50 questions · timed · auto-graded
Solution:
The difference between the highest and lowest value of observations is called 'Range' of observations.
Solution:
In a frequency distribution, ogives are graphical representation of cumulative frequency.
Solution:
Given that the mean of a, b, c, d and e is 28. They are 5 in numbers.
Hence, we have
$\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$
$\Rightarrow\frac{(\text{a}+\text{c}+\text{e})+(\text{b}+\text{d})}{5}=28$
$\Rightarrow\frac{(\text{a}+\text{c}+\text{e})}{5}+\frac{(\text{b}+\text{d})}{5}=28$
But, it is given that the mean of a, c and e is 24. Hence, we have
$\Rightarrow\frac{(\text{a}+\text{c}+\text{e})}{5}=24$
$\Rightarrow\text{a}+\text{c}+\text{e}=72$
Then We have
$\frac{72}{5}+\frac{(\text{b}+\text{d})}{5}=28$
$\Rightarrow\frac{\text{b}+\text{d}}{5}=28-\frac{72}{5}$
$\Rightarrow\frac{\text{b}+\text{d}}{5}=28-14.4$
$\Rightarrow\frac{\text{b}+\text{d}}{5}=13.6$
$\Rightarrow\text{b}+\text{d}=68$
$\Rightarrow\frac{\text{b}+\text{d}}{2}=\frac{68}{2}$
$\Rightarrow\frac{\text{b}+\text{d}}{2}=34$
Hence, the mean of b and d is 34.
Solution:
The given class intervals are 10-20, 20-30. In these class intervals the value 20 is lies in the class interval 20-30.Hence, the correct choice is (b).
Solution:
Add the values corresponding to the height of the bar from 60 to 100
10 + 5 + 3 + 3 = 21
54
Solution:
Arranging the points in an ascending order,
We have:
22, 34, 39, 45, 54, 56, 68, 78, 84
Here, n = 10, Which is even
$\therefore\ $median = mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{10}{2}+1\Big)^\text{th}$ terms
= mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{12}{2}\Big)^\text{th}$ term
= mean of 5th and 6th terms
$=\frac{1}{2}(54+54)$
$\frac{1}{2}\times108$
$=54$
Solution:
If $\overline{\text{X}}$ be the mean of the n observations q X1, ......Xn Then we have
$\overline{\text{X}}=\frac{1}{\text{n}}\sum\limits_{\text{i}=1}^{\text{n}}\text{X}_\text{i}$
$\Rightarrow\sum\limits_{\text{i}=1}^{\text{n}}\text{X}_\text{i}=\overline{\text{X}}$
Let $\overline{\text{X}}$ be the mean of n Values Xi,.....Xn. so we have
$\overline{\text{X}}=\frac{1}{\text{n}}\sum\limits_{\text{i}=1}^{\text{n}}\text{X}_\text{i}$
$\Rightarrow\sum\limits_{\text{i}=1}^{\text{n}}\text{X}_\text{i}=\text{n}\overline{\text{X}}$
The sum of the deviations of n values Xi,......Xn from their mean $\overline{\text{X}}$ is
$(\text{x}_1-\overline{\text{X}})+(\text{x}_2-\overline{\text{X}})+.....+(\text{x}_\text{n}-\overline{\text{X}})$
$=\sum\limits_{\text{i}=1}^{\text{n}}(\text{x}_\text{i}-\overline{\text{X}})$
$=\sum\limits_{\text{i}=1}^{\text{n}}\text{x}_\text{i}-\sum\limits_{\text{i}=1}^{\text{n}}\overline{\text{X}}$
$=\text{n}\overline{\text{X}}-\text{n}\overline{\text{X}}$
$=0$
Solution:
Let the numbers be x1, x2 ....., xn
Now the new numbers after decrasing every number by 8: (x1 - 8), (x2 - 8).... (xn - 8)
New mean $=\frac{(\text{x}_1 - 8)+(\text{x}_2 - 8)+...+(\text{x}_\text{n} - 8)}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+...+\text{x}_\text{n}-\text{8n}}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+...+\text{x}_\text{n}}{\text{n}}-8$
New mean = mean -8
Hence, mean is decreased by 8.
