Question 1073050
Without a calculator, this is difficult, because even the approximation of 50% is not too accurate.
One could use a normal approximation which barely makes the assumptions valid.
E(x)=5.6, and V(x)=5.6*0.44=0.2464, almost 0.25, so sd is 0.5
5 would be about 1 sd to the left (1.2 sd exactly) and about 0.31 or a little more
at least 6 would be everything to the right of z=0.8, which is a little less than 0.75.
less than 4 would be z< 3.1 sd and this would be expected to be near 0. The true values are considerably different:
The exact number is 10C5(0.56)^5(0.44)^5
This is not too dissimilar from getting 5 heads in 10 coin tosses.
That is 252*(1/1024) or about a quarter.  The exact answer is 0.2289
For at least 6, it would more, but the values of 5,4,3,2,1,0 would not be relevant 0.2427+0.1765+0.0843+0.0238+0.003, or 0.5276
For less than 4, it would be 3,2,1,0 and those probabilities are roughly 175/1024, a little less, since the probability of 0.44 is less than 0.50 that I am using to estimate.
0.0002+0.0035+0.0198+0.0672+=0.0907