Respuesta :
Answer:
(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([tex]\mu=24.003,\sigma^{2} = 0.006^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ 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 [tex]\leq[/tex] 23.92)
P(X < 24) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{24-24.003}{0.006}[/tex] ) = P(Z < -0.5) = 1 - P(Z [tex]\leq[/tex] 0.5)
= 1 - 0.69146 = 0.30854
P(X [tex]\leq[/tex] 23.92) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{23.92-24.003}{0.006}[/tex] ) = P(Z [tex]\leq[/tex] -13.83) = P(Z [tex]\geq[/tex] 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 [tex]\leq[/tex] 23.92)
P(X < 24.018) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{24.018-24.003}{0.006}[/tex] ) = P(Z < 2.5) = 0.99379
P(X [tex]\leq[/tex] 23.92) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{23.92-24.003}{0.006}[/tex] ) = P(Z [tex]\leq[/tex] -13.83) = P(Z [tex]\geq[/tex] 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-24.003}{0.006}[/tex] ) = 0.005
P(Z > [tex]\frac{x-24.003}{0.006}[/tex] ) = 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.;
[tex]\frac{x-24.003}{0.006}[/tex] = 2.5758
[tex]x[/tex] - 24.003 = [tex]2.5758 \times 0.006[/tex]
[tex]x[/tex] = 24.003 + 0.01545 = 24.018 ≈ 24
So, the diameter that will be exceeded by only 0.5% of shafts is 24.018 mm.