Solution:
Mean of 100 items = 64
Sum of 100 items = 64 × 100 = 6400
Correct sum = (6400 + 36 + 90 - 26 - 9) = 6491
Correct mean $=\frac{6491}{100}=64.91$
Solution:
The given frequency varies from 14 to 112.
So the class intervals are:
13-22, 23-32, 33-42, 43-52, 53-62, 63-72, 73-82, 83-92, 93-102, 103-112.
Number of class interval = 10.
Solution:
The class width is the difference between the upper- or lower-class limits of consecutive classes.
In this case, class width equals to the difference between the lower limits of the first two classes.
Let, W be the class width
W = 21 - 1 = 20
So class width is 20
Solution:
Mid value $=\frac{\text{Lower}\ \text{limit}+\text{Upper}\ \text{limit}}{2}$
$\Rightarrow\text{m}=2\text{m}-\text{L}$
$\therefore$ Upper class boundry of the class = 2m - L.
Solution:
Class size is the difference between two consecutive values of the class mark.
Here, the difference between two consecutive class mark is 6.
i.e., 34 - 28 = 6
Solution:
Since, 3 Median = 2 Mean + Mode
$\therefore$ 3 × 33 = 2 Mean + 45
⇒ 2 Mean = 99 - 45
⇒ 2 Mean = 54
⇒ Mean = 27
Solution:
Let sum of all the 30 observations be x.
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.
$\frac{\text{x}}{30}=12$
$\text{x}=360$
$360-25=335$
Solution:
In a histogram the class intervals or the groups are taken along the horizontal axis or X−axis.
| Class | 10-20 | 20-30 | 30-40 |
| Cumulative frequency | 5 | 14 | 25 |
Solution:
A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Subtract the previous cumulative frequency (c.f.) from the cumulative frequency of the current class.
So frequency of the class interval 20 - 30 is 14 - 5 = 9
15
Solution:
Arranging the marks in an ascending order,
We have:
14, 14, 14, 14, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20
Clearly, 15 occurs maximum number of times.
Hence, mode = 15
Solution:
We know that algebraic sun of deviations from mean is zero.
$\sum\limits_{\text{a}=1}^\text{b} (\text{x}_\text{t}-\bar{\text{x}})=(\text{x}_1-\bar{\text{x}})+(\text{x}_2-\bar{\text{x}})+(\text{x}_3-\bar{\text{x}})+\ ...\ +(\text{x}_\text{n}-\bar{\text{x}})$
$= (\text{x}_1+\text{x}_2+\text{x}_3+... +\text{x}_\text{n})- \text{n}\bar{\text{x}}$
$\Rightarrow\sum\limits_{\text{t}-1}^\text{b}\text{x}_\text{i}-\text{n}\bar{\text{x}}=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}=0$ $\bigg[\because\sum\limits_{\text{i}=1}^\text{n} \text{x}_\text{i}=\text{n}\bar{\text{x}}\bigg]$
Hence, (b) is correct answer.
Solution:
The mode in a list of numbers refers to the integers that occur most number of times.
In the given list both 8 and 9 occur two times.
So the value of x will decide the mode
If x = 8, then the mode will be 8
If x = 9, then the mode will be 9
Hence, the difference between the two modes is 1.
Solution:
The most common measures of central tendency are mean, median and mode.
Standard deviation is a measure of the dispersion of a set of data from its mean. It is calculated as the square root of variance.
Hence standard deviation is not a measure of central tendency.
Solution:
Most Frequent value is called mode.
| Marks | Less than 10 | Less than 20 | Less than 300 | Less than 40 |
| Cumulative frequency | 3 | 17 | 37 | 92 |
Solution:
A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Less than 30 has the class interval 20-30. Frequency of this class interval will be corresponding to.
| Marks | Cumulative frequency | Class | Frequency |
| Less than 10 | 3 | 1-10 | 3 |
| Less than 20 | 17 | 10-20 | 14 |
| Less than 30 | 37 | 20-30 | 20 |
| Less than 40 | 92 | 30-40 | 55 |
Solution:
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.
