im puzzled by your thread title. You are given a lognormal distribution, not a weibull or a poisson distribution.
your method in (A) looks right, i haven't checked the arithmatic.
Exercise 3:
The lifetime of a roller bearing follows a lognormal distribution. The mean and standard
deviation of the corresponding normal distribution are 8 months and 2.5 months respectively.
b) Calculate the probability that a bearing will last more than 1 year.
OK I tried the following two ways, which one is correct:
A) P(X > 12) = 1 - P(X < 12) = 1 - P(e^w < 12)
= 1 - P(w < ln(12))
= 1 - P(Z<((ln(12)-8)/2.5)
= 1 - P(Z< -2.21)
= 1 - 0.014
= 0.986
B) F(X) = 1 - exp [ -(x/δ)^β ]
P(X > 12) = 1 - F(8)
= exp[-(12/8)^2.5]
= e^-2.756
= 0.064
it depends on the shape of the distribution. This one is incredibly skewed:
plot 1/(2.5 *x*\sqrt(2 pi) ) *e^(((lnx -8)^2)/44), 0<x<132000 - Wolfram|Alpha
The graph continues for far beyond the range i have plotted as well, before the values become ignorably small.
OK, this is so helpful, I feel bad asking all these questions, I hope you don't mind though.
The width of a door frame is normally distributed with a mean of 61cm and a standard
deviation of 0.32cm. The width of a door is normally distributed with a mean of 60.64cm and
standard deviation of 0.16cm. Assume independence.
c) What is the probability that the door does not fit in the frame?
I tried to use P(Y<0.36) but i wasnt sure if i was meant to use a different way
The width of a door frame is normally distributed with a mean of 61cm and a standard
deviation of 0.32cm. The width of a door is normally distributed with a mean of 60.64cm and
standard deviation of 0.16cm. Assume independence.
a) Determine the mean and standard deviation of the difference between the width of the
frame and the width of the door.
b) What is the probability that the width of the frame minus the width of the door exceeds
0.64cm?
c) What is the probability that the door does not fit in the frame?
a) Frame = mean 61, sd 0.32
Door = mean 60.64, sd - 0.16
Y = X1-X2
E(Y) = C*Mew1 - C*Mew3 = 61-60.64 = 0.36cm
V(Y) = C^2*Sigmasquared1 - C^2*sigmasquared2 = 0.32^2 + 0.16^2 = 0.128
b) P(Y > 0.64) = P ((Y - mewy/sigma y) > (0.64 - 0.36)/0.128)
P(Y > 0.64) = P(z> 2.1875)
= 0.986
c) P( Y less than or equal to 0.36) ??? I don't really get this part, is it similar to B?