The A.M. of a 50 set of numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is

38.5

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37.5

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36.5

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Solution

The correct option is B 37.5
$G i v e n, \frac{\sum x_{i}}{50} = 38, ∴ \sum x_{i} = 1900$
$N e w v a l u e o f \sum x_{i} = 1900 - 55 - 45 = 1800, n = 48$
$∴ N e w m e a n = \frac{1800}{48} = 37.5$

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The A.M. of a 50 set of numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is

The correct option is B 37.5Given, ∑xi50=38,∴∑xi=1900 New value of ∑xi=1900−55−45=1800,n=48 ∴ New mean=180048=37.5

The correct option is B 37.5
$G i v e n, \frac{\sum x_{i}}{50} = 38, ∴ \sum x_{i} = 1900$
$N e w v a l u e o f \sum x_{i} = 1900 - 55 - 45 = 1800, n = 48$
$∴ N e w m e a n = \frac{1800}{48} = 37.5$