CBSE Class 10 Maths HOTs Statistics Set 05

Question. Find the median of 6, 8, 9, 10, 11, 12 and 13.
Answer: Total number of terms \( = 7 \)
The middle term \( = \frac{1}{2}(7 + 1) = 4\text{th} \)
Median = Value of the 4th term = 10.
Hence, the median of the given series is 10.

Question. Find the mean of 21, 22, 23, 24, 25, 26, 27, and 28.
Answer: Total number of terms \( = 8 \)
Median \( = \text{Value of } \frac{1}{2} \left[ \frac{8}{2} \text{th term} + \left( \frac{8}{2} + 1 \right) \text{th term} \right] \)
\( = \text{Value of } \frac{1}{2} \text{[4th term + 5th term]} = \frac{1}{2} [24 + 25] = \frac{49}{2} = 24.5 \)

Question. Find the median of the following distribution :

Answer: We have,

Here, the median class is 400 - 500 as \( \frac{44}{2} \), i.e., 22 belongs to the cumulative frequency of this class interval.
Lower limit of median class = \( l = 400 \)
Width of the class interval = \( h = 100 \)
Total frequency = \( N = 44 \)
Cumulative frequency preceding median class frequency = \( C = 8 \)
Frequency of Median-class = \( f = 20 \)
Median \( = l + h \left( \frac{\frac{N}{2} - C}{f} \right) = 400 + 100 \left( \frac{\frac{44}{2} - 8}{20} \right) = 400 + 100 \left( \frac{22 - 8}{20} \right) = 400 + 100 \left( \frac{14}{20} \right) \)
\( = 400 + 70 = 470 \)
Hence, the median of the given frequency distribution is 470.

Question. Compute the median from the marks obtained by the students of class X.

Answer: First we shall convert the given inclusive series into an exclusive (continuous) series.
Adjustment factor \( = \frac{1}{2} \) [Lower limit of second class - Upper limit of first class]
\( = \frac{1}{2} [50 - 49] = \frac{1}{2} = 0.5 \)
For the conversion of the series we subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Since, \( \frac{100}{2} \) belongs to the cumulative frequency (65) of the class interval 69.5 - 79.5, therefore 69.5 - 79.5 is the median class.
Lower limit of the median class \( = l = 69.5 \)
Width of the class interval \( = h = 10 \)
Total frequency \( = N = 100 \)
Cumulative frequency preceding median class frequency \( = C = 35 \)
Frequency of the median class \( = f = 30 \)
Median \( = l + h \left( \frac{\frac{N}{2} - C}{f} \right) = 69.5 + 10 \left( \frac{\frac{100}{2} - 35}{30} \right) \)
\( = 69.5 + 10 \left( \frac{50 - 35}{30} \right) = 69.5 + 10 \left( \frac{15}{30} \right) = 69.5 + 5 = 74.5 \)
Hence, the median of given frequency distribution is 74.50.

Question. An incomplete frequency distribution is given as follows :

Given that the median value is 46, determine the missing frequencies using the median formula.
Answer: Let the frequency of the class 30 - 40 be \( f_1 \) and that of 50 - 60 be \( f_2 \).

From the last item of the third column, we have
\( 150 + f_1 + f_2 = 229 \)
\( \implies \) \( f_1 + f_2 = 229 - 150 \)
\( \implies \) \( f_1 + f_2 = 79 \) ... (1)
Since, the median is given to be 46, the class (40 - 50) is median class.
Therefore, \( l = 40, C = 42 + f_1, N = 229, h = 10 \)
Median \( = 46, f = 65 \)
\( \implies \) Median \( = l + \frac{h \left( \frac{N}{2} - C \right)}{f} = 46 \)
\( 46 = 40 + 10 \left( \frac{\frac{229}{2} - (42 + f_1)}{65} \right) \)
\( \implies \) \( 6 = \frac{10}{65} \left( \frac{229}{2} - 42 - f_1 \right) \)
\( \implies \) \( 6 = \frac{2}{13} \left( \frac{229 - 84 - 2f_1}{2} \right) \)
\( \implies \) \( 78 = 229 - 84 - 2f_1 \)
\( \implies \) \( 2f_1 = 229 - 84 - 78 \)
\( \implies \) \( 2f_1 = 67 \)
\( \implies \) \( f_1 = \frac{67}{2} = 33.5 \approx 34 \)
Putting the value of \( f_1 \) in (1), we have
\( 34 + f_2 = 79 \)
\( \implies \) \( f_2 = 45 \)
Hence, \( f_1 = 34 \) and \( f_2 = 45 \).

Question. Recast the following cumulative table in the form of an ordinary frequency distribution and determine the median

Answer:

Since \( \frac{655}{2} \) belongs to the cumulative frequency 465 of the class interval (10 - 15), therefore (10 - 15) is the median class.
Lower limit of the median class \( = l = 10 \)
Width of the class interval \( = h = 5 \)
Total frequency \( = N = 655 \)
Cumulative frequency preceding median class frequency \( = C = 224 \)
Frequency of median class \( = f = 241 \)
Median \( = l + h \left( \frac{\frac{N}{2} - C}{f} \right) = 10 + 5 \left( \frac{\frac{655}{2} - 224}{241} \right) = 10 + 5 \left( \frac{327.5 - 224}{241} \right) \)
\( = 10 + \frac{5 \times 103.5}{241} = 10 + 2.147 = 12.147 \)
Hence, the median of given frequency distribution is 12.147.

Question. If the median of the distribution given below is 28.5, find the values of \( x \) and \( y \).

Answer: Here, it is given that median in 28.5.
and \( n = \sum f_i = 60 \)
We now prepare the following cumulative frequency table :

Here \( n = 60 \). So, \( \frac{n}{2} = 30 \)
Since the median is given to be 28.5, thus the median class is (20 - 30).
\( \therefore l = 20, h = 10, f = 20 \) and \( cf = 5 + x \)
\( \therefore \text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( \implies \) \( 28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10 \)
\( \implies \) \( 28.5 = 20 + \frac{25 - x}{2} \)
\( \implies \) \( 57 = 40 + 25 - x \)
\( \implies \) \( x = 65 - 57 = 8 \)
Also, \( 45 + x + y = 60 \)
So, \( 45 + 8 + y = 60 \) [ \( \because x = 8 \) ]
or \( y = 60 - 53 = 7 \)
Hence, \( x = 8 \) and \( y = 7 \).

SOLVED EXAMPLES

Question. If mean = 60 and median = 50, find mode.
Answer: We have, Mean = 60, Median = 50
Mode = 3 Median - 2 Mean = 3(50) - 2(60) = 30

Question. If mode = 70 and mean = 100, find median.
Answer: We have, Mode = 70, Mean = 100
Median = Mode + \( \frac{2}{3} \) (Mean - Mode)
\( = 70 + \frac{2}{3} (100 - 70) = 70 + 20 = 90 \)

Question. If mode = 400 and median = 500, find mean.
Answer: Mean = Mode + \( \frac{3}{2} \) (Median - Mode)
\( = 400 + \frac{3}{2} (500 - 400) = 400 + \frac{3}{2} (100) = 400 + 150 = 550 \)

Question. For the following distribution.

The modal class is :
(a) 10 - 20
(b) 20 - 30
(c) 30 - 40
(d) 50 - 60
Answer: (c) 30 - 40

Question. If mode of a data is 45, mean is 27, then median is :
(a) 30
(b) 27
(c) 33
(d) None of the options
Answer: (c) 33

Question. For a symmetrical distribution, which is correct
(a) Mean > Mode > Median
(b) Mean < Mode < Median
(c) Mode = \( \frac{\text{Mean + Median}}{2} \)
(d) Mean = Median = Mode
Answer: (d) Mean = Median = Mode

Question. The relation connecting the measures of central tendencies is :
(a) mode = 2 median - 3 mean
(b) mode = 3 median - 2 mean
(c) mode = 2 medians - 3 mean
(d) mode = 3 median + 2 mean
Answer: (b) mode = 3 median - 2 mean

Question. The mean and median of a data are 14 and 15 respectively. The value of mode is
(a) 16
(b) 17
(c) 13
(d) 18
Answer: (b) 17

Question. If mode of the following data is 7, then value of \( k \) in 2, 4, 6, 7, 5, 6, 10, 6, 7, 2k + 1, 9, 7, 13 is :
(a) 3
(b) 7
(c) 4
(d) 2
Answer: (a) 3

Question. If mode = 80 and mean = 110 then the median is:
(a) 110
(b) 120
(c) 100
(d) 90
Answer: (c) 100

Question. The upper limit of the median class of the following distribution is :

(a) 17
(b) 17.5
(c) 18
(d) 18.5
Answer: (a) 17

Question. The measure of central tendency is given by the \( x \) coordinate of the point of intersection of the more than ogive and less than ogive is :
(a) Mean
(b) Median
(c) Mode
(d) All of the options
Answer: (b) Median

Question. Construction of a cumulative frequency table is useful in determining the
(a) mean
(b) median
(c) mode
(d) all of the options three measures
Answer: (b) median

Question. The measures of central tendency which can't be found graphically is :
(a) mean
(b) median
(c) mode
(d) none of the options
Answer: (a) mean

