M.C.Q - Page 3 - MATHS STD 9 Questions - Vidyadip

Questions · Page 3 of 6

M.C.Q

Question 1011 Mark

Write the correct answer in the following:
A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data:
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.
The frequency of the class 310-330 is:

Answer

6

Solution:
The observation corresponding to class 310–330 (330 not included in this interval) are 310, 310, 320, 319, 318, 316, i.e., 6 observations.

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Question 1021 Mark

The number of times a particular item occurs in a given data is called its:

Answer

Frequency.

Solution:
The number of times a particular item occurs in a given data is called the frequency of the item. Hence, the correct choice is (b).

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Question 1031 Mark

The class-mark of the class 130-150 is:

Answer

140

Solution:
Class mark $=\frac{130+150}{2}=\frac{280}{2}=140$

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Question 1041 Mark

The smallest of three consecutive even integers is 32. Then, the mean of the three integers is:

Answer

34

Solution:
32 is the smallest even integer.
So three consecutive even integers are 32, 34 and 36
The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
$\frac{32+34+36}{3}=34$

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Question 1051 Mark

The following marks were obtained by the students in a test:
81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62
The range of the marks is:

Answer

33

Solution:
The marks obtained by the students are 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79 and 62.
The highest and lowest marks are 95 and 62 respectively. Therefore, the range of marks is
95 - 62
= 33
Hence, the correct option is (d).

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Question 1061 Mark

Median of the following observations, arranged in an ascending order is 22. If the numbers are 8, 11, 13, 15, x + 1, x + 3, 30, 35, 40, 43. Then, the value of x is:

Answer

20

Solution:
The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
For even number of observations, median is calculated as average of two middle number
$22=\frac{(\text{x}+1)+(\text{x}+3)}{2}$
$44=2\text{x}+4$
$40=2\text{x}$
$\text{x}=20$

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Question 1071 Mark

Write the correct answer in the following:
The range of the data: 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is:

Answer

26

Solution:
Maximum value of the variate = 32
And the minimum value of the variate = 6
Range = Maximum value of the variate-Minimum value of the variate = 32 - 6 = 26

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Question 1081 Mark

The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. The correct mean is:

Answer

39.4

Solution:
Calculted mean of 50 students = 39
$\therefore\ $Calculated sum of these numbers = 39 × 50 = 1950
Correct sum of these numbers = 1950 - (wrong term) + (correct term)
= 1950 - 23 + 43
= 1970
$\therefore\ $the corrected mean $=\frac{1970}{50}$
$=39.4$

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Question 1091 Mark

If the less than ogive and the more than ogive intersect at (32, 48), then the median of the data is:

Answer

32

Solution:
If the less than ogive and the more than ogive intersect at (32, 48), then the median of the data is 32. Because on the graph, the point of the x-axis, where less than ogive and more than ogive intersects, is the median. Therefore, the Median of the data is 32.

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Question 1101 Mark

A frequency polygon is constructed by plotting frequency of the class interval and the:

Answer

Mid value of the class.

Solution:
Frequency polygon is the plot of frequencies vs. the mid values of the classes.

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Question 1111 Mark

There are 50 numbers. Each number is subtracted from 53 and the difference between the mean of the numbers so obtained is found to be -3.5. The mean of the given number is:

Answer

56.5

Solution:
Let the mean of the initial sequence is x.
Given that, after subtracting 53 from each number, the difference between the means is 3.5
So, x - 53 = 3.5
Mean of the number is x = 53 + 3.5 = 56.5

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Question 1121 Mark

For a given data, the difference between the maximum and minimum observation is known as its.

Answer

Range

Solution:
Difference between maximum and minimum value of observation is called as range.

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Question 1131 Mark

The mean of 50 observations is 39. If one of the observations which was 23 was replaced by 43, the resulting mean will be:

Answer

39.4

Solution:
The mean of 50 observations is 39.
So sum of these 50 observations is 50 × 39 = 1950
After replacing the observation value 23 by 43,
Sum becomes 1970
So the mean is $\frac{1970}{50}=39.4$

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Question 1141 Mark

$\text{Mode}+\frac{2}{3}(\text{Mean - Mode})=$

Answer

Median

Solution:
Since, 3 Median = Mode + 2 Mean
$\Rightarrow\text{Median}=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}$
$=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}-\frac{2}{3}\text{Mode}+\frac{2}{3}\text{Mode}$
$\text{Median}=\text{Mode}+\frac{2}{3}(\text{Mean - Mode})$

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Question 1151 Mark

Which one of the following is not a measure of central tendency?

Answer

Variance

Solution:
Median and Mode are the most common measures of central tendency.
These may be considered depending on the type of data and data distribution
Variance measures how far the data set is spread out and is not a measure of central tendency.

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Question 1161 Mark

In a histogram, which of the following is proportional to the frequency of the corresponding class?

Answer

Area of the rectangle.

Solution:
In, Histogram each rectangle is drawn, where width equivalent to class interval and height equivalent to the frequency of the class.
Since class interval are same across the distribution table, area of the rectangle is corresponding to frequency or height of the rectangle.

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Question 1171 Mark

The width of each of five continuous classes in a frequency distribution is 5 and the lower-class limit of the lowest class is 10. The upper-class limit of the highest class is:

Answer

35

Solution:
The classes are 10-15, 15-20, 20-25, 25-30, 30-35 so that upper limit of the highest class is 35.

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Question 1181 Mark

A data is such that its maximum value is 75 and range is 20, then the minimum value is:

Answer

55

Solution:
Difference between the maximum and minimum value of the observations is called as range.
Let, minimum value be 'x'
75 - x = 20
So, x = 55

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Question 1191 Mark

Write the correct answer in the following:
The class marks of a frequency distribution are given as follows: 15, 20, 25, .... The class corresponding to the class mark 20 is:

Answer

17.5 - 22.5

Solution:
The class mark are 15, 20, 25, ….
The size of each class interval is 25 - 20 = 20 - 15 = 5
Hence, the class interval corresponding to the class mark 20 is,
(20 - 2.5) - (20 + 2.5) i.e., 17.5 - 22.5.
So, (b) is the correct answer.

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Question 1201 Mark

The graph given below shows the frequency distribution of the age of 22 teachers in a school. The number of teachers whose age is less than 40 years is:

Answer

15

Solution:
Add the values corresponding to the height of the bar before 40.
6 + 3 + 4 + 2 = 15.

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Question 1211 Mark

The mean of prime numbers between 30 and 40 is:

Answer

34

Solution:
Prime numbers between 30 and 40 are 31 and 37.
Mean $=\frac{31+37}{2}$
Mean $=34$

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Question 1221 Mark

Mean – Mode =

Answer

3(Mean – Median)

Solution:
Since, Mean – Mode + Mean + 2 Mode = 3 Median
⇒ Mean – Mode = 3 Median – Mean – 2 Mode
= 3 Median – Mean – 2 (3 Median – 2 Mean)
= 3 Median – Mean – 6 Median + 4 Mean
= 3 Mean – 3 Median
= 3 (Mean – Median)

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Question 1231 Mark

To analyse the election results, the data is collected from a newspaper. The data thus collected is known as:

Answer

Secondary data.

Solution:
Secondary data is the readily available data collected by someone else & published in newspapers or journals etc.

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Question 1241 Mark

Median of the following numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3 is:

Answer

6

Solution:
The observations in ascending order can be written as:
3, 4, 4, 5, 6, 7, 7, 7, 12
Median $=\Big(\frac{9+1}{2}\Big)\text{th}$ term = 5th term = 6

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Question 1251 Mark

The following marks were obtained by the students in a test:
81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62
The range of the marks is.

Answer

33

Solution:
Range of observations = Highest observation - Lowest observation
= 95 - 62 = 33

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Question 1261 Mark

Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is:

Answer

15

Solution:
We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20
From above, we see that 15 occurs most frequently i.e., 5 times.
Hence, the mode of the given data is 15.

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Question 1271 Mark

What is class size of interval 10, 12, 14, 16, 18?

Answer

2

Solution:
Difference between the successive values of the class is called the class size.

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Question 1281 Mark

In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is:

Answer

7

Solution:
Given,
Mid value of the class = 10
Width of each class = 6
Now,
Let the lower limit be x.
We know,
Upper limit = Lower limit + class size
= x + 6
Also,
Mid value $=\frac{\text{x}+\text{x}+6}{2}$
$=\frac{2\text{x}+6}{2}=\text{x}+3$
$\Rightarrow\text{x}+3=10$
$\Rightarrow\text{x}=7$
Thus, the lower limit is 7.

