N(W, 4) (normal with mean W and variance 4)

2kg is 1 sigma, so we are asking that the probability of z=(R-W)/sigma(c) What is the probability that the reading is 2kg more than what the person actually weighs?

(which has a standard normal distribution) be greater that 1. So we look

1 up in the cumulative standard normal distrinution to find

P(z<=1) = 0.8413,

then:

P(z>1) = 1-0.8413 = 0.1587

RonL