Ask question

Ask question

All categories

3 years ago

15

An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc

h and a standard deviation of 0.07 inch. Assume that the shaft diameter follows a Gaussian distribution. What percentage of the diameter of the total shipment of shafts will fall between 2.39inch and 2.60 inch?

1 answer:

Vadim26 [7]3 years ago

7 0

Answer: 78.89%

Explanation:

Given : Sample size : n= 1200

Sample mean : $\overline{x}=2.45$

Standard deviation : $\sigma=0.07$

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, $z=\dfrac{x-\mu}{\sigma}$ , the z-value corresponds to 2.39 will be :-

$z=\dfrac{2.39-2.45}{0.07}\approx-0.86$

z-value corresponds to 2.60 will be :-

$z=\dfrac{2.60-2.45}{0.07}\approx2.14$

Using the standard normal table for z, we have

P-value = $P(-0.86$

$=P(z$

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

You might be interested in

How are scientific discoveries used in engineering design?

Citrus2011 [14]

Answer:

one is technology

6 0

3 years ago

(d) Suppose two students are memorizing a list according to the same model dL dt = 0.5(1 − L) where L represents the fraction of

Neporo4naja [7]

Answer:

Rate of learning =0

Explanation:

Please see attachment

8 0

3 years ago

The iron-blade plow allowed humans to increase food production during<br> what age?

lilavasa [31]

Answer:

The Iron Age

Explanation:

4 0

3 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

5 0

3 years ago

A 46.0-g meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then a 210.0-g weight is hung wit

Anna71 [15]

Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.

<h3>What is clockwise torque?</h3>

The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.

A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.

Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.

The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.

When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.

You can calculate the torque's magnitude using

$\begin{displaymath}\tau =rF_{\bot }=rF\sin \theta .\end{displaymath}$

To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.

Clockwise torque due to 100 g ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm

Clockwise torque due to 200 g ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm

Clockwise torque due to stick mass ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm

Counter-clockwise torque due to normal force ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm

Learn more about torque

brainly.com/question/1544595

#SPJ4

7 0

1 year ago