Solomon S1-Paper B

Qu7) The volume of liquid in bottles of sparkling water from one producer is believed to be normally distributed with a mean if 704ml and a variance of 3.2ml^2.
Calculate the probabilty that a randoly chosen bottle from this producer contains
(a) more than 706ml,
(b) between 703 and 708ml.

I have tried them, and got for part (a) p=0.3686, contrary to the paper's answer which states the p=0.1314

And for part (b) i got p=0.7005, and the paper states the answer as 0.6998.

---
Please, attempt to include relevant information that may lead me to understand where i went wrong or where the paper went wrong :biggrin:

Reply 1

Simba

"Qu7) The volume of liquid in bottles of sparkling water from one producer is believed to be normally distributed with a mean if 704ml and a variance of 3.2ml^2.
Calculate the probabilty that a randoly chosen bottle from this producer contains
(a) more than 706ml,
(b) between 703 and 708ml."

a) Let X ~ N(704, 3.2).

P(X > 706) = P(Z > (706 - 704)/(3.2^(1/2))) = P(Z > 1.118...) = 1 - P(Z < 1.118...) = 1 - 0.868224... = 0.13177... = 0.132 (3 s.f.)

b) P(703 < X < 708) = P((703 - 704)/(3.2^(1/2)) < Z < (708 - 704)/(3.2^(1/2))) = P(-0.5590... < Z < 2.2360...) = P(Z < 2.2360...) - P(Z < -0.5590...) = P(Z < 2.2360...) - (1 - P(Z < 0.5590...)) = P(Z < 2.2360...) + P(Z < 0.5590...) - 1 = 0.987326... + 0.711925... - 1 = 0.69925... = 0.699 (3 s.f.)

Generally, if you're pretty close to the answer (as in part b), then they will give it as correct :smile: