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(D) 10

First we have to arrange the marks (out of 10) obtained by 28 students in a Mathematics test.

0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 10, 10.

The number of students who scored marks less than 4 is 10.

(i) Observations:

Observation is the value at a particular period of a particular variable.

(ii) Raw data:

Raw data is the data collected in its original form.

(iii) Frequency of an observation:

Frequency of an observation is the number of times a certain value or a class of values occurs.

(iv) Frequency distribution:

Frequency distribution is the organization of raw data in table form with classes and frequencies.

(v) Discrete frequency distribution:

Discrete frequency distribution is a frequency distribution where sufficiently great numbers are grouped into one class.

(vi) Grouped frequency distribution:

Grouped frequency distribution is a frequency distribution where several numbers are grouped into one class.

(vii) Class-interval:

Class interval is a group under which large number of data is grouped to analyse its Range and Distribution.

(viii) Class-size:

Class size is the difference between the upper and the lower values of a class.

(ix) Class limits:

Class limits are the smallest and the largest observations (data, events, etc.) in a class.

(x) True class limits:

True class limits are the actual class limits of a class.

(i)

(ii) High score is 100.

(iii) Lowest score is 37

(iv) Range = 100 – 37 = 63

(v) Number of students failed = 2. (37 , 39)

(vi) Number of students scored more than 75 are = 8. 

(vii) Those observations are = 51 , 54 , 57 

(viii) number of people scored less than 50 = 5. (37, 39, 44, 48, 48)

(i) Observations:- 

Data obtains by the observer from the given problem is called observations. 

(ii) Raw data:- 

Data collected in its original form is called Raw Data. 

(iii) Fequency of an observation:- 

Number of times a certain value occurs is termed as frequency. 

(iv) Frequency distribution:- 

Organisation of Raw data in tabular form is called Frequency Distribution. 

(v) Discrete frequency distribution:- 

Frequency Distribution in which Raw Data is ungrouped is called Discrete frequency distribution. 

(vi) Grouped frequency distribution:- 

A frequency Distribution in which Raw data is grouped in specific class.

(vii) Class - interval:- 

Class interval is a group under which large number of data is grouped to analyze its Range and Distribution. 

(viii) Class - size:- 

Class – size is defined as the difference between upper and lower boundaries of any class. 

(ix) Class limits:- 

Class limits works as separation of one class in a grouped frequency distribution from another. 

(x) True class limits:- 

The Exact class limits of frequency distribution are called True class limits.

Probability = \(\frac{1}{2}\)

Given data: 4,5, 11, 8. 

Sum of observations = 4 + 5 + 11 + 8 = 28 

Number of observations = 4 

∴ Arithmetic Mean = \(\frac {Sum\,of\, observations}{Number\,of\,observations}\) = 28/4 = 7

Given data, ₹300, ₹450, ₹700, ₹650, ₹400, ₹750, ₹900 and ₹850. 

Sum of observations = 300 + 450 + 700 +650 +400 + 750 + 900 + 850 = 5000 

Number of observations = 8 

∴ Arithmetic Mean = \(\frac {Sum\,of\, observations}{Number\,of\,observations}\) = 5000/8 = ₹625

∴ Arithmetic Mean of amount donated = ₹625

Correct option is D) 1200

Correct option is : (c) KLMNO

A BC DEF GHIJ KLMNO 

So, in the series next word is KLMNO.

The angle at the centre of a circle is 360°.

The correct option is (b) 10.

Total number of outcomes when a dice is thrown
= 6 (1, 2, 3, 4, 5, 6)

(i) Total number of favorable outcomes = 3 (2, 3, 5)

∴Required Probability = 3/6 = 1/2

(ii) Total number of favorable outcomes = 3 (1, 4, 6)

∴Required Probability = 3/6 = 1/2

(iii) Total number of favorable outcomes

= 3 (4, 5, 6)

∴Required Probability = 3/6 = 1/2

(iv) Total number of favorable outcomes = 5 (1, 2, 3, 4, 5)

∴Required Probability = 5/6

(v) Total number of favorable outcomes = 3 (1, 3, 5)

∴Required Probability = 3/6 = 1/2

The correct option is (b).

