A particular manufacturing design requires a shaft with a diameter between 23.92 and 24.018 mm. The manufacturing process yields shafts with diameters normally distributed, with a mean of 24.003 and standard deviation of .006.
a) for this process what is the proportion of shafts with a diameter between of 23.92 and 24.00 mm.
b) The probability that the shaft is acceptable is ___
c) The diameter that will be exceeded by only .5% of shafts is __

(a) More than 30.85%

(b) More than 99.38%

(c) Diameter that will be exceeded by only 0.5% of shafts is 24.018 mm.

Step-by-step explanation:

We are given that the manufacturing process yields shafts with diameters normally distributed, with a mean of 24.003 and standard deviation of .006.

Let X = shafts with diameters

So, X ~ N()

The z score probability distribution is given by;

          Z = ~ N(0,1)

(a) Probability that the proportion of shafts with a diameter between 23.92 and 24.00 mm = P(23.92 mm < X < 24 mm)

   P(23.92 < X < 24) = P(X < 24) - P(X 23.92)

   P(X < 24) = P( < ) = P(Z < -0.5) = 1 - P(Z 0.5)

                                                     = 1 - 0.69146 = 0.30854

   P(X 23.92) = P( ) = P(Z -13.83) = P(Z 13.83)

                                                                                        = Less than 0.0005%

Therefore, P(23.92 < X < 24) = 0.30854 -  Less than 0.0005% = More than 0.308535 or More than 30.85%

(a) Probability that the shaft is acceptable = P(23.92 mm < X < 24.018 mm)

   P(23.92 < X < 24.018) = P(X < 24.018) - P(X 23.92)

   P(X < 24.018) = P( < ) = P(Z < 2.5) = 0.99379

   P(X 23.92) = P( ) = P(Z -13.83) = P(Z 13.83)

                                                                                        = Less than 0.0005%

Therefore, P(23.92 < X < 24.018) = 0.99379 -  Less than 0.0005% = More than 0.993785 or More than 99.38%

(c) We have to find the diameter that will be exceed by only 0.5% of shafts, which means ;

   P(X > x) = 0.005

   P( > ) = 0.005

    P(Z > ) = 0.005

Now in the z table the critical value of X which have an area greater than 0.005 is 2.5758, i.e.;

            = 2.5758

       - 24.003 =

                      = 24.003 + 0.01545 = 24.018 ≈ 24

So, the diameter that will be exceeded by only 0.5% of shafts is 24.018 mm.

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step-by-step explanation: