There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is :
a. 46.5
b. 49.5
c. 53.5
d. 56.5
Solution:
It is given that
n = 50
\(Mean\: \overline{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}\overline{x}=\frac{1}{50}\times \sum_{i=1}^{50}x_{i} \) …. (1)
\(\sum_{i=1}^{50}x_{i} =50\overline{x}\)
By subtracting 53 from each observation, we get x̄new
x̄new = [(-x1 + 53) + (-x2 + 53) + …. + (-x50 + 53)]/50
Substituting the values
-3.5 = [-(x1 + x2 + ….. + x50) + (53 + 53 + ….. + 50 times)]/50
-3.5 × 50 = -(x1 + x2 + ….. + x50) + 53 × 50
So we get
\(\sum_{i=1}^{50}x_{i} \) = 2650 + 175 = 2825
We know that
Mean of 50 observations = 1/50 \(\sum_{i=1}^{50}x_{i} \)
Substituting the value
= 1/50 × 2825
= 56.5
Therefore, the mean of the given numbers is 56.5.
✦ Try This: There are 20 numbers. Each number is subtracted from 23 and the mean of the numbers so obtained is found to be -2.5. The mean of the given numbers is :
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 14
NCERT Exemplar Class 9 Maths Exercise 14.1 Problem 19
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is : a. 46.5, b. 49.5, c. 53.5, d. 56.5
Summary:
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is 56.5
☛ Related Questions:
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