Guide 5 Chapter 10 Page 172

[kpkanupriyakhmi]

Q 14:- Set X consists of 100 bags of rice with an average weight of 90 pounds........

[n00bpron00bpron00b]

Hi Tommy/kpkanupriyakhmi,

I am not quite sure about this problem, here is my approach.

If the average wt of 100 bags is 90 ( mean m = 90)
Standard Deviation of the weights is 8

So possible range

...66-74-82-90(mean)-98-106-114....

Bag A weights 2 standard deviation below the avg. wt = 88

Bag B weights 5 standard deviation above the avg. wt = 95

Quantity A :

Twice the difference between the weight of bag B and the weight of bag A

2(95-88) = 2(7) = 14

Quantity B:

Range of weights of the bags of rice in set X

Total bags = 100
avg wt = 90
sum = 100*90 = 90,00

How are we supposed to find conflicting range values for such a large number to prove "D" given the time constraint ?

if the lower benchmark is say 82 (82 x 95 bags) = 7790
higher benchmark is say 242 (i just kept adding standard deviation of 8 to get 242) 242 x 5 = 1210

7790 (95 bags) + 1210 (5 bags) = 9,000 (100 bags)

Range = 242 - 82 = 160 . Quantity B greater

Similarly to find range value that lies below 14 to prove "D"

[tommywallach]

Hey Guys,

Remember that set X is actually a defined set here, not a variable term.

So we know that Bag A is at 74 (two SDs below 90 -- you interpreted this as 2 POUNDS below, Noob, but every SD here is 8 pounds).

Bag B is thus 95.

Quantity A: 21 * 2 = 42

Quantity B: No way to know. Standard Deviations don't tell you the total range, but they allow you to group the data into sets.

Thus, the answer is (D).

-t

[n00bpron00bpron00b]

Hi Tommy,

Understood my mistake. Thanks a lot.

- noob

[tommywallach]

Glad to help!