Statistics Test

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Statistics Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

If the mean of group A of 10 observations is 18, and that of group B of 12 observations is 23, then what will be the combined mean of the two groups?

A
20.73

B
1.87

C
20.27

D
2.86

Solution

Combined mean = =
Question 2
1 / -0

The arithmetic mean of 5, x, 7 and 3y is 15 and the arithmetic mean of 3x, 5, 2 and y is 18. What is the arithmetic mean of x and y?

A
12.234

B
14.125

C
15.678

D
16.678

Solution

= 15
x + 3y = 48 ... (i)
= 18
3x + y = 65 ... (ii)

Adding (i) and (ii),
4x + 4y = 113
Or x + y = 113/4 = 28.25
Mean = = 14.125
Question 3
1 / -0

The following data has been arranged in ascending order. If their median is 93, what is the value of x?

47, 52, 70, 82, x, x + 2, 98, 100, 105, 112

A
92

B
94

C
91

D
95

Solution

n = 10 (even)
∴ Median = 93 = 93 = 186 = 2(x + 1)
x + 1 = 93
x = 92
Question 4
1 / -0

Find the modes of the following data:

63, 1, 5, 71, 63, 5, 85, 90, 1, 71, 56, 71, 90, 1, 5, 8

A
1, 5, 71

B
63, 90, 5

C
90, 1, 5

D
63, 90, 8

Solution

The modes are 5, 1, 71 (as they occur maximum number of times, i.e. three times).
Question 5
1 / -0

The given data shows the distance (in km) between the residence and the work place of each of 20 doctors.

3, 4, 5, 11, 3, 18, 6, 7, 12, 11, 6, 6, 9, 21, 20, 17, 16, 12, 11, 7

The range of the given data is ______.

A
12

B
19

C
18

D
10

Solution

Range of data is defined as the difference between two extreme observations of distribution
Or range = maximum value - minimum value.
In the given data, the maximum value = 21 and the minimum value = 3.
So, range = 21 - 3 = 18
Question 6
1 / -0

Directions: In the following question, three lists of data are given.

List I: 80, 81, 82, 83, 84, 85
List II: 85, 85, 85, 85, 85, 85
List III: 50, 60, 70, 85, 90, 95

The order of data lists I, II and III, arranged in ascending order of their median is:

A
I, II, III

B
I, III, II

C
II, III, I

D
III, I, II

Solution

List I: 80, 81, 82, 83, 84, 85

Median = = 82.5

List II: 85, 85, 85, 85, 85, 85

Median = = 85

List III: 50, 60, 70, 85, 90, 95

Median = = 77.5

Hence, the correct order is III, I, II.
Question 7
1 / -0

What is the mode of the following data, if the students are standing according to the marks they scored in the class?

51, 58, 56, 96, 85, 75, 87, 23, 25, 56, 57, 66, 69, 85, 45, 96, 52, 43, 85, 99, 92, 47, 59 and 66.

A
85

B
75

C
69

D
66

Solution

Arrange the data in ascending order,
23, 25, 43, 45, 47, 51, 52, 56, 56, 57, 58, 59, 66, 66, 69, 75, 85, 85, 85, 87, 92, 96, 96 and 99
Since 85 comes three times.
So, 85 is the mode.

Question 8

1 / -0

Find the mode of the fee paid by the students of the college.

Fee paid	No. of the student
Rs. 20000	12
Rs. 30000	17
Rs. 40000	11
Rs. 50000	21
Rs. 60000	15
Rs. 70000	19
Rs. 80000	16

Rs. 60,000

Rs. 50,000

Rs. 40,000

Rs. 30,000

Solution

The arithmetic mode of the given data is Rs. 50 000 as the highest frequency 21 is associated with Rs. 50,000.

Question 9
1 / -0

The mean of 20 observations is p. If each observation is multiplied by 3 and then 1 is added to it, the new mean obtained will be

A
3p + 20

B
p + 20

C
3p + 1

D
3(p + 1)

Solution

Mean of 20 observations is p.
So, sum of the observations will be 20p.
When each observation is multiplied by 3, the mean will become (3 × 20p)/20
When 1 is added to each observation, new mean will be (20 + 60p)/20
So, new mean obtained after multiplying each observation by 3 and adding 1 to it = (20 + 60p)/20 = 3p + 1.
Question 10
1 / -0

If the median of a set of 7 distinct observations is 40.5 and each of the largest 3 observations of the set is increased by 17.5, then what will be the change in the median?

