A scientist is weighing each of $30$ fishes. Their mean weight worked out is $30 g m$ and a standard deviation of $2 g m$ . Later, it was found that the measuring scale was misaligned and always under reported every fish weight by $2 g m$ . The correct mean and standard deviation of fishes are :

28, 2

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28, 4

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32, 2

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32, 4

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Solution

The correct option is B $32, 2$
Given that mean weight of fishes is $30 g m$

Standard deviation is $2 g m$

Given that every fish weight is increased by $2 g m$

Therefore, over all mean weight is $30 + 2 = 32 g m$

Standard deviation is not effected by the addition of a constant. Therefore standard deviation remains same.

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A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a standard deviaiton of 2 gm. Later, it was found that the measuring scale was misaligned and always under-reported every fish weight by 2gm. The correct mean and standard deviation (in gm) of fishes are respectively:

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A scientist is weighing each of 30 fishes. Their mean weight worked out is 30gm and a standard deviation of 2gm. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2gm. The correct mean and standard deviation of fishes are :

The correct option is B 32,2Given that mean weight of fishes is 30gmStandard deviation is 2gmGiven that every fish weight is increased by 2gmTherefore, over all mean weight is 30+2=32gmStandard deviation is not effected by the addition of a constant. Therefore standard deviation remains same.

A scientist is weighing each of $30$ fishes. Their mean weight worked out is $30 g m$ and a standard deviation of $2 g m$ . Later, it was found that the measuring scale was misaligned and always under reported every fish weight by $2 g m$ . The correct mean and standard deviation of fishes are :