The A.M. of a set of 50 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

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36.5

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37.5

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38.5

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Solution

The correct option is C
37.5

$w e h a v e, \frac{\sum x_{i}}{50} = 38, ∴ \sum x_{i} = 1900 N e w v a l u e o f \sum x_{i} = 1900 - 55 - 45 = 1800 a n d n = 48 ∴ n e w m e a n = \frac{1800}{48} = \frac{450}{12} = \frac{225}{6} = 37.5.$

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The A.M. of a set of 50 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 C 37.5 we have,∑xi50=38, ∴ ∑xi=1900New value of ∑xi=1900−55−45=1800 and n=48∴ new mean=180048=45012=2256=37.5.

The correct option is C
37.5

$w e h a v e, \frac{\sum x_{i}}{50} = 38, ∴ \sum x_{i} = 1900 N e w v a l u e o f \sum x_{i} = 1900 - 55 - 45 = 1800 a n d n = 48 ∴ n e w m e a n = \frac{1800}{48} = \frac{450}{12} = \frac{225}{6} = 37.5.$