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Question 1291 Mark

If x is the mean of $x_1, x_2, ..... x_n, y$ is the mean of $y_1, y_2, ... x_n,....y_n,$ then $z$ the mean of $x_1, x_2, ....x_{n,} y_1, y_2, ....yn$ is equal to:

Answer

Since $x¯$ and $y¯$ are two numbers, though being means, their arithmetic mean is given by: $\text{z}=\frac{\text{x and y}}{2}$

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Question 1301 Mark

The median and mode of distribution are 20 and 18, then the mean is:

Answer

21

Solution:
3 Median = 2 × Mean + Mode
⇒ 3 × 20 = 2 × Mean + 18
⇒ 2 × Mean = 60 - 18 = 42
⇒ Mean = 21

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Question 1311 Mark

More than’ cumulative frequency table for a given data is as follows: Then, the frequency of the class interval 70-80 is:

Marks	More than 89	More than 79	More than 69	More than 59
Cumulative frequency	8	18	30	65

Answer

12

Solution:
A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Subtract cumulative frequency of class more than 70 from the next cummulative Frequency of class more than 69.
30 − 18 = 12

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Question 1321 Mark

Write the correct answer in the following:
Median of the following numbers, 4, 4, 5, 7, 6, 7, 7, 12, 3 is:

Answer

6

Solution:
First, we arrange the given numbers in ascending order is,
3, 4, 4, 5, 6, 7, 7, 7 and 12
Here, n = 9
Since, n is odd, so we use the formula for median,
Now, Median $=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}$
$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{observation}$ [Put n = 9]
$=\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation}$
$=5^{\text{th}}\text{observation}=6$

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Question 1331 Mark

The marks obtained by 17 students in a mathematics test (out of 100) are given below:
91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49.
Find the range of the data.

Answer

54

Solution:
Highest Marks = 100
Lowest Marks = 46
Range of data = 100 - 46 = 54

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Question 1341 Mark

In a frequency distribution, the mid-value of a class is 60.5 and the width of the class is 10. The lower limit of the class is:

Answer

55.5

Solution:
upper limit - lower limit = class width
upper limit - lower limit = 10
$\frac{\text{(upper limit + lower limit)}}{2}=\text{mid}-\text{value}$
upper limit + lower limit = 2 × 60.5
upper limit + lower limit = 121
By solving the above two equations, we get
upper limit = 65.5
Lower limit = 55.5

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Question 1351 Mark

The range of the data
12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is:

Answer

26

Solution:
We have:
Maximum value = 32
Minimum value = 6
We know:
Range = Maximum value - minimum value
= 32 - 6
= 26

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Question 1361 Mark

Find the median of the given data: 7, 8, 7, 7, 9, 10, 13.

Answer

8

Solution:
Arrange the given data in ascending order.
7, 7, 7, 8, 9, 10, 13
For odd number (n) of observation median $=\Big[\frac{(\text{n+1)}}{2}\Big]\text{th}$ term,
Here n = 7 so median $=\Big[\frac{{(7+1)}}{2}\Big]\text{th}$ term = 4th term that is 8,
Hence median = 8

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Question 1371 Mark

The median of the numbers $4, 4, 5, 7, 6, 7, 7, 12, 3$ is:

Answer

Arranging the points in an ascending order,
We have:
$3, 4, 4, 5, 6, 7, 7, 7, 12$
Here$, n = 9,$ Which is odd
$\therefore\ $median score = value of $\frac{1}{2}(9+1)^\text{th}$ terms
$=$ value of $\Big(\frac{1}{2}\times10\Big)^\text{th}$ term
$=$ value of $5^{th}$ term
$= 6$

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Question 1381 Mark

If $x$ is the mean of $x_1, x_2, ..., X_n$ then for $\text{a}\neq0,$ the mean of $\text{ax}_1,\ \text{ax}_2,\ ...,\ \text{ax}_\text{n},\ \frac{\text{x}_1}{\text{a}},\ \frac{\text{x}_2}{\text{a}},\ ...,\ \frac{\text{x}_\text{n}}{\text{a}}$ is:

Answer

Mean of $ax_1, ax_2, ..., ax_n$, is $\text{a}\bar{\text{x}}$
Mean of $\frac{\text{x}_1}{\text{a}},\ \frac{\text{x}_2}{\text{a}},\ ...,\ \frac{\text{x}_\text{n}}{\text{a}}$ is $\frac{1}{\text{a}}\bar{\text{x}}$
so the their mean is $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}.$

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Question 1391 Mark

In the class intervals 10-20, 20-30, the number 20 is included in:

Answer

20-30

Solution:
This is the continuous from of frequency distribution. Here, the upper limit of each class is excluded, while the lower limit is included. So, the number 20 is included in the class interval 20-30.

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Question 1401 Mark

A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data :
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236. The frequency of the class 310-330 is:

Answer

6

Solution:
The class interval is 250 - 270, 270 not included.
It means that the class is continuous.
Also, the data can be tabulated as follows:

Thus, frequency of the class 310 - 330 is 6 as can be seen from the table.