A
14

B
0

C
1.5

D
17.5

Solution

The median is the middle of a sorted list of numbers.
When the number of observations is odd, the median is the middle value.
When the number of observations is even, the median is the average of the two middle values.
Here, 7 is the odd number and median is 40.5.
So, when we increase the last three values by 17.5 there will be no change in the median as it is the 4^th term.
Therefore, net change is zero.
Question 11
1 / -0

The following data has been arranged in ascending order. If the median of the data is 7, then what is the value of m?

2, m, m + 2, 10

A
6

B
5

C
7

D
3

Solution

Median =
7 =
14 = 2m + 2 2m = 14 - 2 = 12
m = = 6

Question 12

1 / -0

Find the difference between the ranges of the given data sets.

Set 1	89	73	84	91	87	77	94
Set 2	78	67	85	69	70	73	56

6

7

8

9 Solution

Range of the 1^st data set = 94 - 73 = 21
Range of the 2^nd data set = 85 - 56 = 29
Difference between the ranges = 29 - 21 = 8

Question 13
1 / -0

The mean of 14 terms in a series is 7.5. If two terms are excluded from the series, the mean increases by 1.5. Now, which of the following two numbers should be added to the series to get the original mean back?

A
0.5 and 0.25

B
1.5 and 0.5

C
1.5 and 1.25

D
1.5 and 1.5

Solution

Mean =

The sum of 14 terms = 7.5 × 14 = 105
The sum of 12 terms after 2 terms are excluded = (7.5 + 1.5) × 12 = 108
The sum of excluded terms = 108 - 105 = 3 = (1.5 + 1.5)
Question 14
1 / -0

The average marks scored by Ajay in certain number of tests were 84. He scored 100 marks in his last test. If his average score of all these tests was 86, then the total number of tests he appeared in was

A
8

B
7

C
9

D
10

Solution

x₁ + x₂ + x₃ + … + x_n = 84n

n = 7

Total number of tests = 7 + 1 = 8

Question 15

1 / -0

What is the value of p if the mean of the given data is 18?

x_i	13	15	17	19	21	23
f_i	8	2	3	4	5p	6

4

2

0

1 Solution

x_i	f_i	f_ix_i
13	8	104
15	2	30
17	3	51
19	4	76
21	5p	105p
23	6	138
Total	23 + 5p	399 + 105p

Mean () =
18 =
414 + 90p = 399 + 105p
p = 1 [1]

Question 16
1 / -0

The mean and median are to be calculated for 50 items. The mean is found to be 70 and the median is found to be 68. But later on it is found that due to some error in the data, the largest item was 150 instead of 140. Find the corrected mean and median, respectively.

A
69.8 and 68

B
70.2 and 68

C
70.8 and 69

D
71.2 and 70

Solution

We know,
Mean =
Now,

Therefore,
New corrected mean

But median doesn't depend on the largest value item, so the median will remain unchanged.
Hence, this is the required solution.
Question 17
1 / -0

Five years ago, the average age of A and B was 15 years. Today, the average age of A, B and C is 20 years. How old will C be after 10 years?

A
35 years

B
40 years

C
30 years

D
50 years

Solution

Five years ago, (A + B) = 15 × 2 = 30 years
Now, (A + B) = 30 + 5 × 2 = 40 years
Now, (A + B + C) = 20 × 3 = 60 years
Now, C = 60 - 40 = 20 years
After 10 years, C = 20 + 10 = 30 years

Question 18

1 / -0

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states. Find the mean percentage of female teachers.

Percentage of Female Teachers	20	30	40	50	60	70	80
Number of States	6	11	7	4	4	2	1

35.98

36.57

38.91

39.71 Solution

Mean =

Mean =

=

Question 19
1 / -0

The average age of a class of 30 students and a teacher decreases by 0.5 year, if we exclude the teacher. If the overall average is 14.5 years, then find the age of the teacher.

