9. The boom is intended to support two vertical loads, F₁ and F₂. F₁ and F₂ independently follow normal distributions as F₁ ~ N(1500, 50²) lb and F₂ ~ N(600, 25²) lb. (1) What is the distribution of the maximum load of cable CB; (2) If the cable CB can sustain a maximum load of 1400N before it fails, what is the probability that the cable may fail?

Solution

(1)

From above equation, we have

With F₁ ~ N(1500, 50²) lb and F₂ ~ N(600, 25²) lb,

Thus, the distribution of load of cable CB is: T_max ~ N(1308.4, 34.06²) lb. Ans.

(2) The probability of failure is

Ans.