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(c) Mode

Mode is the observation that occurs most frequently in the data.

True.

The possible outcomes when the die is thrown are 1, 2, 3, 4, 5 and 6.

If both dice get 6, then sum of the number is 12.

Central tendency

Mean, median and mode are the representative values of a group of observations. They are also called the measures of central tendency of the data.

Correct option is B) Median

Here, the number of observations n is 10. Since n is even,

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

63 = x + (x + 2)/2

63 = (2x + 2)/2

63 = 2 (x + 1)/2

63 = x + 1

x = 63 – 1

x = 62

False.

The measures of central tendency lies between the maximum and minimum values of data.

From the question is given that,

Probability of choosing a white ball is = 2/9

The total number of balls in the bag is = 36

Number of white ball chosen = (2/9) × 36

= 8 white balls

Then, the number of yellow balls = Total balls in bag – number of white balls

= 36 – 8

= 28 yellow balls

(b) 5/26

Probability = Number of Vowels/ Total number of alphabets

= 5/26

False.

When the given data is arranged in ascending (or descending) order, then the middle most observation is the median of the data.

So, if the extreme observations on both the ends of a data arranged in ascending order are removed, the median is not get affected.

(i) From the Bar Graph we infer that the viewers for cricket is most and athletics is least.

(ii) Cricket is most popular.

(iii) Watching the sport is more preferred than participating.

Given n = Number of observations = 200

Mean = sum of observations/ number of observations

50 = sum of observations/ 200

Sum of the observations = 50 x 200 = 10,000.

Thus, the incorrect sum of the observations = 50 x 200

Now,

The correct sum of the observations = Incorrect sum of the observations – Incorrect observations + Correct observations

Correct sum of the observations = 10,000 – (92 + 8) + (192 + 88)

Correct sum of the observations = 10,000 – 100 + 280

Correct sum of the observations = 9900 + 280

Correct sum of the observations = 10,180.

Therefore correct mean = correct sum of the observations/ number of observations

= 10180/200

= 50.9

(a) 1300 (b) 300  (c) 4,5,6,7,8 (d) 7 (e) 8 (f) False 

1.Frequency distribution table

2.pictograph and bar graph

3.9

True.

Since either coin can turn up Head (H) or Tail (T), are the possible outcomes.

From the question it is given that, 1 – unit length = 10 crore,

So, bar of length 6 unit = 6 × 10

= 60 crore

Then, number of units will represent 75 crore = 75/10

= 7.5 units

On the scale of 1-unit length = 10 crore, the bar of length 6 units will represent 60 crore and of 7.5 units will represent 75 crore.

On the scale of 1 unit length = 10 crore, the bar of length 6 units will represent 60 crore and of 7.5 units will represent 75 crore.

Correct option is  C) 75

Correct option is  B) 20

(d) 4

Head Head, Tail Tail, Tail Head, Head Tail.

Correct option is  D) 80

(i) Cars sold by dealer A = 6 x 50 = 300

Cars sold by dealer D = 4 x 50 = 200 ,

∴ A sold more cars than dealer D by = 300 – 200 = 100

∴A has sold 100 more cars than dealer D.

(ii) No. of cars = 23

Scale = 50 cars

∴Total no. of cars = 23 x 50 = 1150 cars Ans.

(1) Students in science = \(\frac{Central\,angle\,of\,the\,corresponding\,sector\times\,Total\,students}{360^\circ}\)

1000 = \(\frac{100^\circ\times\,Total students}{360^\circ}\)

∴ Total students = 3600

(2) Students in arts = \(\frac{Central\,angle\,for\,arts\times\,Total\,students}{360^\circ}\) = \(\frac{120^\circ\times\,3600}{360^\circ}\) = 1200

∴ Ratio of students in science and arts = 1000:1200 = 5:6

(b) 3/5

Probability = Number of times head appeared / Total number of times coin is tossed

= 120/200

= 12/20 … [divide both numerator and denominator by 4]

= 3/5

(a) What is the most common mode of transport?

By observing the given pie chart the most common mode of transport is Bus i.e. its central angle is 120^o.

(b) What fraction of children travel by car?

By observing the given pie chart, central angle of the children travel by car is 90^o.

Then,

The fraction of children travel by car,

= (90^o/360^o)

= 9/36 … [divide both numerator and denominator by 9]

= ¼

(c) If 18 children travel by car, how many children took part in the survey?

Let us assume total number of children took part in the survey be y

Number of children travel by car = (¼) of total number of children

18 = ¼ × y

Y = 18 × 4

Y = 72

So, 72 children took part in the survey.

(d) How many children use taxi to travel to school?

Central angle of children use taxi to travel to school = (360^o – (120^o + 60^o + 90^o + 60^o)

= (360^o – 330^o)

= 30^o

Then,

Out of 72 children number of children use taxi to travel to school = (30^o/360^o) × 72

= 6

(e) By which two modes of transport are equal number of children travelling?

By observing the given pie chart, cycle and walk are the two modes of transport are equal number of children travelling. Because, central angle of the two modes are same i.e. 60^o.

Correct Answer is  24 cars

(i) Scale = 5 cars.

