# Ball bearings

One bearing is selected from the shipment of ball bearings. It is known from previous deliveries that the inner bearing radius can be considered as a normal distribution of N (µ = 0.400, σ2 = 25.10^−6). Calculate the probability that the selected radius will exceed 0.405 for the selected bearing.

Result

p =  0.159

#### Solution:

$µ = 0.400 \ \\ σ^2 = 25 \cdot \ 10^{ -6 } \ \\ σ = \sqrt{ 25 \cdot \ 10^{ -6 } } = 0.005 \ \\ \ \\ p = N(x>0.405,µ,σ) \ \\ \ \\ \ \\ p = 0.1587 = 0.159$

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