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CBSE Class 9 Mathemaics
Important Questions
Chapter 14
Statistics
2 Marks Quetions
1. The mean of 10 numbers is 20, If 5 is subtracted from every number, what will be the
new mean.
Ans. 15
2. Find the mean of first 10 even natural no.
Ans. 11
3. Calculate the mean for the following distribution.
Ans. 7.025
4. Find the median of 37, 31, 42, 43, 46, 25, 39, 45, 32
Ans. 39
5. Find the mode of following series.
25, 23, 22, 22, 24, 27, 27, 25, 23, 22, 26, 32
Ans. 22
6. If the median of a series of data is 3 and mean is 2 then find the mode.
Ans. 5
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7. If the mean of 5 observation x, x + 4, x + 8, x + 12, x + 16 is 13, find the mean of the
observations?
Ans.
The given set of 5 observations are 5, 9, 13, 17, 21
8. The class marks of the observations are 17, 21, 25, 29, 33, 37, 41, 45. Find the class
intervals.
Ans. Class marks are 17, 21, 25, 29, 33, 37, 41 and 45
Class size = 21 – 17 = 25 – 21 = 4 and Half of class size =
So, Class intervals are:
9. The value of up to 15 decimal places is: 3. 419078023195679
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(i) List the digits from 0 to 9 & make frequency distributions of the digit after the
decimal points.
(ii) What are the most * the least frequently occurring digits?
Ans. (i) Frequency distribution table
(ii) Most frequency occurring digits = 9 & least frequently occurring digits = 2, 3, 4, 5, 6, 8
10. A random survey of the number of children of various age grout playing in the park
was found:
Draw a histogram to represent the data above?
Ans. Since the class intervals are not of equal width; we calculate the adjusted frequencies
[AF] for histogram. Minimum class size [CS] = 1
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12. Find the mode of the following data: 14, 25, 14, 14, 25, 24, 20, 28, 18, 20.
Ans. Arranging the given data in ascending order: 14, 14, 14, 18, 20, 20, 24, 25, 25, 28
We observe that the value 14 occurs most frequently i.e. 3 times in the given set of the
observations.
Mode is 14
13. Find the median of 5, 7, 10, 9, 5, 12, 15, 12, 18, 20. If 9 is replaced by 14, what will be
the new median.
Ans. The given observations arranged in ascending order: 5, 5, 7, 9, 10, 12, 15, 18, 20
Here, n = 10 [even number]
So,
When 9 is replaced by 14, we get 5, 5, 7, 10, 12, 12, 14, 15, 18, 20
Now 5
th observation = 6
th observation = 12
So, New Median =
14. The average mark of boys in an examination is 68 & that of girls in 89. If the average
mark of all candidates in that examination is 80, find the ratio of the no. of boys to the
number of girls that appeared in the examinations.
Ans. Let number of boys be x & that of girls be 4.
Total marks of boys = = 68x
& Total marks of girls = = 89y
Hence total marks for boys & girls = 68x + 89y ----------- (1)
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Hence, the excluded number is 60
16. The median of the observation 11, 12, 14, 18, x+2, x+4, 30, 32, 35, 41, arranged in
ascending order is 24. find the value of x.
Ans. Number of observation, n=10
Since n is even,
Hence, x =21
17. Find the median of the following data: 25, 34, 31, 23, 22, 26, 35, 28, 20, 32,
Ans. Arranging the data in ascending order, we get
20, 22, 23, 25, 26, 28, 31, 32, 34, 35
Hence, the no. of observation n=10 (even)
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Form a cumulative frequency table with class interval of length 20.
Ans.
22. For the following data, draw a histogram and a frequency polygon:
Ans.
23. If is the mean of observation , then prove that the mean of
is where a is any real number.
Ans. We have
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24. The mean of 16 numbers is 8. If 2 is added to every number, what will be new mean?
Ans.
= 128 .....(i)
New numbers are
Be the mean of new numbers. Then,
[Using (i)]
25. Calculate the mean from the given data:
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Ans.
26. The following table gives the mark scored by 50 students in an entrance
examination:
From this table find:
(i) the less than series and
(ii) the more than series.
Ans. (i) Less than cumulative frequency table.
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(ii) More than cumulative frequency table.
27. Find the sum of the deviations of the various values 3, 4, 6, 8, 14 from their mean.
Ans. Recall that the deviations of the values x1
, x2
, x3 ......... xn about A are
X1
,-A, x2
-A, x3
-A .......... Xn–A.
Let be the deviations of the values 3, 4, 6, 8, 14. Then,
=
Now sum of the deviations of the values 3, 4, 6, 8, and 14. From their mean = 7 is given by
(3-7) + (4-7) + (6-7) + (8-7) + (14-4) = -4-3-1+1+7 =0
28. The mean of 40 observations was 200. It was detected on rechecking that the value of
65 was wrongly copied as 25 for computation of mean. Find the correct mean.
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Ans. Hence n=40 =200
So =
=8000
Incorrect value of
Now correct value of incorrect value of
=8000-25+65
=8040
=201
29. It is the mean of n observation ,
then prove that
i.e. the algebraic sum of deviations from mean is zero.
Ans. We have
Now,
=0