all are normal distributions

1) Sample size = 6. mean of 10 and variance of 2. Calculate the probability that 2 of the values are less than 9 and the other 4 are greater than nine.

I can find the probability of 1 being greater than or less than 9. That would just be (8-10)/(sqrt 2) right?

2)The sizes of claims are normal random variables. Mean = 1800. SD=400. Find probability that 2 random claims differ by more than 500.

3) n = 80 they are independent. indiv mean = 574 with sd = 186. P[average claim <565]

P[Z<565-574/(Sqrt(186^2/80)). that gave me -.433 as my Z and then i took 1-.6664 (chart value of .433). I got .3336 as my answer.

I got (sqrt(186^2/80) as my variance for each average term. the math was 1/80^2 * var(s)= 80*Var/(80^2). Var=SD^2 so that gave me (sqrt(186^2/80)

4) X = rate of interest. mean=.06 SD=.015 Amount in account after year => Y=10,000 e^x Find E[x] after a year