$\frac{4+7+\text{x}+8+9+10}{5}=8$
$\frac{(38+\text{x})}{6}=8$
$\text{x}=48-38=10$
Solution:
Given $\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{2}\bigg)^2=(\text{M})^2$
Taking cube on both sides
$\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{2}\bigg)^3=(\text{M})^3$
$\bigg(\text{x}+{\frac{1}{\text{x}}}\bigg)^3=(2\text{M})^3$
$\bigg(\text{x}^2+3\text{x}\times{\frac{1}{\text{x}}}\Big(\text{x}+\frac{1}{\text{x}}\Big)+\frac{1}{\text{x}^3}\bigg)=(2\text{M})^3$
$\bigg(\text{x}^3+{\frac{1}{\text{x}^3}}\bigg)={8\text{M}^3-3}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Divide by 2 on both sides to get mean
$\frac{\bigg(\text{x}^3+{\frac{1}{\text{x}^3}}\bigg)}{2}={4\text{M}^3-\frac{3}{2}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\frac{\bigg(\text{x}^3+{\frac{1}{\text{x}^3}}\bigg)}{2}={4\text{M}^3}-{3\text{M}}$
Solution:
We are given frequency distribution 15, 20, 25, 30 ....
Class size = 20 - 15 = 5
Class marks = 20
Lower limit $=\Big(20-\frac{5}{2}\Big)$
$=\frac{35}{2}=17.5$
Upper limit $=\Big(20+\frac{5}{2}\Big)$
$=\frac{45}{2}=22.5$
Thus, the required class is 17.5-22.5.
| Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 180 | 150 | 160 | 170 | 150 | 190 |
Solution:
$\frac{150}{1000}=\frac{3}{20}$
Solution:
Class mark $=\frac{\text{Upper}\ \text{limit}+\text{Lower}\ \text{limit}}{2}$
$=\frac{120+100}{2}$
$=110$
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.
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).
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.
$\frac{32+34+36}{3}=34$
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).
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
$22=\frac{(\text{x}+1)+(\text{x}+3)}{2}$
$44=2\text{x}+4$
$40=2\text{x}$
$\text{x}=20$
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
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.
Solution:
Frequency polygon is the plot of frequencies vs. the mid values of the classes.
Solution:
Class mark $=\frac{130+150}{2}=\frac{280}{2}=140$
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
Solution:
Difference between maximum and minimum value of observation is called as range.
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 $\frac{1970}{50}=39.4$
Solution:
Since, 3 Median = Mode + 2 Mean
$\Rightarrow\text{Median}=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}$
$=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}-\frac{2}{3}\text{Mode}+\frac{2}{3}\text{Mode}$
$\text{Median}=\text{Mode}+\frac{2}{3}(\text{Mean - Mode})$
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.
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.
| Class intervai | 5-10 | 10-15 | 15-25 | 25-45 | 45-75 |
| Frequency | 6 | 12 | 10 | 8 | 15 |
Solution:
The adjusted frequency for the class 25 - 45 is
$= \frac{\text{Frequency of the class }}{\text{Class width }}\times \text{Minimum width }=\frac{8}{20} \times5=2$
Solution:
Median = 4
Mode = 2
$\text{Mean}=\frac{2+2+4+5+12}{5}=\frac{25}{5}=5$
Hence, (Mean = 5) > (Mode = 2)
Solution:
Abcissac are the class marks of the classes.
Solution:
Class mark $=\frac{\text{Upper limit + lower limit}}{2}=\frac{120+100}{2}=110$
Solution:
The empirical Relation between mean, median and mode is:
Mode = 3 Median - 2 mean.
| Wages(in Rs) | No of workers |
| More than 140 | 12 |
| More than 130 | 27 |
| More than 120 | 60 |
| More than 110 | 105 |
| More than 100 | 124 |
| More than 90 | 141 |
| More than 80 | 150 |
Solution:
| Wages(in Rs) | No of workers |
| 140–150 | 12 |
| 130–140 | 15 |
| 120–130 | 33 |
| 110–120 | 45 |
| 100–90 | 19 |
| 90–100 | 17 |
| 80–90 | 9 |
Therefore, the number of workers having a wage range (in Rs.) 110-120 is 45
Solution:
The given data is 7, 5, 13, x and 9. They are 5 in numbers.
The mean is
$\frac{7+5+13+\text{x}+9}{5}=\frac{34+\text{x}}{5}$But, it is given that the mean is 10. Hence, we have
$\frac{34+\text{x}}{5}=10$
⇒ 34 + x = 50
⇒ x = 50 - 34
⇒ x = 16