When a dice is thrown, the 6 possible outcomes are 1, 2, 3, 4, 5, 6.

i) Education 

ii) Food 

iii) Total income is represented by 360° = Rs. 9,000 

food represented by 120 = \(\frac {120}{360}\)× 9000 = Rs. 3000 

iv) The expenditure on food is represented by 120 = Rs. 3000 

Then expenditure on education is represented by  \(\frac {60}{120}\)× 3000 = Rs. 1500  

False.

The composite numbers in a dice are 4 and 6.

Probability = total number of composite numbers/ Total numbers

= 2/6 … [divide numerator and denominator by 2]

= 1/3

The prime numbers in a dice are 2, 3 and 5.

Probability = total number of prime numbers/ Total numbers

= 3/6 … [divide numerator and denominator by 3]

= ½

True.

Each outcome or a collection of outcomes in an experiment makes an Event.

In throwing a die the number of possible outcomes is 6.

i) Mean expenditure =  \(\frac {Total\, expenditure}{No. \,of\, persons}\)

= \(\frac {16+17+21}{3} = \frac {54}{3} \) = Rs 18

ii) Amount spent = 3 times 

I.e., 3 × 16; 3 × 17; 3 × 21

= Rs. 48; Rs. 51; Rs. 63

Now the average = \(\frac {48+51+63}{3} = \frac {162}{3} \)

= Rs. 54 = 3 x original average

iii) After 50% discount the amount spent is Half of the actual amount = \(\frac {16}{2},\frac{17}{2},\frac {21}{2}\) 

= Rs.8; Rs. 8.50; Rs. 10.50 

Now the average =  \(\frac {8+8.50+10.50}{3} = \frac {27.00}{3} \) = Rs 9 = \(\frac {Original \,average}{2}\)

iv) The change In the observations is also carried out in the mean.

Given data : 

2, 3, 7, 5, 3, 2, 6, 7, 1,2. 

By arranging the numbers with same values together 

1, 2, 2, 2, 3, 3, 5, 6, 7, 7. 

As 2 occurs more frequently than other observations in the data. 

∴ Mode = 2

(i) Outcome of getting a number 6 from ten separate slips is one. Therefore, probability of getting a number 6 = 1/10

(ii) Numbers less than 6 are 1, 2, 3, 4 and 5 which are five. So there are 5 outcomes. Therefore, probability of getting a number less than 6 = 5/10 = 1/2

(iii) Number greater than 6 out of ten that are 7, 8, 9, 10. So there are 4 possible outcomes. Therefore, probability of getting a number greater than 6 = 4/10 = 2/5

(iv) One digit numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 out of ten. Therefore, probability of getting a 1-digit number = 9/10

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

Average of first five prime numbers = \(\frac {Sum \, of\, the\, numbers}{Numbers\, of\, primes}\)

= \(\frac {2+3+5+7+11}{5} = \frac {28}{5} = 5.6\) 

Correct option is  D) No mode

(d) As we know that when a coin is tossed, it has two possible outcomes, head or tail.

1, 2, 3, 4, 5, 6

When a die is rolled, then the six possible outcomes are 1,2, 3, 4, 5 and 6.

The total outcomes of tossing a coin are 2.

First ten natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10

∴ Average = \(\frac {Sum\,of\,the\,numbers}{Number}\)

= \(\frac {1+2+3+4+5+6+7+8+9+10}{10}= \frac {55}{10}\) = 5.5

∴ The mean of the first ten natural numbers = 5.5.

First 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 

In the given observations there is no repeated number. 

So, the given data has no mode.

First ten even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean = (2 +4 +6 +8 +10 +12 + 14 +16 +18 +20)/10

= 110/10

= 11

head, tail 

On tossing a coin one time, the outcome’can-be head or tail.

The data can be arranged in a tabular form using Tally marks.

(i) For 540 marks, central angle = 360° 

For 1 mark, central angle = 360/ 540

For 105 marks, central angle = 360/540 x 105 = 70° 

Hence the student scored 105 marks in Hindi. 

(ii) Central angle = 360° for 540 marks, 

For 1 mark, central angle = 360°/ 540 

For 90 marks, central angle = 540/360 × 90 marks = 135 marks. 

Thus, the student gets 135 marks in Mathematics. From part (i) we get that the student gets 105 marks in Hindi. 

Difference in marks = 135 – 105 = 30 

Hence, the student gets 30 more marks in Mathematics than in Hindi.

First ten where numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. 

Maximum value = 9 

Minimum value = 0 

Range = Maximum Value – Minimum Value 

= 9 – 0 = 9 

∴ Range = 9

a part

A pie chart is used to compare a part to a whole.

First 5 whole number are 0, 1, 2, 3 and 4.

Mean = (0+1+2+3+4)/5  

= 10/2 

= 5

(a) Pie chart

A geometric representation showing the relationship between a whole and its parts is pie chart.

True.

Pie chart shows the relationship between a whole and its parts.

First we need to find total marks. 

So, 

Marks scored in mathematics = \(\frac{Central\,angle\,of\,sector\times\,Total\,Marks}{360^\circ}\)

135 = \(\frac{90^\circ\times\,Total\,Marks}{360^\circ}\)

∴ Total Marks = 540

Marks scored in Hindi = \(\frac{Central\,angle\,of\,sector\times\,Total\,Marks}{360^\circ}\) = \(\frac{60\,\times\,540}{360^\circ}\) = 90 marks

Similarly, marks scored in science = \(\frac{76\,\times\,540}{360^\circ}\) Marks = 114 marks

Marks scored in social science = \(\frac{72\,\times\,540}{360^\circ}\) Marks = 108 marks

Marks scored in English = \(\frac{62\,\times\,540}{360^\circ}\) Marks = 93 marks

In order to find the highest and lowest marks, we have to arrange the marks in ascending order as follows:

39, 48, 56, 75, 76, 81, 84, 85, 90, 95

(i) Clearly, the highest mark is 95 and the lowest is 39.

(ii) The range of the marks obtained is: (95 – 39) = 56.

(iii) From the following data, we have

Mean marks = Sum of the marks/ Total number of students

Mean marks = (39 + 48 + 56 + 75 + 76 + 81 + 84 + 85 + 90 + 95) ÷ 10

= 729 ÷ 10

= 72.9.

Hence, the mean mark of the students is 72.9.

Given Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days

Average number of study hours = sum of hours/ number of days

Average number of study hours = (4 + 5 + 3) ÷ 3

= 12 ÷ 3

= 4 hours

Thus, Ashish studies for 4 hours on an average.

Given runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100

Mean score = total sum of runs/number of innings

The mean score = (58 + 76 + 40 + 35 + 48 + 45 + 0 + 100) ÷ 8

= 402 ÷ 8

= 50.25 runs.

(i) Primary data 

(ii) Secondary data 

(iii) Primary data 

(iv) Secondary data

i) Maximum temperature = 33C 

ii) Average temperature = \(\frac {Sum\,of\,the\,temperatures}{No.\,of\,observations}\)

= \(\frac {26+27+30+30+32+33+32}{7} = \frac {210}{7}\) 

∴ The average temperature of the week 30°C.

Mean = Sum of observations ÷ Number of observations

Mean = (x + x + 2 + x + 4 + x + 6 + x + 8) ÷ 5

Mean = (5x + 20) ÷ 5

Mean = 5 (x + 4) ÷ 5

Mean = x + 4

First six multiples of 4 are 4, 8, 12, 16, 20, 24

Mean = Sum of all observations/ Number of observations

= 84/6

= 14

The first five multiples of 3 are 3, 6, 9, 12 and 15.

Mean of first five multiples of 3 are = (3 + 6 + 9 + 12 + 15) ÷ 5

= 45 ÷ 5

= 9

\(\overline X\)

We know that

\(\overline X\)= sum of observations/ number of observations

\(\overline X\)= sum of weights of babies/ number of babies

\(\overline X\)= (3.4 + 3.6 + 4.2 + 4.5 + 3.9 + 4.1 + 3.8 + 4.5 + 4.4 + 3.6) ÷ 10

\(\overline X\)= (40) ÷ 10

\(\overline X\)= 4 kg

Here, the number of observations n is 9.

Since n is odd, the median is the n+12th observation, i.e., the 5^th observation.

As the numbers are arranged in the descending order, we therefore observe from the last.

Median = 5^th observation.

=> 25 = 2x – 8

=> 2x = 25 + 8

=> 2x = 33

=> x = (33/2)

x = 16.5

(i) Descending order = 5, 4, 3, 3, 2, 2, 1, 0 

(ii) Descending order = 11.5, 10.6, 9.1, 8.3, 5.6, 3.7