Question

3. A mix beverage machine releases a certain amount of syrup into a chamber where it is mixed with carbonated water. The amount of syrup follows a normal distribution with a mean of 1.29 fl.oz. and a standard deviation 0.016 fl.oz.

a. Find the probability that a syrup amount does not exceed 1.33 fl.oz. (10)

b. Find the probability that a syrup amount is less than 1.29 fl.oz. (10)

c. Find the probability that a syrup amount exceeds the mean value by more than two standard deviations. (10)

d. Find the syrup amount so that the probability that it is exceeded is 5%. (10, bonus)

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	1.29
std deviation   =σ=	0.016

a)

probability =P(X<1.33)=(Z<(1.33-1.29)/0.016)=P(Z<2.5)=0.9938

b)

probability =P(X<1.29)=(Z<(1.29-1.29)/0.016)=P(Z<0)=0.5

c)

probability =P(X>1.322)=P(Z>(1.322-1.29)/0.016)=P(Z>2)=1-P(Z<2)=1-0.9772=0.0228

d)

for 95th percentile critical value of z=		1.645
therefore corresponding value=mean+z*std deviation=1.29+1.645*0.016 =			1.3163