Answer:
a) The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.
b) The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.
Step-by-step explanation:
Given : The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.
To find : What is the probability that
(a) the total gross sales over the next 2 weeks exceeds $5000;
(b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?
Solution :
Let
and
denote the sales during week 1 and 2 respectively.
a) Let
Assuming that
and
follows same distribution with same mean and deviation.




So, 





The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.
b) The probability that sales exceed teh 2000 and amount in at least 2 and 3 next week.
We use binomial distribution with n=3.





Let Y be the number of weeks in which sales exceed 2000.
Now, 
So, 



The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.
4a^2+7ab-b^2
Step-by-step explanation:
Add the first two equations together:
3a^2-ab-2b^2
+ 2a^2+5ab-3b^2
---------------------------
5a^2+4ab-5b^2
Subtract that answer from the remaining trinomial:
5a^2+4ab-5b^2
- a^2-3ab-4b^2
--------------------------
4a^2+7ab-b^2
The answer will be a because if you add 70 +98+83+66+41 divided by 5 is 325.2 and I'd u add 81 and divide it by 6 then it will equal 371.5 so it will increase