for the mean, which is:
$$ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Total number of observations}} $$
Calculating the sum of the observations: $$ 56 + 34 + 89 + 13 = 192 $$
Since there are four numbers in this data set, the total number of observations is $4$. Inserting these values into the formula gives:
$$ \text{Mean} = \frac{192}{4} = 48 $$
Thus, the correct option is D. 48.
Following table shows the frequency of students who have scored more than 10 marks in Maths. However, there is a score of 40 which is left out. To which class should this score be mapped to?
Class | Frequency |
---|---|
$10-20$ | 12 |
$20-30$ | 2 |
$30-40$ | $x$ |
$40-50$ | $y$ |
(A) $40-50$
(B)$30-40$
(C) either a or b
(D) neither a nor b
The correct option is (A) $40-50$.
Explanation: When an observation falls exactly on the boundary between two class intervals, the convention is to include it in the higher class interval. In this case, a score of $40$ should be assigned to the class interval $40-50$, rather than the class interval $30-40$. Thus, the student scoring $40$ is categorized in the $40-50$ range.
The graphical representation of a cumulative frequency table is called a histogram.
A) True
B) False
The correct answer is B) False.
The graphical representation of a cumulative frequency table is known as an ogive, while a histogram represents a grouped frequency distribution.
If the sum of the deviations of a set of values $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ measured from 50 is -10, and the sum of deviations of the values from 46 is 70, then the mean is $\qquad$
A) 49
B) 49.5
C) 49.75
D) 50
The correct option is B
Given the deviations of a set of values $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ from 50 and 46, we can formulate two equations based on these sums:
-
The sum of deviations from 50: $$ (x_{1} - 50) + (x_{2} - 50) + \ldots + (x_{n} - 50) = -10 $$ Simplifying, we get: $$ \sum_{i=1}^n x_i - 50n = -10 \quad \text{(i)} $$
-
The sum of deviations from 46: $$ (x_{1} - 46) + (x_{2} - 46) + \ldots + (x_{n} - 46) = 70 $$ Simplifying, we get: $$ \sum_{i=1}^n x_i - 46n = 70 \quad \text{(ii)} $$
By equating the expressions for $ \sum_{i=1}^n x_i $ from equations (i) and (ii), we arrive at: $$ -10 + 50n = 70 + 46n $$ Simplifying this, we find: $$ 4n = 80 \ n = 20 $$
Plugging $ n = 20 $ back into equation (i): $$ \sum_{i=1}^n x_i = -10 + 50n = -10 + 50 \times 20 = 990 $$
Thus, the mean of these values is calculated as: $$ \text{Mean} = \frac{\sum_{i=1}^n x_i}{n} = \frac{990}{20} = 49.5 $$
Therefore, the mean of the dataset is 49.5.
Complete the following sentence: "Mode is the $\quad$
A. Most frequent value
B. Least frequent value
C. Middle most value
D. Average value
The correct answer is A. Most frequent value.
The mode is determined by organizing the data to count the frequency of each result. Therefore, the mode is the value that appears most frequently in a data set.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate by step deviation method.
Literacy rate (in percentage) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
---|---|---|---|---|---|
Number of cities | 3 | 10 | 11 | 8 | 3 |
A) 66.44
B) 69.43
C) 67.39
D) 71.55
The mean literacy rate can be calculated using the step deviation method. Below we'll breakdown the process:
Identify the class size, $h$: Since the difference between the upper and lower class limit is constant for all classes, we have: $$ h = 55 - 45 = 10 $$
Calculate the class mark ($x_i$) for each class, which is the midpoint of each class: $$ x_i = \frac{\text{upper limit} + \text{lower limit}}{2} $$
Choose an assumed mean, $a$, to simplify the calculations. Here, let's take $a = 70$. This choice aims to minimize the values of $d$ (deviation from assumed mean).
Calculate $d_i = x_i - a$ and $u_i = \frac{d_i}{h}$ for each class:
Class Interval Frequency ($f_i$) Class mark ($x_i$) $d_i = x_i - a$ $u_i = \frac{d_i}{h}$ 45-55 3 50 -20 -2 55-65 10 60 -10 -1 65-75 11 70 0 0 75-85 8 80 10 1 85-95 3 90 20 2 Calculate the mean ($\bar{x}$): $$ \bar{x} = a + \left(\frac{\sum (f_i u_i)}{\sum f_i}\right) \times h $$ Plugging in the values: $$ \bar{x} = 70 + \left(\frac{3(-2) + 10(-1) + 11(0) + 8(1) + 3(2)}{35}\right) \times 10 = 70 + \left(\frac{-6-10+0+8+6}{35}\right) \times 10 $$ $$ = 70 + \left(\frac{-2}{35}\right) \times 10 = 70 - 0.57 \approx 69.43 $$ Thus, the mean literacy rate is about 69.43, corresponding to option $\mathbf{B}$.
The number of cars manufactured each day of a week are 2, 4, 6, 8, 10, 5, 7. Find the standard deviation of the number of cars manufactured using the Direct method.
A) 2.83
B) $\quad$ 3.45
C) 6.35
D) 8.23
Solution
The correct answer is A) 2.83.
First, we tabulate the values of $x_i$ (number of cars manufactured each day) and $x_i^2$ (square of each day's output).
$x_i$ | $x_i^2$ |
---|---|
2 | 4 |
4 | 16 |
6 | 36 |
8 | 64 |
10 | 100 |
5 | 25 |
7 | 49 |
Sum ($\sum x_i$) | 42 |
Sum ($\sum x_i^2$) | 294 |
Next, we calculate the standard deviation: $$ \sigma = \sqrt{\frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2} $$
Where:
- $n = 7$ (number of days)
- $\sum x_i = 42$
- $\sum x_i^2 = 294$
Plugging in the values: $$ \sigma = \sqrt{\frac{294}{7} - \left(\frac{42}{7}\right)^2} $$ $$ \sigma = \sqrt{42 - 36} = \sqrt{6} = 2.83 $$
Therefore, the standard deviation is 2.83.
The middle most observation in a data set is called Mean.
True
False
The correct option is B: False
The middle most observation in a data set is called the Median, not the Mean. The Median represents the value that lies in the center of a data set when it is arranged in ascending or descending order.
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