Question

The life of a toy is normally distributed. Suppose 92.51% of the items lives exceeding 2,160 hours and 3.92% have lives exceeding 17,040 hours. Find the mean and the standard deviation.

Answer 1

From standard normal tables, we have:
P(Z < -1.440) = 0.0749,

Therefore P(Z > -1.440) = 1 = 0.0749 = 0.9251

Also, from standard normal tables, we have:
P(Z < 1.760) = 0.9608, therefore P(Z > 1.760) = 1 - 0.9608 = 0.0392

Therefore, we have here:
Mean - 1.44Std Dev = 2160
Mean + 1.760Std Dev = 17040

Subtracting the first equation from the second one, we get here:
(1.76 + 1.44)Std Dev = 17040 - 2160
3.2 Std Dev = 14880
Std Dev = 4650

Mean = 2160 + 1.44*Std Dev = 2160 + 1.44*4650 = 8856

Therefore 8856 is the required mean here and 4650 is the standard deviation.