Probability

1.
(a)
Normal table used :
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

(i)
The required probability
= P(x > 5.2)
= P(z > [5.2-5]/0.14)
= P(z > 1.429)
= 1 - P(z < 1.429)
= 1 - 0.9235
= 0.0765

(ii)
P(4.85 < x < 5.15)
= P([4.85-5]/0.14 < x < [5.15-5]/0.14)
= P(-1.071 < x < 1.071)
= P(x < 1.071) - P(x < -1.071)
= 0.8579 - 0.1421
= 0.7158

(iii)
P(z < 2.054) = 98%
P(z ≥ 2.054) = 2%
P(x ≥ [5+2.054*0.14]) = 2%
P(x ≥ 5.29) = 2%
Expected minimum weight = 5.29 grams

(b)
A: arrive within specified time
A': not arrive within specified time

P(A') = 2% = 0.02
P(A) = 1 - 0.02 = 0.98

(i)
The required probability
= P(1A'+9A)
= 10C1 * 0.02 * 0.98⁹
= 0.1667

(ii)
The required probability
= P(10A) + P(1A'+9A)
= 0.98¹⁰ + 10C1 * 0.02 * 0.98⁹
= 0.9838