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An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The specification limits under which the ball bearings can operate are 0.74 inch lower) and 0.76 inch (upper). Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed with a mean of 0.753 inch and a standard deviation of 0.004 inch. What is the probability that a ball bearing diameter will be a. Between the target and the actual mean? b. Above the upper specification limit? c. Below the lower specification limit? d. Above which value in diameter will 93 percent of the ball bearings be? (5 points) (5 points) (5 points (5 points)Please type your answer.

math   Statistics-And-Probability   
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actual mean μ =0.753, and standard deviation, σ From that the given information, targeted mean-0.75 P(0.75 sx s 0.753) -P 004 (0.75-0.753 x-0.753 0.753-0.753 0.004 0.004 -0.003 0.004 -P(-0.75szs0.00) = P(0.00)-P(-0.75) From the standard normal table, the area to the left of z =-0.75 is 0.2266 and the area to left of z = 0.00 is 0.5000. P(0.75 sx s 0.753) P(0.00)-P(-0.75) = 0.5000-0.2266 -0.2734

From that the given information, upper specification limits 0.76, μ-0.753, and σ-0.004 Consider, P(x 0.76)-1-P(xs0.76) x-0.753 0.76-0.753) 1-PI-004 0.004 0.007 0.004 -1-P(z s1.75) From the standard normal table, area to the left of z 1.75 is 0.9599 P(x 0.76)-1-P(Z s1.75) -1-0.9599 0.0401

The probability is obtained below: -0.753, and ơ-074 From that the given information, the lower specification limits 0.74, Consider, P(x s0.74)-p Y-0753 0.004 074-0753 0.004 , Pz-0.013 0.004 -P(zs-3.25) From the standard normal table, the area to the left of z--3.25 is 0.0006 P(x s 0.74)-P(zs-3.25) 0.0006

The value asked for here will be the 7th percentile, since 93% of the ball bearings will have diameters above that. So we will look up .4300 in the Z-table in a backwards manner. The closest area to this is .4306, which corresponds to a Z-value of 1.48. So the X value, once we un-standardize, is 1.48- X-0.753 0.004 -000592 X-0.753 X-0.74708 so 0.74708 in. is the value that 93% of the diameters are above

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