Math statistics question regarding min and max values

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 = 194, p = 0.16

the answer is Minimum: 20.83; maximum: 41.25. but how do i get it. thanks :)

We have n = 194, p = 0.16, and q = 1 - p = 0.84

The mean is np = 194(0.16) = 31.04

The standard deviation is √(npq) = √(31.04)(0.84)

= √26.0736 = 5.106231487

µ - 2σ = 31.04 - 2(5.106231487) = 20.82753703

µ + 2σ = 31.04 + 2(5.106231487) = 41.25246297