Let's see if we have the right idea:
You calculate zhi , zlo (z = (x - μ)/σ ) and use the normal distribution to find the area between the z values: A (-inf to zhi) - A(-inf to zlo). I get .563. Probability of any one outside of limits is 1-P = .437
2 or more out of 6 with a weight outside is the same as 5 or 6 out of 6 inside the range.
Bernoulli Dist:
6C0 (.563)6 + 6C1(.563)5(.437)
( The first term is all 6 inside, 2nd is 5 inside and 1 out which happens in 6 ways.)
Good luck. You may have misunderstood the question as this probability should be < .5