Question

Use the given values of n and p to find the minimum usual value μ-2σ and the maximum usual value μ+2σ. Round your answer to the nearest hundredth unless otherwise noted. n = 2112, p = 3/4A) Minimum: 1478.7; maximum: 1689.3 B) Minimum: 1544.2; maximum: 1623.8 C) Minimum: 1564.1; maximum: 1603.9 D) Minimum: 1623.8; maximum: 1544.2

Answer 1

Here it is given that n = 2112 and p = 3/4.

So, the mean is given by

And the standard deviation is given by

Now, the maximum value is mean plus two standard deviations and the minimum value is mean minus two standard deviations.

So, the max value is

=1584+2*19.9

=1584+39.8

=1623.8

The minimum value is

=1584-2*19.9

=1584-39.8

=1544.2

So, the correct answer is option (B) 1544.2 and 1623.8.