Data Handling Class 7 Worksheet Maths Chapter 4

Ans: (i) - True

A mode is the value(s) that appear most frequently in a data set. If two or more values have the same highest frequency, then the data set can have more than one mode (this is called a bimodal or multimodal distribution).

(ii) - False

The mode is the most frequently occurring value, but in this data set, all values appear only once. Since no value appears more than once, this data set has no mode.
(iii) True. 

The median is the middle value in a data set when arranged in order. If the data set has an even number of values, the median is the average of the two middle values, and this average may not be an original value in the data set.
(iv) - True

The maximum sum when rolling two standard dice (each numbered 1 to 6) is 12 (6 + 6). Therefore, a sum of 14 is impossible.
(v) - False

The mean (or average) is calculated as the sum of all the values divided by the number of values. The mean cannot be less than each of the individual observations; it will always lie between the smallest and largest values. It can be less than some of the values but not all.
(vi) - False

The median is the value that divides the data into two equal parts. The mean is the average of the data values and does not necessarily divide the data into two equal parts.
(vii) - False

 The median is the middle value and remains the same regardless of whether the data is arranged in ascending or descending order, as long as the order is correct.
(viii) - True

 A standard die has faces numbered from 1 to 6. Therefore, rolling a 7 is impossible.
(ix) - False

The mean is the average of all the values in the data set. It may not necessarily be one of the values in the data set. For example, the mean of 1, 2, 3 is 2, which is a data value, but the mean of 1, 2, 4 is 2.33, which is not in the data set.
(x) - True

The mode is the value(s) that appears most frequently in the data set, so it must always be one of the values in the data set.

Ans:  Mean = sum of all values / number of values  

[36+35+50+46+60+55]/6

Ans: (i) - Range
(ii) - Bar graph
(iii) - array
(iv) - 0
(v) - equally likely
(vi) -  Mean, median and Mode
(vii) - mean
(viii) - outcomes
(ix) - probability
(x) - 0

Ans:  Let the fifth number be "x"

 Mean = sum of all values / number of values  

25= [16+26+31+32+x]/5

transposing  5 to L.H.S

25*5 =105 + x 

125 =105 + x 

transposing  105 to L.H.S

125-105 = x

20 = x

Ans: Arranging the given data with the same value together, we get
22, 23, 24, 25, 26, 26, 26, 28
Here, 26 occurs the greatest number of times i.e. 3 times
Thus, the required mode = 26.

Ans: Let us arrange the given data in increasing order,
6, 8, 10, 12, 14
n = 5 (odd)
Median = (n + 1/2)^th term = 3^rd term = 10
Thus, the required median = 10.

Ans: Number on a die = 1, 2, 3, 4, 5, 6

n(S) = 6
Prime numbers = 2, 3, 5
n(E) = 3
Probability = n(E) / n(S) = 3 / 6 = 1 / 2
Thus the required probability = 1 / 2.

Ans: Five multiples of 3 are 3, 6, 9, 12 and 15

Mean =
= 45 / 5
= 9

Ans: First five prime numbers are: 2, 3, 5, 7 and 11

Mean =
= 28 / 5
= 5.6

Ans: Total number of balls = 5 + 9 = 14 balls
(a) Number of white ball = 5
n(E) = 5
Probability of getting white ball = n(E)/n(S) = 5 / 14
(b) Number of red balls = 9
n(E) = 9
Probability of getting white ball = n(E)/n(S) = 9 / 14

Ans: Greatest number 30
Smallest number = 10
Range = 30 – 10 = 20

Ans: Average = Sum of the numbers / Total number of terms

= 8 + 13 + 15 / 3
= 36/3
= 12

Ans: Arranging the given data in increasing order, we get
6, 9, 10, 12, 13, 14, 18, 19, 20, 25
n = 10 (even)

Thus, the required median = 13.5

Ans:

(i) 
(ii) In the year of 2012, maximum number of books i.e. 575 were sold.
(iii) Minimum number of books i.e. 150 were sold in the year 2008.

Ans: In the given frequency distribution table, we find that the observation 7 has maximum frequency, i.e., 20
Hence, the required mode = 7.