Can someone help me on this play
2 answers:
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Answer:
Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .
Step-by-step explanation:
We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;
Mean, = 1.3 cm and Standard deviation,
= 0.01 cm
Also, since distribution is normal;
Z = ~ N(0,1)
Let X = range of diameters
So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)
P(X < 1.33) = P( <
) = P(Z < 3) = 0.99865
P(X <= 1.27) = P( <
) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865
= 0.00135
P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .
Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .