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Chapter 18: Problem 720

Mean of n observations is \(\mathrm{m}\) and sum of \(\mathrm{n}-3\) observations is b then mean of remaining 3 observations is (a) \(n m+b\) (b) \([(\mathrm{nm}-\mathrm{b}) / 3]\) (c) \([(n m+b) / 3]\) (d) \(n m-b\)

Short Answer

Expert verified
The mean of the remaining three observations is \(\boxed{[(nm-b)/3]}\).

Step by step solution

01

Find the sum of n observations

First, we need to find the sum of all n observations. To do this, we can use the formula for the mean: mean = (sum of observations) / number of observations We are given that the mean of n observations is m, so we can write this equation as: \( m = \frac{sum\,of\,n\,observations}{n} \) Now, we can solve for the sum of n observations: \( sum\,of\,n\,observations = n\times m \)
02

Find the sum of the remaining three observations

To find the sum of the remaining three observations, we subtract the sum of the n-3 observations (which is given as b) from the sum of all n observations. So, we have: \( sum\,of\,remaining\,3\,observations = (n\times m) - b \)
03

Find the mean of the remaining three observations

Now that we have the sum of the remaining three observations, we can find their mean by dividing the sum by the number of remaining observations (which is 3). So, we get: \( mean\,of\,remaining\,3\,observations = \frac{sum\,of\,remaining\,3\,observations}{3} \) Substituting the expression for the sum of the remaining three observations from Step 2, we get: \( mean\,of\,remaining\,3\,observations = \frac{(n\times m) - b}{3} \) So, the mean of the remaining three observations is: \( \boxed{[(nm-b)/3]} \) Hence, the correct answer is (b).

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Most popular questions from this chapter

Two dice are rolled one after the other. The probability that the number on the first is smaller than the number on the second is (a) \((1 / 2)\) (b) \((7 / 18)\) (c) \((3 / 4)\) (d) \((5 / 12)\)

If \(n\) integers taken at random are multiplied together, then the probability that the last digit of the product is \(1,3,7\) or 9 is (a) \(\left(2^{n} / 5^{n}\right)\) (b) \(\left[\left(4^{n}-2^{n}\right) / 5^{n}\right]\) (c) \(\left(4^{n} / 5^{n}\right)\) (d) \((2 / 5)\)

If the mean deviation about the median of the observations a, \(2 a, \ldots \ldots . .50 a\) is 50 then \(|a|=\) (a) 2 (b) 3 (c) 4 (d) 5

Mean of sequence \(1,2,4,8,16 \ldots \ldots .2^{n-1}\) is (a) \(\left[\left(2^{n}-1\right) / n\right]\) (b) \(\left[\left(2^{n-1}-1\right) /(n-1)\right]\) (c) \(\left[\left(2^{n}+1\right) / n\right]\) (d) \(\left[\left(2^{n}-1\right) /(n-1)\right]\)

The probability of having at least one tail in 4 throws with a coin is (a) \((15 / 16)\) (b) \((1 / 16)\) (c) \((1 / 4)\) (d) \((1 / 8)\)
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