(ii) First week

(iii) Second week

(iv) Five years

(v) 110 cars

(vi)  35, 30, 25, 20

First Week

The data can be put in the form of frequency distribution in the following manner:

Mean number of children per family = sum of total number of children / total number of families

= (2 + 4 + 3 + 4 + 2 + 3 + 5 + 1 + 1 + 5) ÷ 10

= 30 ÷ 10

= 3

Thus, on an average there are 3 children per family in the locality.

Bar graph showing pass percentage of different classes from VI to X. 

Steps of construction: 

1. Draw two mutually perpendicular lines on a graph sheet – one horizontal (X-axis) and one vertical (Y-axis) mark. Number of class on the X-axis and their pass percentage on the Y-axis. 

2. Take the 10 as scale on the Y-axis i.e., 1 cm = 10% 

3. Calculate the lengths or heights of the bars by dividing the pass percentage with the scale. 

Class VI = 65 ÷ 10 = 6.5 cm 

Class VII = 75 ÷ 10 = 7.5 cm 

Class VIII = 85 ÷ 10 = 8.5 cm

Class IX = 60 ÷ 10 = 6 cm 

Class X = 80 ÷ 10 = 8 cm 

4. Draw rectangular vertical bars of same width (I cm) on the X-axis with their calculated heights. 

Hence, required vertical bar graph was constructed.

(i) D has the maximum value of 350

(ii) A + D = 250 + 350 = 600

B + E = 300 + 275 = 575

Hence A + D is greater.

The required frequency table will be as shown below :

Writing these numbers in descending order we get:

6.1, 6.0, 5.9, 5.9, 5.8, 5.8, 5.7, 5.7, 5.6, 5.5, 5.5, 5.4, 5.3, 5.2, 5.1

(i) Highest value = 6.1

(ii) Lowest value = 5.1

(iii) Range of values = Highest value – lowest value = 6.1 -5.1 = 1.0

(i) Ascending order = 5, 7, 10, 11, 12, 15, 16 

(ii) Ascending order = 4.7, 5.6, 5.9, 6.3, 9.8, 12.3

 In a bar graph, each bar (rectangle) represents only one value of the numerical data. True

To represent the population of different towns using bar graph, it is convenient to take one unit length to represent one person. False

False.

Population of towns are large numbers, then we may need a different scale.

True.

By the definition, pictographs and bar graphs are pictorial representations of the numerical data.

Pictographs and bar graphs are pictorial representations of the numerical data. True

(i) Data : The word data means information in the form of numerical figures.

(ii) Frequency of an observation: The number of times a particular observation occurs is called its frequency.

(i) Food.

(ii) Charity and Education.

(iii) Others.

(b) 144^o

Distribution of protein in the skin = 1/10

Distribution of protein in the bones = 1/6

Distribution of protein in the Muscles = 1/3

Central angle of the sector representing skin, muscles and bones,

= 1/10 + 1/6 + 1/3

= (3 + 5 + 10)/30

= 18/30 × 360^o

= 216^o

Then,

Central angle of the sector representing hormones enzymes and other proteins,

= 360^o – 216^o

= 144^o

(d) 96^o

Distribution of protein in the skin = 1/10

Distribution of protein in the bones = 1/6

Central angle of the sector representing skin and bones,

= 1/10 + 1/6

= (3 + 5)/30

= 8/30 × 360^o

= 96^o

(a) 108^o

The following pie chart gives the distribution of constituents

in the human body = 16% + 14% = 30%

= (30/100) × 360^o

= 108^o

(d) 2 : 1

From the chart,

Distribution of protein in the muscles = 1/3

Distribution of protein in the bones = 1/6

Then,

Ratio of distribution of proteins in the muscles to that of proteins in the bones,

= 1/3 : 1/6

= (1/3) × (6/1)

= (1/1) × (2/1)

= 2/1

= 2 : 1

Arranging the data in ascending order such that same values are put together, we get:

10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 18

Here, clearly, 14 occurs the most number of times.

Therefore, 14 is the mode of the given data.

(a) 2/5

Probability = Number of cards having 2/ Total number of cards

= 2/5

Arranging the data in ascending order such that same values are put together, we get:

5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

Here, n = 15

Therefore median = ((n+1)/2)^th term

Median = value of 8^th term

Median = 20

Here, clearly, 20 occurs most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.

Yes, the median and mode of the given data are the same.

Calculation of mean

Mean = Σ f_i x_i/ Σ f_i

= 332/25

= 13.28

Here, n = 25, which is an odd number. Therefore,

Therefore median = ((n+1)/2)^th term

Median = value of 13^th term

Median = 13

Now, by using empirical formula we have,

Mode = 3 Median – 2 Mean

Mode = 3 (13) – 2 (13.28)

Mode = 39 – 26.56

Mode = 12.44.

(i) lower limit = 50.

(ii) Class marks of class 40 – 50 = \(\frac{40+50}{2}\) = 45.

Class marks of class 50 – 60 = \(\frac{50+60}{2}\) = 55

(iii) Class size = 40 – 30 = 10, or 60 – 50 = 10.

(a) LPG (b) 10 (c) 5000