Ten voltage measurements in a circuit have produced the following values: 49.3, 50.1, 48.9, 49.2, 49.7, 50.5, 49.9, 49.3, 49.8, 50.2 If the standard deviation for the above data set is 0.50 V, Which of the values can be rejected:

48.9 and 50.5 are rejected.

Step-by-step explanation:

Since the standard deviation shows the spread of the values of the voltages, we need to find the mean of the values and use the standard deviation to find the minimum and maximum voltage allowed.

So mean μ = Σx/N where x = individual values of voltage and N = number of values = 10

So, μ = Σx/N

= (49.3 + 50.1 + 48.9 + 49.2 + 49.7 + 50.5 + 49.9 + 49.3 + 49.8 + 50.2)/10

= 496.9/10

= 49.69

≅ 49.7 V

So, the acceptable values of voltage are in the range μ ± σ where σ = standard deviation = 0.50 V.

So, the acceptable values of voltage are in the range 49.7 ± 0.50 V

So, minimum value = 49.7 V - 0.50 V = 49.2 V

and maximum value = 49.7 V + 0.50 V = 50.2 V

So any voltage outside the range 49.2 V ≤ x ≤ 50.2 V is rejected.

There are two values outside this range which are 48.9 and 50.5. So these two values are rejected.

x=6 and x=3

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