In: Statistics and Probability
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.
b. Find the probability that a syrup amount is less than 1.29 fl.oz.
c. Find the probability that a syrup amount exceeds the mean value by more than two standard deviations.
d. Find the syrup amount so that the probability that it is exceeded is 5%.
Solution :
Given that ,
mean = = 1.29
standard deviation = = 0.016
a) P( x< 1.33) = P[(x - ) / < (1.33 - 1.29) / 0.016 ]
= P(z < 2.50)
Using z table,
= 0.9938
b) P( x< 1.29) = P[(x - ) / < (1.29 - 1.29) / 0.016 ]
= P(z < 0)
Using z table,
= 0.5000
c) P(x > 1.322) = 1 - p( x< 1.322)
=1- p P[(x - ) / < (1.322 - 1.29) / 0.016]
=1- P(z < 2.00)
Using z table,
= 1 - 0.9772
= 0.0228
d) Using standard normal table,
P(Z > z) = 5%
= 1 - P(Z < z) = 0.05
= P(Z < z) = 1 - 0.05
= P(Z < z ) = 0.95
= P(Z < 1.645 ) = 0.95
z = 1.645
Using z-score formula,
x = z * +
x = 1.645 * 0.016 + 1.29
x = 1.32 fl.oz