Question

Statistics - Please help. Show work. Thank you!

6. Assume the life of a roller bearing follows a Weibull distribution with parameters ß = 2 and 8 = 7,500 hours. (a) Determine the probability that the bearing will last at least 8000 hours. (b) Verify your answer using R. (c) Determine the expect value and the variance.

Answer 1

ANSWER::

6Q)

Let XV Weibullimi B The probability density funcHon f(x) = ) at cé(carp); x70 xro B>0 ; otherwise The distribution function is Fx (u) = foxsu) - do ė (278) mean of x- B vanance of x = ß2 lita - (lita)'). 1 Now, je B=2, 8=7000 9-2202 = 7000 last . a probability that the bearing will 8000 hours. » P(x78000) = L-P(x=8000) El- 1 t ė (8000/7000) - (8/7) = ē 1.3061 = 0.2709 cs Scanned with CamScanner

b) By using R - isoftware p x > 8000) = 1 - P(x = 8009 - 1-PWeibull ( 8000, shap=2, scale=1000) = 1- 0.72913 = 0.27087 = 0.2709 c) mean 04 x = mean B 5+1 - 7000 +1 = 7000 I 11 = 70001 E = 3500 (JT) = 6203.5885 variance of x = ( 3+1 - +)%) = (7000)? ( 2 - (3/2)2) = (70002) (1- (IT) = 10 515 490 Scanned with CamScanner

   (OR) TRY THIS ANSWER

6Q. Assume the life of a roller bearing follows a Weibull distribution with parameters

β = 2 and

δ = 7,500 hours.

Solution:

a) The probability that a bearing lasts at least 8000 hours:

P(X ≥ 8000) = 1 - e​​​​​​-(X/δ)^β

P(X ≥ 8000) = 1 - e​​​​​​-(8000/7500)^2

P(X ≥ 8000) = 1 - e​​​​​​-(1.138)

P(X ≥ 8000) = 1 - 0.3205

P(X ≥ 8000) = 0.6795

Therefore, the probability that a bearing lasts at least 8000 hours would be 0.6795.

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