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NCERT Class 10 Mathematics Statistics Formative Assessment CBSE Board Sample Problems Part 2
Multiple-choice Questions
(3) While computing mean of grouped data, we assume that the frequencies are
(a) Centred at the upper limits of the classes
(b) Centred at the lower limits of the classes
(c) Centred at the class-marks of the classes
(d) Evenly distributed over all the classes
Solution (c)
Question 4
Which of the following is the measure of central tendency?
(a) Mean
(b) Mode
(c) Median
(d) Range
Solution (a) , (b) , (c)
Question 5
For drawing a frequency polygon of a continuous frequency distribution, we plot the Points whose ordinates are the frequencies of the respective classes and abscissae are respectively:
(a) Class marks of the classes
(b) Upper limits of preceding classes
(c) Lower limits of the classes
(d) Upper limits of the classes
Solution (a)
Question 6
Find the mean of 32 numbers given mean often of them is 12 and the mean of other 20 is 9. And last 2 number is 10
(a) 10
(b) 12
(c) 13
(d) 14
Solution (a)
Mean of 10 number = 12
Sum of these 10 numbers = 120
Mean of 20 number = 9
Sum of these 20 numbers = 180
Mean of 2 number = 10
Sum of these 2 numbers = 20
Mean of 32 number
Question 7
The median and mean of the first 10 natural numbers.
(a) 5.5,5.5
(b) 5.5,6
(c) 5,6
(d) None of these
Solution (a)
Mean = 5.5
Median is mean of 5 and 6th term, So 5
Question 8
Anand says that the median of 3, 14, 19,20, 11 is 19. What doesn՚t the Anand understand about finding the median?
(a) The dataset should be ascending order
(b) Highest no in the dataset is the median
(c) Average of lowest and highest is the median
(d) None of these
Solution (a)
Question 9
The following observations are arranged in ascending order:
If the median is 63, find the value of x.
(a) 62
(b) 64
(c) 60
(d) None of these
Solution (a)
Median is mean of 5 and 6 term
So
Question 10
The mean of 20 observations was 60. It was detected on rechecking that the value of 125 was wrongly copied as 25 for computation of mean. Find the correct mean
(a) 67
(b) 66
(c) 65
(d) None of the above
Solution
Let x be the sum of observation of 19 numbers leaving 125,
Then
Now
Subtracting
Question 11
Compute the Median for the given data
Class interval | 100 - 110 | 110 - 120 | 120 - 130 | 130 - 140 | 140 - 150 | 150 - 160 |
Frequency | 6 | 35 | 48 | 72 | 100 | 4 |
Solution
First, calculate the Class mark and cumulative frequency of the data
Class interval | 100 - 110 | 110 - 120 | 120 - 130 | 130 - 140 | 140 - 150 | 150 - 160 |
Frequency | 6 | 35 | 48 | 72 | 100 | 4 |
Class Mark | 105 | 115 | 125 | 135 | 145 | 155 |
Cumulative Frequency | 6 | 41 | 89 | 161 | 261 | 266 |
We have N = 266, So , Cumulative frequency first greater than 133, lies in class 130 - 140
So median class is 130 - 140
Now
Median is calculated as
Where
l = lower limit of median class,
n = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming class size to be equal)
Here
Substituting these
Match the column
A histogram | Is the diagram showing a system of connections |
Or interrelations between two or more things by using bars | |
Discontinuous Frequency Distribution | A frequency distribution in which the upper limit of one class coincides from the lower limit of the succeeding class |
Continuous Frequency Distribution | Is the bar graph such that the area over each class interval is proportional to the relative frequency of data within this interval. |
Ogive is the graph of | Is a set of adjacent rectangles whose areas are proportional to the frequencies of a given continuous frequency distribution? |
lower⟋upper limits and cumulative frequency |