1. X~B(400, 2/5)
But since the number is way too big, and (2/5)(400) > 50, we can use a normal approximation.
Mean = np = 2/5 x 400 = 160
Variance = npq = 2/5 x 3/5 x 400 = 96
Therefore, X~N(160,96)
You will want to find P(X<165). Note to use continuity correction as you are approximating from binomial.
z= (x-u)/sd = (164.5-160) / rt96 = 0.4593
Using the normal dist. table; a z value of 0.4593 corresponds to 0.6768, and theres your answer
- 2nd question
i) Probability of getting an orange disc = 1/3
Probability of not getting an orange disc = 2/3
5 picked out, with replacement, so P(X=0) = (2/3)^5 = 0 .132
ii) there are 10x3 discs in total that end with a 6.
so 5C2 (number of possible combinations in which the discs are picked out) x (30/300)^2 x (270/300)^3 = 0.0729
iii) Again, apply the probability. 10 orange discs ending with 6 out of 300 total discs = 10/300
5C2 x (10/300)^2 x (290/300)^3 = 0.01
iv) Mean = np, variance = npq
where n=5
p= 1/3
q=2/3
Pop numbers in and you should get mean= 5/3 and variance = 10/9
Hope that helps
good luck for your exam!