Answer to Question #330660 in Statistics and Probability for mau

Question

Assumed that the fill amount in 2-liter soft drink bottles is normally distributed, with a mean of 2.0 liters and a standard deviation of 0.05 liter. If bottles contain less than 95% of the listed net content (1.90 liters, in this case), the manufacturer may be subject to a penalty by the state office of consumer affairs. Bottles that have a net content above 2.10 liters may cause excess spillage upon opening. What proportion of the bottles will contain (a) between 1.90 and 2.00 liters? (b) between 1.90 and 2.10 liters? (c) below 1.90 liters or above 2.10 liters? (d) At least how much soft drink is contained in 99% of the bottles? (e) 99% of the bottles contain an amount that is between which two values(symmetrically distributed) around the mean?

Answer 1

X~$N(2, 0.05^2)$

a) $P(1.9<X<2)=P(1.9<N(2,0.05^2)<2)=P(1.9<2+0.05Z<2)=P(-2<Z<0)=0.5-0.02275=0.47725$

b) due to the symmetry of the normal distribution, $P(1.9<X<2.1)=2P(1.9<X<2)=2*0.47725=0.9545$ .

c) $P(X<1.9\,or\,X>2.1)=1-P(1.9<X<2.1)=1-0.9545=0.0455$

d) $P(X<a)=0.99\implies P(Z<{\frac {a-2} {0.05}})=0.99\implies {\frac {a-2} {0.05}}=2.29\implies a=2.1145$ litres.

e) let's find upper bound

$P(X<b)=0.995\implies P(Z<{\frac {b-2} {0.05}})=0.995\implies {\frac {b-2} {0.05}}=2.58\implies b=2.129$ litres.

Thus, due to the symmetry, the lower bound would be c=2-(2.129-2)=1.871 litres.

So, the sought interval is (1.871, 2.129).