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Question (1):
Find the median of : 83, 37, 70, 29, 45, 63, 41, 70, 30, 54
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Answer:
29, 30, 37, 41, 45, 54, 63, 70, 70, 83
Median = Middle-most score
Median = 49.5
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Question (2):
The median of prime numbers between 51 and 80.
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Answer:
53, 59, 61, 67, 71, 73, 79
Median = Middle-most score
Median = 67
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Question (3):
The median of the first ten prime numbers.
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Answer:
The first 10 prime numbers ={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
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Question (4):
The following is in ascending order: 12, 14, 17, 20, 22, x, 26, 28, 32, 36
i) If the median of the data is 23, find x.
ii) In the above data, if 32 is changed to 23, find the new median.
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Answer:
12, 14, 17, 20, 22, x, 26, 28, 32, 36
i) Median = 23
46 = 22+x
x = 24
ii) Replacing 32 by 23.
12, 14, 17, 20, 22, 23, 24, 26, 28, 36
Median = 22.5
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Question (5):
The weights of 10 students (in kg) are:
55, 51, 60, 38, 52, 42, 49, 63, 47, 35
i) Find the median weight.
ii) If the weight 63 kg is replaced by 36 kg, find the new median weight.
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Answer:
i) 35, 38, 42, 47, 49, 51, 52, 55, 60, 63
Median = 50
ii) If 63 kg is replaced by 36 kg.
35, 36, 38, 42, 47, 49, 51, 52, 55, 60
Median = 48kg
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Question (6):
The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result.
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Answer:
The sum of 31 results = 60 x 31 = 1860
The sum of first 16 results = 58 x 16 = 928
The sum of second 16 results = 62 x 16 = 992
 16th result = (928 + 992) - 1860 = 60
16th result = 60
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Question (7):
Find the mean  for the following distribution:
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Answer:
Mean=7.025
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Question (8):
Find the value of 'p', if the mean of the following distribution is 7.5.
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Answer:
Mean=7.5
303 + 9p = 7.5(41 + p)
303 + 9p = 307.5 + 7.5p
9p - 7.5p = 307.5 - 303
1.5p = 4.5
p = 3
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Question (9):
Find the median of : 15, 8, 14, 20, 13, 12, 16.
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Answer:
8, 12, 13, 14, 15, 16, 20
n = 7 (odd)
Median = 14
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Question (10):
Find the mode of : 43, 42, 44, 40, 48, 45, 40, 40
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Answer:
40, 40, 40, 42, 43, 44, 45, 48
Since 40 is the most repeated score,
Mode = 40
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Question (11):
Find the mean of the following frequency distribution:
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Answer:
Mean = 25
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Question (12):
The data on number of patients in a hospital in a month are given below. Find the average number of patients attending the hospital in a day.
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Answer:
Mean = 28.67
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Question (13):
Find the missing frequency (p) in the following distribution, whose mean is 54.
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Answer:
Given Mean = 54
Cross - multiplying, we get
54(92 + p) = 5160 + 30p
4968 + 54p = 5160 + 30p
54p - 30p = 5160 - 4968
24p = 192
p = 8
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Question (14):
A TV station broadcasts  hours each day on an average. A programme analysis shows.
Represent the above data by (a) a bar chart (b) a pie chart.
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Answer:
(a) Bar chart
(b) Pie chart
Calculation of Angles.
Drama:Angle at the centre 
Documentary: Angle 
Sports: Angle 
News: Angle 
Others: Angle 
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Question (15):
The total votes polled (to the nearest 100) in a certain by - election were distributed among the various candidates as follows:
Represent the above data by (a) a bar graph, (b) a pie graph
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Answer:
(a) Bar graph
(b) Pie Graph
Calculation of Angles:
A: Angle at the centre  
B: Angle at the centre  
C: Angle at the centre  
D: Angle at the centre  
E: Angle at the centre  
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Question (16):
The daily average price of Rank Xerox's share is listed below for a week.
Represent the data by a bar graph.
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Answer:
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Question (17):
The rainfall over an island for a year is recorded below.
Present this information by a bar chart.
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Answer:
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Question (18):
The monthly expenses of two families are as follows:
Illustrate the data by 'double bar' graph
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Answer:
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Question (19):
Areas in million square km of oceans of the world are:
Draw (a) a bar graph (b) a pie chart
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Answer:
(a) Bar graph
(b) Pie chart
Working:
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Question (55):
Three coins are tossed simultaneously 250 times with the following frequencies of different out comes.
If these coins are tossed gain what is the probability of getting 2 heads?
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Answer:
Let E denote the event of getting 2 heads.
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Question (20):
A die is thrown 50 times. 26 times the number which was shown on the top face was even. If the die is thrown again, what is the probability of getting an odd number?
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Answer:
Total number of trials =50
Number of times an even number was got = 26
 Number of times an odd number was got= 24
P(getting an odd number)  |
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Question (21):
The runs scored by a cricketer in six innings are given below:
What is the probability that he would score a half-century in an innings?
