A manufacturer guarantees a product for 1 year. The lifespan of the product after it is sold is given by the probability density function below, where t is time in months. f(t) = { 0.008e ^ -0.009t if t >= 0 0 otherwise What is the probability that a buyer chosen at random will have a product failure (A) During the warranty? (B) During the second year after purchase? (A) What is the probability that the product will fail within one year? (Round to three decimal places as needed.)
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To do this, we need to integrate the probability density function from 0 to 12: $$P(0 \leq t \leq 12) = \int_{0}^{12} 0.008e^{-0.009t} dt$$ Now, we can integrate: $$P(0 \leq t \leq 12) = -\frac{8}{9}e^{-0.009t} \Big|_0^{12}$$ $$P(0 \leq t \leq 12) = Show moreā¦
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A manufacturer guarantees a product for 1 year. The time to failure of the product after it is sold is given by the probability density function $$ f(t)=\left\{\begin{array}{ll} .01 e^{-.01 t} & \text { if } t \geq 0 \\ 0 & \text { otherwise } \end{array}\right. $$ where $t$ is time in months. What is the probability that a buyer chosen at random will have a product failure (A) During the warranty period? (B) During the second year after purchase?
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