In: Statistics and Probability
The thickness of a cardboard sheet has a normal distribution with a mean of 3 cm and a standard deviation of 0.1 cm. 60 of these sheets are stacked on top of each other to transport to the next operation in the process.
a) Determine the probability distribution for the total height of the stack of 60 cardboard sheets. Give the name of the distribution and its parameter(s).
b) The trailer used to transport the cardboards sheets has a capacity for a stack of height 181 cm. What is the probability that all 60 sheets won’t fit?
c) Upon further study, it is decided the thickness of the cardboard is better modeled by a gamma distribution with k = 900 and v = 300. Now what is the probability that the capacity of the trailer is exceeded? (An approximate probability is fine.)
a)
The sum of 'n' Normal distribution N(, ) is a normal distribution with mean n and variance n.
Let H be the height of the stack of 60 cardboard sheets. Then H ~ N( = 60 * 3 = 180, = 60 * 0.12 = 0.6)
The distribution is Normal distribution with mean 180 cm and variance of 0.6 cm2.
b)
Standard deviation of H is = 0.7746
Probability that all 60 sheets won’t fit = P(H > 181) = P[Z > (181 - 180)/0.7746]
= P[Z > 1.29]
= 0.0985
c)
The sum of 'n' Normal distribution Gamma(k, v) is a Gamma distribution with parameters nk and v
Thus, H ~ Gamma(k = 60 * 900 = 54000, v = 300)
Mean = k/v = 54000 / 300 = 180
Variance = k/v2 = 54000 / 3002 = 0.6
Using Normal approximation of Gamma distribution, H ~ N( = 180, = 0.6)
Probability that all 60 sheets won’t fit = P(H > 181) = P[Z > (181 - 180)/0.7746]
= P[Z > 1.29]
= 0.0985