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Answer:
Number of times he has scored an half-century = 4
P(his scoring a half century) 
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Question (22):
750 families with 2 children were selected randomly, and the following data were recorded:
Compute the probability of a family chosen at random, having i) no by ii) 1 boy iii) 2 boys. Also check whether the sum of these probabilities is 1.
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Answer:
E1 : the family has no boy
E2 : the family has one boy
E3 : the family has two boys
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Question (23):
A bag contains red, blue and green balls. One ball is drawn at random from this bag, its colour is noted and then the ball is put back into the bag. This is repeated 125 times and the observations are recorded as follows:
What is the probability that the ball drawn is not blue?
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Answer:
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Question (24):
A person fires 50 times at a target and hits the target 30 times. When he fires the next time at the target. What is the probability that he hits the target?
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Answer:
The probability of hitting the target  |
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Question (25):
A dice is thrown 500 times and the frequencies of the outcomes and noted as follows:
When the dice is thrown again what is the probability of a prime number showing up?
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Answer:
Total number of trials = 500
Number of trials in which a prime number showed up = 48 + 82 + 96
= 226
P(getting a prime number)  |
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Question (26):
In the EXIT POLL results given by a TV Channel in the past few years, it was observed that 65 out of 80 times, the predictions proved correct. What is the probability that the next EXIT POLL result given by the TV channel will be correct? What is the probability that the result will be in correct?
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Answer:
E : The exit poll result is correct
Total number of trials = 80
Number of times the result is correct = 5
Probability of the result being incorrect
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Question (27):
A survey was conducted about the distance travelled each day by the employees of a company to their place of work and the result is given below:
If an employee is chosen at random, what is the probability
1) that he travels less than 10 KM?
2) that he travels 20 km or more?
3) that he travels less than 20 km but not less than 10 km?
Is the sum of these probabilities equal to one?
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Answer:
E1 : the employee travels less than 10 KM
E2: the employee travels 20 km or more but not less than 10 km
Total number of trials = 30 + 172 + 180 + 58 + 35 + 25 = 500
Number of trials in which the outcomes is E1
= 30 + 172 = 202
Number of trials in which the outcome is E2
= 35 + 25 = 60
Number of trials in which the outcome is E3
= 180 + 58 = 238
P(E1) + P(E2) + P(E3) = 0.404 + 0.12 + 0.476
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Question (28):
The marks of a student in 10 tests are given below:
If 60% and above is I class, what is the probability that the student gets I class in any test?
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Answer:
Total number of tests = 10
Number of tests in which the student gets I class = 6
Probability of getting I class
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Question (29):
A study was conducted on a group of 1000 people (randomly selected) in a city to find a relation between age and the number of visits to a hospital in a year. The data obtained is in the following table.
If a person chosen at random from the city, find the probability of his
1) being between 10 and 30 years and visiting and visiting the hospital thrice in the year
2) being below 60 years and visiting the hospital twice.
3) visiting the hospital only once
4) being between 30 and 60 years of age and visiting the hospital more than twice
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Answer:
1) Total number of people = 1000
Number of people who are between 10 and 30 years of age whose number of visits to the hospital is 3 = 57.
P(the person is between 10 and 30 years of age and whose Number of visits to the hospital is 3) 
2) Number of people below 60 years visiting the hospital twice
=26 + 60 + 36 =122
P(the person is below 60 years and visits the hospital twice)
3) Number of people who visit the hospital only once
= 16 + 86 + 52 + 12 = 166
P (the person visits the hospital only once) 
4) Number of people between 30 - 60 years visiting the hospital more than twice
= 64 + 50 + 50 = 164
P (the person is between 30 - 60 years and visits the hospital more than twice)
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Question (30):
A card is drawn from a well-shuffled pack of cards and its suit is noted. (spade, clubs, hearts or diamond). This trial is repeated 400 times and the Number of times each suit is drawn is given below.
When a card is drawn at random what is the probability that it is
1) a diamond
2) a black card
3) not a spade
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Answer:
1) Total number of trials = 400
Number of trials in which a diamond showed up = 76
P (the card drawn is a diamond)
2) Number of times a black card showed up = 96 + 108 = 204
P (the card drawn is black)
3) Number of times a card other than a spade showed up = 108 + 120 + 76 = 304
P (the card is not a spade)
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Question (31):
The monthly wages of 200 workers in a factory is given by the following table:
If one worker in the factory is chosen at random, what is the probability that
1) his wage is in the range Rs. 3500 - 4500?
2) his wage is Rs. 4000 or above?
3) Give two events in the content, one having probability 0 and the other having probability 1.
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Answer:
Total number of workers = 200
1) Number of workers whose wage is between Rs. 3500 and Rs. 4500
= 63 + 42 = 105
P (the worker's wage is between Rs. 3500 and Rs. 4500)
2) Number of workers with wage above Rs. 4000
= 42 + 19 = 61
P (the worker's wage is above Rs. 4000)
3) Let E1 be the event "the wage of the worker is less than Rs. 3000".
