What is the net force on the loop from the wire if the loop carries a clockwise current i2 = 50 a?
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Answer:
The percentage of legal quarters that will be rejected is 1.954 %.
Explanation:
Here we are required to find
P(X<5.48 or X>5.82)
We have μ = 5.67 and σ = 0.07 where the range is between 5.48 and 5.82 we each z-score given by
Therefore for 5.48 we have = -2.714
and for 5.82 we have = 2.143
Therefore we have the area of interest on the normal distribution chart given by
P(Z<-2.714) + P(Z > 2.143) = P(Z<-2.714) + (1 - P(Z < 2.143))
+ 0.00336 + 1 - 0.98382 = 0.01954
= 0.01954 × 100 = 1.954 %
Therefore 1.954 percentage of legal quarters will be rejected.