A
36.5 years

B
32.5 years

C
29.5 years

D
28.5 years

Solution

Age of the teacher = Total age of students and teacher - Total age of students
= 31(14.5) – 30(14)
= 449.5 - 420
= 29.5 years
Question 20
1 / -0

Three years ago, the average age of a family of 5 members was 17 years. On addition of a child to the family, the present average age of the family is still the same. The present age of the child is

A
3 years

B
2 years

C
1 year

D
None of these

Solution

3 years ago, total age of 5 members = 17 × 5 = 85
Present age of 6 members = 85 + 3 × 5 + x = 100 + x
Or, = 17
Or, x = 102 - 100
Or, x = 2 years
So, the present age of the child is 2 years.

Question 21

1 / -0

Match the following:

Column A	Column B
(i) A histogram is a graphical representation of a grouped frequency distribution with	(A) class mark
(ii) The middle value of a class interval is called	(B) continuous class
(iii) When the data is collected at the first time by an investigator, it is called	(C) raw data
(iv) When the data is compiled in the same form and order in which it is collected, it is known as	(D) primary data

(i) - A, (ii) - B, (iii) - C, (iv) - D

(i) - B, (ii) - A, (iii) - D, (iv) - C

(i) - A, (ii) - D, (iii) - C, (iv) - B

(i) - D, (ii) - C, (iii) - B, (iv) - A

Solution

(i) A histogram is a graphical representation of a grouped frequency distribution with continuous class.
(ii) The middle value of a class interval is called class mark.
(iii) When the data is collected at the first time by an investigator, it is called primary data.
(iv) When the data is compiled in the same form and order in which it is collected, it is known as raw data.

Question 22

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Fill in the blanks with the help of the given table.

(i) The mean of a certain set of observations is 20. If each observation is divided by 2, then the new mean of the set of observations is P .
(ii) Q is that value of the given number of observations which divides it into exactly two parts.
(iii) The difference between the highest and lowest values in a given data is known as R of the data.
(iv) The points scored by John in 7 games of darts are 8, 10, 12, 13, 13, 13 and 5. The median of the given observations is S .

	P	Q	R	S
A	20	Mode	width	13
B	10	Median	range	12
C	15	Mean	frequency	5
D	10	Mode	width	8

A

B

C

D

Solution

(i) Mean of the set of observations = 20
If each observation is divided by 2, then:
New mean = 20/2 = 10

(ii) Median is that value of the given number of observations which divides it into exactly two parts.

(iii) The difference between the highest and lowest values in a given data is known as range of the data.

(iv) By arranging the data in ascending order, we get
5, 8, 10, 12, 13, 13, 13
The middle number is 12, which is the median of the data.

Question 23
1 / -0

For 14 consecutive days, the number of persons booked for violating the speed limit of 40 km/hr was as follows:

59, 52, 58, 61, 68, 57, 62, 50, 55, 62, 53, 54, 51, 56

Find :
(i) Mean
(ii) Median
(iii) Mode

A
(i) - 57, (ii) - 56, (iii) - 57

B
(i) - 55, (ii) - 56.5, (iii) - 62

C
(i) - 55 , (ii) - 56.5, (iii) - 62

D
(i) - 57, (ii) - 56.5, (iii) - 62

Solution

14 observations in ascending order:

50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 62, 68

(i) Mean =
Mean = Mean = 57

(ii) Median = = 56.5
(iii) Mode = 62
Question 24
1 / -0

Directions For Questions

Directions: Study the given information and answer the following question.

Sales and profit figures of company ABC Ltd.
(All figures in Rs. lakh)

...view full instructions

What were the average sales per month?

A
Rs. 134 lakh

B
Rs. 135 lakh

C
Rs. 136 lakh

D
Rs. 137 lakh

Solution

Average sales per month of company = = Rs. 137 lakh
Question 25
1 / -0

Directions For Questions

Directions: Study the given information and answer the following question.
Sales and profit figures of company ABC Ltd.
(All figures in Rs. lakh)

...view full instructions

For the period from January to June, the percentage of net profit to sales was

A
15.2%

B
15.4%

C
15.6%

D
15.8%

Solution

From January to June, the percentage of net profit to sales = = 15.2%