g Find the required probabilities using the exponential density function f(t) = 1 λ e−t/λ, [0, [infinity]). The time t (in years) until failure of a component in a machine is exponentially distributed with λ = 3.5. Find the probabilities that the lifetime of a given component will be less than 1 year, more than 2 years but less than 4 years, and at least 5 years. (Round your answers to three decimal places.) (a) less than 1 year (b) more than 2 years but less than 4 years (c) at least 5 years

(a) less than 1 year =0.751488

(b) more than 2 years but less than 4 years = 0.2458

(c) at least 5 years= 0.760331

Step-by-step explanation:

f(t) = 1 λ e−t/λ, [0, [infinity])

First we calculate the probability for an exponential random variable X with parameter λ

P (X= t) = ∫ 1 λ e−t/λ, dt

P (x=t) =  e−t/λ,

λ = 3.5

Now P (X< 1 ) = ∫ 1 λ e−t/λ, dt        [the limits are (-∞ to 1)]

                         = e−1/3.5= 0.751488

P ( 2<X< 4 ) =  ∫ 1 λ e−t/λ, dt         [ the limits are (2 to 4)]

                     =e−2/3.5- e−4/3.5

                  = e-0.57142-e-1.142857  

                  = 0.5647-0.31890

                  = 0.2458

P (at least 5)  = 1- P (x=5)

                   = 1-∫ 1 λ e−t/λ, dt           [ the limits are (∞ to 5)]

                     = 1- e−5/3.5

                     = 1 - e-1.4285

                     = 1-0.239

                        = 0.760331

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )