The normal distribution described by this problem is #N(0.400, 0.001^2)#.
The range of acceptable bearing sizes in this problem is given by #0.399 +- 0.0015#. That equates to the interval #[0.3975. 0.4005]#.
To determine the percentage, it will be helpful for us to convert these two endpoints from this normal distribution into z-scores on the standard normal distribution #N_s (0, 1^2)#. We can then use z-score tables (or a z-score calculator) to determine what the area under the standard normal distribution curve is between those two z-scores.
Recall the z-score formula:
#z = (x-mu)/sigma#
Thus:
#z_0.3975 = z_1 = (0.3975-0.400)/0.001 = -2.5#
#z_0.4005 = z_2 = (0.4005-0.400)/0.001 = 0.5#
Next, we turn to a z-score table to find the cumulative from #-oo# area measures for each of the z-scores. Once we have those, we simply subtract #z_1# from #z_2# to find the area between those two z-scores:
#P(Z < z_1) = P(Z < -2.5) = 0.0062#
#P(Z < z_2) = P(Z < 0.5) = 0.6915#
#{:(P(z_1 < Z < z_2),=, P(-2.5 < Z < 0.5)),
(,=, P(Z < 0.5) - P(Z < -2.5)),
(,=,0.6915-0.0062),
(,=,0.6853):}#
68.53% of bearings can be used on the machine.