Question. Which of the following is not a measure of central tendency :
(a) Mean
(b) Median
(c) Range
(d) Mode
Answer: (c) Range

Question. A data has 25 observations (arranged in descending order). Which observation represents the median ?
(a) 12th
(b) 13th
(c) 14th
(d) 15th
Answer: (b) 13th

Question. For a given data with 50 observations the 'Less than ogive' and the 'more than ogive' intersect at (15.5, 20). The median of the data is :
(a) 4.5
(b) 20
(c) 50
(d) 15.5
Answer: (d) 15.5

Question. For a given data with 40 observations the less than ogive and the more than ogive intersect at (20.5, 15). The median of the data is :
(a) 5.5
(b) 20.5
(c) 15
(d) 40
Answer: (b) 20.5

Question. The Classmark of a Class Interval is
(a) Lower limit + Upper limit
(b) Upper limit - Lower limit
(c) \( \frac{1}{2} \) (Lower limit + Upper limit)
(d) \( \frac{1}{4} \) (Lower limit + Upper limit)
Answer: (c) \( \frac{1}{2} \) (Lower limit + Upper limit)

Question. One of the methods of determining mode is
(a) Mode = 2 Median - 3 Mean
(b) Mode = 2 Median + 3 Mean
(c) Mode = 3 Median - 2 Mean
(d) Mode = 3 Median + 2 Mean
Answer: (c) Mode = 3 Median - 2 Mean

Question. The arithmetic mean of 1, 2, 3, ..., \( n \) is
(a) \( \frac{n - 1}{2} \)
(b) \( \frac{n + 1}{2} \)
(c) \( \frac{n}{2} \)
(d) \( \frac{n}{2} + 1 \)
Answer: (b) \( \frac{n + 1}{2} \)

Question. Mode is
(a) least frequent value
(b) middle most value
(c) most frequent value
(d) None of the options
Answer: (c) most frequent value

Question. The algebraic sum of the deviations of a frequency distribution from its mean is
(a) always positive
(b) always negative
(c) 0
(d) a non - zero number
Answer: (c) 0

Question. Which of the following cannot be determined graphically?
(a) Mean
(b) Median
(c) Mode
(d) None of the options
Answer: (a) Mean

Question. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are

The number of athletes who completed the race in less then 14.6 seconds is :
(a) 11
(b) 71
(c) 82
(d) 130
Answer: (c) 82

Question. Consider the following distribution :

the frequency of the class (30 - 40) is
(a) 3
(b) 4
(c) 48
(d) 51
Answer: (b) 4

Question. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(a) mean
(b) median
(c) mode
(d) all of the options
Answer: (b) median

Question. For the following distribution :

The sum of lower limits of the median class and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Answer: (b) 25

Question. Consider the following frequency distribution:

The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Answer: (b) 17.5

Question. For the following distribution :

the modal class is
(a) 10 - 20
(b) 20 - 30
(c) 30 - 40
(d) 50 - 60
Answer: (c) 30 - 40

Question. Consider the data :

The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0
(b) 19
(c) 20
(d) 38
Answer: (a) 0

Question. Average of first ten prime numbers is :
(a) 12.6
(b) 12.9
(c) 13.9
(d) 14.9
Answer: (b) 12.9

Question. In the following distribution :

the number of families having income range (in Rs.) 16000 - 19000 is
(a) 15
(b) 16
(c) 17
(d) 19
Answer: (d) 19

Question. Consider the following frequency distribution of the heights of 60 students of a class :

The sum of the lower limit of the modal class and upper limit of the median class is
(a) 310
(b) 315
(c) 320
(d) 330
Answer: (a) 310

Question. In the formula \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \), for finding the mean of grouped data \( d_i \)'s are deviations from \( a \) of
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid points of the classes
(d) frequencies of the class marks
Answer: (c) mid points of the classes

Question. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes
(b) centred at the classmarks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Answer: (b) centred at the classmarks of the classes

Question. If \( x_i \)'s are the mid points of the class intervals of grouped data, \( f_i \)'s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( \sum (f_i x_i - \bar{x}) \) is equal to
(a) 0
(b) -1
(c) 1
(d) 2
Answer: (a) 0

Question. In the formula \( \bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right) \), for finding the mean of grouped frequency distribution, \( u_i = \)
(a) \( \frac{x_i + a}{h} \)
(b) \( h(x_i - a) \)
(c) \( \frac{x_i - a}{h} \)
(d) \( \frac{a - x_i}{h} \)
Answer: (c) \( \frac{x_i - a}{h} \)

Question. Mean of 11 observations is 50. If the mean of first six observations is 49 and that of last six observations is 52, then the sixth observation will be :
(a) 55
(b) 56
(c) 57
(d) 58
Answer: (b) 56