Question 1137234
Rather than using a normal approximation, I would use a binomial, since np is <5
11% of 48 is 5.28, so want probability 6 or more are defective or 1-probabilty 5 or fewer are defective
0: 0.938^48=0.0463
1:48*0.062*0.938^47=0.1469
2: 48C2*0.062^2*0.938^46=0.2283
3: 48C3*0.062^3*0.938^45=0.2313
4: 48C4*0.062^4*0.938^44=0.1720
5: 0.1001
sum is 0.8786
so answer of question is 0.1214.

Fewer than 2% is fewer than 1 since 2% is 0.96
That is 0.938^48 or 0.0463 from above.