The percentage of marks obtained by students of a class in mathematics are:
64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.
We have :
The numbers of children in 10 families of a locality are:
2, 4, 3, 4, 2, 3, 5, 1, 1, 5. Find the mean number of children per family.
The mean number of children per family =
Thus, on an average there are 3 children per family in the locality.
The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
We have:
n = The number of observations = 100, Mean = 40
⇒ 40 ×100 = Sum of the observations
Thus, the incorrect sum of the observations = 40 x 100 = 4000.
Now,
The correct sum of the observations = Incorrect sum of the observations  Incorrect observation + Correct observation
⇒ The correct sum of the observations = 4000  83 + 53
⇒ The correct sum of the observations = 4000  30 = 3970
∴
The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.
We have:
So, sum of the five numbers = 5 ×27 = 135.
Now,
The mean of four numbers =
So, sum of the four numbers = 4 ×25 = 100.
Therefore, the excluded number = Sum of the five numbers  sum of the four numbers
⇒ The excluded number = 135  100 = 35.
The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.
We have:
Let the weight of the seventh student be x kg.
Thus, the weight of the seventh student is 61 kg.
The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean?
Let x_{1}, x_{2}, x_{3}...x_{8} be the eight numbers whose mean is 15 kg. Then,
x_{1}, x_{2}, x_{3}...x_{8} = 15 x 8
⇒ x_{1}, x_{2}, x_{3}...x_{8} = 120
Let the new numbers be 2x_{1} , 2x_{2}, 2x_{3}, ...2x_{8. }Let M be the arithmetic mean of the new numbers.
Then,
The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Let x1,x2,x3,x4 & x5x1,x2,x3,x4 & x5 be five numbers whose mean is 18. Then,
18 = Sum of five numbers ÷ 5
∴ Sum of five numbers = 18 × 5 = 90.
Now, if one number is excluded, then their mean is 16.
So,
16= Sum of four numbers ÷ 4
∴ Sum of four numbers = 16 × 4 = 64.
The excluded number = Sum of the five observations  Sum of the four observations
∴ The excluded number = 90  64
∴ The excluded number = 26.
The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.
n = Number of observations = 200
⇒Sum of the observations = 50 × 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 = 10180.
The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.
We have:
Mean = Sum of five numbers ÷ 5
⇒ Sum of the five numbers = 27 x 5 = 135.
Now, New mean = 25
25 = Sum of six numbers ÷÷ 6
⇒ Sum of the six numbers = 25 x 6 = 150.
The included number = Sum of the six numbers  Sum of the five numbers
⇒The included number = 150  135
⇒The included number = 15.
The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.
Let x_{1}, x_{2}, x_{3}, ... x_{75} be 75 numbers with their mean equal to 35. Then,
The new numbers are 4x_{1}, 4x_{2}, 4x_{3}, ...4x_{75. }Let M be the arithmetic mean of the new numbers. Then,
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