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Question 1411 Mark

A, B, C are three sets of values of x:

A: 2, 3, 7, 1, 3, 2, 3

B: 7, 5, 9, 12, 5, 3, 8

C: 4, 4, 11, 7, 2, 3, 4

Which one of the following statements is correct?

Answer

Mean, Median and Mode of A are equal.

Solution:

A: 1, 2, 2, 3, 3, 3, 7

B: 3, 5, 5, 7, 8, 9, 12

C: 2, 3, 4, 4, 4, 7, 11

$\text{Mean of A}=\frac{1+2+2+3+3+3+7}{7}$
$=\frac{21}{7}=3$
$\text{Mean of B}=\frac{3+5+5+7+8+9+12}{7}$
$=\frac{49}{7}=7$
$\text{Mean of C}=\frac{2+3+4+4+4+7+11}{7}$
$=\frac{35}{7}=5$
Mode of A = 3; Median of A = 3
Mode of B = 5; Median of B = 7
Mode of C = 4; Median of C = 4
(Mean of A = 3) $\neq$ (Mode of C = 4)
(Mean of C = 5) $\neq$ (Median of B = 4)
(Median of B = 7) $\neq$ (Mode of A = 3)
Mean of A = 3, Mode of A = 3, Median of A = 3

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Question 1421 Mark

The mean of $n$ observations is $x$ If the first item is increased by $1,$ second by $2,$ third by $3$ and so on, then the new mean is:

Answer

If the first item is increased by $1, $second by $2,$ third by $3$ and so on,
So the new sequence thus formed is $ x_1 + 1, x_2 + 2, x_3 + 3 ...... x_n + n$
So the new mean is increased by $\frac{\text{n}+1}{2}$
So the new mean is $\text{x}+\frac{\text{n}+1}{2}$
Where $x$ is the mean of first $n$ numbers.

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Question 1431 Mark

The given data is 3, 5, 6, 7, 5, 4, 7, 5, 6, x, 8 and 7. Then the value of x for which the mode of the above data will be 7, is:

Answer

7

Solution:
The mode in a list of numbers refers to the integers that occur most number of times.
In the given list 5 occur three times and 7 occurs three times.
For 7 to be mode of the list it should occur more number of times than any other number
So x should be 7.
Mode = 7

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Question 1441 Mark

Write the correct answer in the following:
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is:

Answer

56.5

Solution:
Given, n = 50, then mean $\bar{\text{x}}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$
Then, $\bar{\text{x}}=\frac{1}{50}\times\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}$
$\Rightarrow\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}=50\bar{\text{x}}$
Now, subtract each observation from 53, we get a new mean say $\bar{\text{x}}_\text{new}$
$\therefore\ \bar{\text{x}}_\text{new}=\frac{(-\text{x}_1+53)+(-\text{x}_2+53)\ +\ .....\ +(-\text{x}_{50}+53)}{50}$
$\Rightarrow-3.5=\frac{-(\text{x}_1+\text{x}_2\ +\ .....\ +\text{x}_{50})+(53+53\ +\ ....\ +50\text{times})}{50}$
$\Rightarrow-3.5\times50=(-\text{x}_1+\text{x}_2+\ ....\ +\text{x}_{50})+53\times50$
$\therefore$ Mean of 50 observation $=\frac{1}{50}\sum\limits^{50}_{\text{i}=1}\text{x}_{\text{i}}$
$=\frac{1}{50}\times2852=56.5$ $\Bigg[\because\text{ mean}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}\Bigg]$
Hence, the mean of given number is 56.5

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Question 1451 Mark

If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observation is:

Answer

$11\frac{1}{3}$

Solution:
Mean of 5 observations = 9
$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observation}}$
$\Rightarrow9=\frac{\text{x}+\text{x}+3+\text{x}+5+\text{x}+7+\text{x}+10}{5}$
$\Rightarrow9=\frac{5\text{x}+18}{5}$
$\Rightarrow45=5\text{x}+18$
$\Rightarrow5\text{x}=20$
$\Rightarrow\text{x}=4$
So, the mean of the last three observations
$=\frac{\text{x}+5+\text{x}+7+\text{x}+10}{3}$
$=\frac{4+5+4+7+4+10}{3}$
$=\frac{34}{3}$
$=11\frac{1}{3}$

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Question 1461 Mark

Which one of the following is not a measure of central value?

Answer

Range.

Solution:
The difference between the highest value and the lowest value in the data set is called Range.

