Question

A scientist is weighing each of $30$ fishes. Their mean weight worked out is $30g$ and a standard deviation is $2g$. Later, it was found that the measuring scale was misaligned and always under-reported every fish weight by $2g$. Find the correct mean and standard deviation (in gram) of fishes.
$A)28,4$
$B)32,2$
$C)32,4$
$D)28,2$

Answer 1

First, we have two linear expressions for one for mean and another for standard deviation.
Mean $({x_i} + \lambda ) = \overline x + \lambda$ , the standard deviation $({x_i} + \lambda )$ remains the same.
We need to find the correct mean and standard deviation of the given question using the given formula.
Complete step-by-step solution:

The weighted mean of each fish is $30$
Standard deviation is $2$ and the error in the measuring weight is $2g$ .
Also, the given that every fish weight is increased (misaligned weight) by the grams as $2g$
Hence the overall mean can be calculated by the old mean addition of the error alignment on the fishes.
Therefore, the mean weight is $30 + 2 = 32g$
Since the standard deviation is unchanged because it will be not affected by the addition of the constant. And thus, the standard deviation will have remained the same as given in the given $2g$
Therefore, we get the corrected mean and standard deviation as $32g,2g$
Thus, the option $B)32,2$ is correct.

Note: The standard deviation and the variance are not changed when each number is either increased or decreased by some constant. But when each variable is multiplied by the constant $\lambda$ , is the new standard deviation and the variance.
Also, we know about how changes in one of mean, variance, or standard deviation can affect the others.
Thus, the corrected mean and standard deviation is found above.