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% .
Answer: It depends on what categories your teacher has and what percent of your grade it is worth. I would recommend an online grade calculator
Answer:
A= 13 B=13.1
Step-by-step explanation:
A-B= -0.1
B-A= 0.1