
Weibull or Poisson?
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

Re: Weibull or Poisson?
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.

Re: Weibull or Poisson?
yeh sorry i got the title totally wrong lol, so A seems correct? In my head i just can't seem to understand how if the mean is 8 months then why would 98.6% be probable to exceed 12 months?

Re: Weibull or Poisson?
the mean of the log(X) is 8 months, that isn't the same as the mean of X.
The mean of X is $\displaystyle e^{8 + \frac{2.5^2}{2}}$ which is more than 60,000 months.

Re: Weibull or Poisson?
Ah thank you so much
for the last part:
Determine the lifetime that will be exceeded by 80% of the bearings.
I got 5.9, i assume that is 5.9 months as an answer?

Re: Weibull or Poisson?
log(X) is greater than 5.9 months 80% of the time
so 80% of the time, X is greater than $\displaystyle e^{5.9}$

Re: Weibull or Poisson?
Yeh they do seem very large, So would my answer be 5.9 months or e^5.9 months?

Re: Weibull or Poisson?

Re: Weibull or Poisson?
Sorry i'm slightly confused.
If The mean of X is more than 60,000 months.
then the lifetime that will be exceeded by 80% of the bearings surely wouldn't be e^5.9 (365 months)?

Re: Weibull or Poisson?
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  WolframAlpha
The graph continues for far beyond the range i have plotted as well, before the values become ignorably small.

Re: Weibull or Poisson?
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

Re: Weibull or Poisson?
X=door
Y=frame
define: A=XY (so door wont fit if A>0)
if X,Y independant and $\displaystyle X \sim N(\mu_x,\sigma_x^2)$ and $\displaystyle Y \sim N(\mu_y,\sigma_y^2)$
then it can be shown that:
$\displaystyle A \sim N(\mu_x\mu_y,\sigma_x^2+\sigma_y^2)$

Re: Weibull or Poisson?
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 = X1X2
E(Y) = C*Mew1  C*Mew3 = 6160.64 = 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?