Since there is no worker whose wage is less than Rs. 3000, P(E1) = 0.
Let E2 be the event: "the wage is not less than Rs. 3000".
As the wages of all the workers considered is greater than or equal to Rs. 3000, P(E2) = 1
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Question (32):
In a cricket match, the number of dot-balls bowled in an over was noted down in the first 30 overs as follows:
What is the probability that an over bowled in the match?
1) has no dot balls
2) has not more than 2 dot balls
3) at least two dot balls?
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Answer:
Total number of overs = 30
1) Number of overs with no dot balls = 7
P (the over has no dot balls) 
2) Number of overs with not more than 2 dot balls
= 7 + 5+ 5 = 17
P (the over has not more than 2 dot balls) 
3) Number of overs with atleast 2 dot balls = 5 + 4 + 3 + 3 + 3 = 18
P (the over has atleast 2 dot balls) 
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Question (33):
100 mangoes are selected at random from each of 5 baskets of mangoes, and the number of mangoes which are spoilt is counted and recorded as follows:
When one such basket is checked what is the probability that it has
1) no spoilt mangoes
2) Atleast 10 spoilt mangoes
3) more than 20 spoilt mangoes
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Answer:
Number of baskets = 5
1) Number of baskets with no spoilt mangoes = 0
P (the basket has no spoilt mangoes) = 0
2) No of basket has no spoilt mangoes = 5
P (the basket has at least 10 spoilt mangoes) =1
3) No of baskets with more than 20 spoilt mangoes = 3
P(the basket has more than 20 spoilt mangoes)
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Question (34):
Consider the following frequency distribution table which gives the heights of 40 students in a class
Find the probability that the height of a student in the class
1) lies in the interval 150 - 155
2) is 145 cm or above 145 cm
3) is below 150 cm
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Answer:
Total Number of students = 40
1) Number of students whose height lies in the interval 150 - 155 cm = 12
P (the height of the student is in the interval 150 - 155) 
2) Number of students whose height is 145 cm or above 145 cm
= 11 +12 + 9 = 32
P (the height of the student is 145 cm or above) 
3) Number of students whose height is below 150 cm= 8 + 11 = 19
P (the height of the student is less than 150 cm) 
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Question (35):
Cards numbered 1 to 10 are placed in a box. One card is drawn and the number noted. This trial is repeated 500 times and the result is tabulated as follows:
If a card is drawn what is the probability that the card no is
1) a prime number
2) an even number
3) not less than 8
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Answer:
Number of trials = 500
1) Number of trials in which the card drawn has a prime number
= 34 + 52 + 54 + 60 = 200
P(the card number is prime) 
2)Number of trials in which the card drawn has an even number
= 34 + 48 + 36 + 62 + 70 = 250
P(the card number is even) 
3)Number of trials in which the card number is greater than 6
= 60 + 62 + 56 + 70 = 248
P (card number is greater than 6) 
4) Number of trials in which the card number is not less than 8
= 62 + 56 + 70 = 188
P (card number is not less than 8) 
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Question (36):
An organisation selected 2400 families at random and surveyed than to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:
Suppose a family is chosen, find the probability that the family chosen is
i) earning Rs. 10000 - Rs. 13000 per month and owning exactly 2 vehicles
ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle
iii) earning less than Rs. 7000 per month and not owning any vehicle
iv) earning Rs. 13000 - 16000 per month and owning more than 2 vehicles
v) owning not more than 1 vehicle.
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Answer:
Total number of families = 2400
i) Number of families earning Rs. 10000 - Rs. 13000,
and owning 2 vehicles = 29
P (the family is earning Rs. 10000 - Rs. 13000 and owns 2 vehicles)
ii) Number of families earning Rs. 16000 or more
and owning 1 vehicle = 579
P (the family is earning Rs. 16000 or more and own 1 vehicle)
iii) Number of families earning less than Rs.7000
and not owning any vehicle = 10
P (the family is earning less than Rs. 7000 and does not own any vehicle)
iv) Number of families Rs. 13000 - Rs. 16000
and owning more than 2 vehicles =25
P (the family is earning Rs. 13000 - Rs. 16000 and owns more than 2 vehicles)
v) Number of families owning not more than 1 vehicle
= 10 + 10 + 1 + 2 + 1 + 160 +305 +535 +469 +579
= 2062
P (the family owns not more than 1 vehicle)
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Question (37):
If E1, E2, E3 cover all possible outcomes of a trial, and P(E1) = 0.7, P (E2) = 0.05. What is the probability of E3?
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Answer:
P(E1) + P(E2) + P(E3) = 1
P(E3) = 1 - (0.7 + 0.05)
= 1 - 0.75 = 0.25
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Question (38):
In a survey conducted among 300 students each of whom could speak either Hindi or English or both, it was found that 200 of them could speak Hindi and 250 of them could speak English. If a student is selected at random, what is the probability that he can speak both the languages?
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Answer:
Total number of students = 300
Number of students who can speak both languages
= 200 + 250 - 300 = 150
P (the student can speak both languages)
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