Statistics And Collection Of Statistical Data(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

STATISTICS AND COLLECTION OF STATISTICAL DATA MCQs

Statistics, Introduction To Statistics, Collection Of Data (11th And 12th Grade)

Total Questions : 105 | Page 7 of 11 pages

Question 61. The A.M. of a set of 50 numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is

36
36.5
37.5
38.5

Discuss Question

Answer: Option C. -> 37.5
:
C

w e h a v e, \frac{\sum x_{i}}{50} = 38, ∴ \sum x_{i} = 1900 N e w v a l u e o f \sum x_{i} = 1900 - 55 - 45 = 1800 a n d n = 48 ∴ n e w m e a n = \frac{1800}{48} = \frac{450}{12} = \frac{225}{6} = 37.5.

Question 62. The coefficient of variation of two series are 58% and 69%. If their standard deviations are 21.2 and 15.6, then their A.Ms are

36.6,22.6
34.8, 22.6
36.6, 24.4
none of these

Discuss Question

Answer: Option A. -> 36.6,22.6
:
A

We know that C . V = \frac{σ \times 100}{¯ x} o r ¯ x = \frac{σ}{c . v} \times 100 ∴ Mean of first series = \frac{21.2 \times 100}{58} = 36.6 Meanof second series = \frac{15.6 \times 100}{69} = 22.6

Question 63. The range of the observations in 3, 8, 4, 9, 16, 19 is

Discuss Question

Answer: Option A. -> 16
:
A
Range = Highest observation- Lowest observation
= 19 − 3
= 16

Question 64. Consider the frequency distribution of the given number

\begin{matrix} V a l u e : & 1 & 2 & 3 & 4 F r e q u e n c y : & 5 & 4 & 6 & f \end{matrix}

If the mean is known to be 3, then the value of f is

Discuss Question

Answer: Option D. -> 14
:
D

M e a n = \frac{1 \times 5 + 2 \times 4 + 3 \times 6 + 4 \times f}{15 + f}

i . e ., 3 = \frac{5 + 8 + 18 + 4 f}{15 + f} \Rightarrow 45 + 3 f = 31 + 4 f

\Rightarrow 45 - 31 = f \Rightarrow f = 14

Question 65. A school has four sections of chemistry in class XII having 40, 35, 45 and 42 students. The mean marks obtained in chemistry test are 50, 60, 55 and 45 respectively for the four sections, the over all average of marks per students is

53
45
55.3
52.25

Discuss Question

Answer: Option D. -> 52.25
:
D
Total number of students

= 40 + 35 + 45 + 42 = 162

Total marks obtained

= (40 \times 50) + (35 \times 60) + (45 \times 55) + (42 \times 45)

= 8465

Overall average of marks per students

= \frac{8465}{162} = 52.25

Question 66. The mean of 10 numbers is 12.5; the mean of the first six is 15 and the last five is 10. The sixth number is

15
12
18
none of these

Discuss Question

Answer: Option A. -> 15
:
A

L e t t h e m e a n o f t h e l a s t f o u r b e A_{2} . T h e n b y t h e f o r m u l a f o r c o m b i n e d m e a n, 12.5 = \frac{6 \times 15 + 4 \times A_{2}}{6 + 4}; o r 125 = 90 + 4 A_{2}; ∴ A_{2} = \frac{35}{4} L e t t h e s i x t h n u m b e r = x; t h e n t a k i n g t h e s i x t h n u m b e r a s a c o l l e c t i o n, t h e c o m b i n e d m e a n o f t h i s c o l l e c t i o n a n d t h e c o l l e c t i o n o f t h e l a s t f i v e i s 10, b y q u e s t i o n . ∴ B y d e f i n i t i o n o f c o m b i n e d m e a n 10 = \frac{1 \times x + 4 \times \frac{35}{4}}{1 + 4}; 50 = x + 35; ∴ x = 15 ∴ s i x t h n u m b e r = 15

Question 67. The variance of the data 2, 4, 6, 8, 10 is:

6
7
8
none of these

Discuss Question

Answer: Option C. -> 8
:
C

M e a n (¯ x) = \frac{\sum x_{i}}{n} = (\frac{(2 + 4 + 6 + 8 + 10)}{5} = 6

V a r i a n c e (σ^{2}) = \frac{\sum (x - ¯ x)^{2}}{n} = \frac{(2 - 6)^{2} + (4 - 6)^{2} + (6 - 6)^{2} + (8 - 6)^{2} + (10 - 6)^{2}}{5} = \frac{(16 + 4 + 0 + 4 + 16)}{5} = \frac{40}{5} = 8

Question 68. The mean deviation of 2, 4, 6, 8, 10 about its mean is

Discuss Question

Answer: Option B. -> 2.4
:
B

M e a n (¯ x) = \frac{\sum x_{i}}{n} = \frac{2 + 4 + 6 + 8 + 10}{5} = 6 ∴ M . D = \frac{\sum | x_{i} - ¯ x |}{n} = \frac{| 2 - 6 | + | 4 - 6 | + | 6 - 6 | + | 8 - 6 | + | 10 - 6 |}{5} = \frac{4 + 2 + 0 + 2 + 4}{5} = \frac{12}{5} = 2.4

Question 69. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations in the set is increased by 2, then the median of the new set

is increased by 2
is decreased by 2
is two times the original median
remains the same as that of the original set

Discuss Question

Answer: Option D. -> remains the same as that of the original set
:
D
We know that median is found when we arrange the observations in ascending or descending order. And if the number of observations are odd then it is the

\frac{n + 1}{2} t h

term.
Since on arranging 9 observations 5th will be the median.
On changing 4 of any side won't affect the median and it'll remain same.

Question 70. If the range of discrete data of n observations is zero, then

all values of data are zero
all values of data are equal to standard deviation
all values of data are equal
the extreme values of data are different

Discuss Question

Answer: Option C. -> all values of data are equal
:
C
Let m be the least value of the discrete data and M be the maximum value of the discrete data.
Then range = M – m = 0 (hypothesis)