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Question 1471 Mark

Write the correct answer in the following:
Let m be the mid-point and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:

Answer

2m - l

Solution:
Let x and y be the lower and upper class limit of a continuous frequency distribution.
Now, mid-point of a class $=\frac{(\text{x}+\text{y})}{2}=\text{m}$ [given]
⇒ x + y = 2m = x + l = 2m
$\big[\therefore$ y = l = upper class limit (given)$\big]$
⇒ x = 2m - l
Hence, the lower class limit of the class is 2m - l.

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Question 1481 Mark

The width of each of nine classes in a frequency distribution is $2.5$ and the lower class boundary of the lowest class $10.6$. Then the upper class boundary of the highest class is:

Answer

Number of classes $= 9$
Lower limit of the lowest class $= 10.6$
Width of each class $= 2.5$
So, Upper limit of the lowest class $= 10.6 + 2.5 = 13.1$
Now, Upper limit of the lowest class $+$ Width of each class $=$ Upper limit of the next class
Thus, we have
Upper limit of the lowest class $ +\ 8\ \times$ width of each class $=$ Upper limit of the highest $(9^{th})$ class
Upper limit of the highest $(9^{th})$ class $= 13.1 + 8 \times 2.5 = 33.1$

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Question 1491 Mark

Let $\bar{\text{x}}$ be the mean of $x_1, x_2, ..., x_n$ and $\bar{\text{y}}$ be the mean of $y_1, y_2, ..., y_n$. If $\bar{\text{z}}$ is the mean of $x_1, x_2, ..., x_n, y_1, y_2, ..., y_n$ then $\bar{\text{z}}=$ ?

Answer

Since $\bar{\text{z}}$ is the mean of $x_1, x_2, ..., x_n, y_1, y_2, ..., y_n,$
$\bar{\text{z}}=\frac{(\text{x}_1+\text{x}_2+_{\dots}+\text{x}_\text{n})+(\text{y}_1+\text{y}_2+{\dots}\text{y}_\text{n})}{2\text{n}}$
Since $\bar{\text{x}}$ is the mean of $x_1, x_2, ..., x_n,$
$\bar{\text{x}}=\frac{\text{x}_1+\text{x}_2+_{\dots}+\text{x}_\text{n}}{\text{n}}$
$\Rightarrow\text{x}_1+\text{x}_2+_{\dots}+\text{x}_\text{n}=\text{n}\bar{\text{x}}$
Since $\bar{\text{y}}$ is the mean of $y_1, y_2, ..., y_n,$
$\bar{\text{y}}=\frac{\text{y}_1+\text{y}_2+_{\dots}+\text{y}_\text{n}}{\text{n}}$
$\Rightarrow\text{y}_1+\text{y}_2+_{\dots}+\text{y}_\text{n}=\text{n}\bar{\text{y}}$
so,
$\bar{\text{z}}=\frac{\text{n}\bar{\text{x}}}{2\text{n}}+\frac{\text{n}\bar{\text{y}}}{2\text{n}}=\frac{\text{n}\bar{\text{x}}+\text{n}\bar{\text{y}}}{2\text{n}}$
$=\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$

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Question 1501 Mark

Let $\bar{\text{x}}$ be thae mean of $x_1, x_2, ......, x_n,$ and $\bar{\text{y}}$ the mean of $y_1 , y_2 , ... , y_n$ . If $\bar{\text{z}}$ is the mean of $x_1 , x_2 , ... , x_n , y_1 , y_2, ... , yn,$ then $\bar{\text{z}}$ is equal to:

Answer

We have $\bar{\text{x}}$ is the mean of $x_1, x_2, ......, x_n,$ and $\bar{\text{y}}$ is the mean of $y_1 , y_2 , ... , y_n$
So, $\bar{\text{x}}=\frac{(\text{x}_1+\text{x}_2+\text{x}_3+\ ....\ +\text{x}_\text{n})}{\text{n}}$
$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+\ .....\ +\text{x}_\text{n}=\text{n}\bar{\text{x}}$
And $\bar{\text{y}}=\frac{(\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n})}{\text{n}}$
$\Rightarrow\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n}=\text{n}\bar{\text{y}}$
If $\bar{\text{z}}$ is the mean of $x_1, x_2, ...., y_1 , y_2 , ... , y_n$ then$,$
_{$\bar{\text{z}}=\frac{\text{n}\bar{\text{x}}+\text{n}\bar{\text{y}}}{\text{n}+\text{n}}=\frac{\text{n}(\bar{\text{x}}+\bar{\text{y}})}{2\text{n}}=\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$}

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