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.

c is the correct answer

step-by-step explanation:

the population of other specie of birds will decline or other specie of birds will find some alternate source of food to survive.

explanation:

according to the competitive exclusion principle, which is also known as gause's law, two specie living in the same niche and feeding on the same resource cannot co-exist stably.

as we have well studied the theory of natural selection which states that only those organisms survive which have better tendency to compete for food and adapt to the environmental conditions.

coming towards the question, what wi ll happen when one species outcompetes the other for the small fish , the other specie will not get enough food and will start to die or decline in population. eventually either other specie needs to find another alternative food source for itself or migrate at some other location where there is less competition otherwise it will just not exist anymore at that niche.

hope it !