(a) P(x <= 5000) = convert to z-distribution: z = (x - 4700)/430

P( z <= 2.3077) = 0.9895 is the probability that the device will wear out in 5000 hours or less.

(b) If the guarantee period is x_0 hours, then the corresponding z-score will be z_0 = (x_0 - 4700)/430 and satisfy the condition P(z <=z_0) = 0.12

From here, z_0 = -1.17499 = (x_0 - 4700)/430 and x_0 = 4700 + 430*(-1.17499) = 4195 hours may be guaranteed under the required condition.

this is for my homework. it says your answers are incorrect...

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