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inna [77]
2 years ago
8

Explain how collecting mean values (X-bar) from samples drawn from a dataset, regardless of the statistical distribution of the

dataset, results in a distribution of those values that is centered on the mean value of the original dataset and is approximately normally distributed depending upon the sample size used. Include the name the theorem that describes this phenomenon, and discuss what it means to treat the X-bar data points derived from sampling as a new random variable. Why is this important in engineering?
Engineering
1 answer:
Pani-rosa [81]2 years ago
8 0

Answer:

answer is given below

Explanation:

Central Limit Theorem: The Central Limit Theorem (CLT) is a statistical theory that gives a sufficiently large sample size with limited variations from the population, the average of all samples from the same population is approximately the same. . In addition, all models follow a nearly normal distribution model.

The given phenomenon is described in the central limit theory. In other words, if we repeatedly take independent random samples of size n from any population, when n is large, the sample distribution is the normal distribution pattern.

mean of the sample means

                    \mu _\bar x = \mu        .............1

and here standard deviation of the sample means is

                      \mu _\bar x = \frac{\sigma }{\sqrt{n}}      .........2

and This theory is found elsewhere in the field of statistics. Although the central limit theory may seem abstract and devoid of any application, this theory is actually important for statistical practice.

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An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening
Viktor [21]

Given:

Pressure, P_{1} = 1300 kPa

Temperature,  T_{1} = 500^{\circ}

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

velocity, v = 40 m/s

A = 1m^{2}

Solution:

For air propertiess at

P_{1} = 1300 kPa

T_{1} = 500^{\circ}

h_{1} = 793kJ/K

v_{1} = 0.172\frac{m^{3}}{kg}

and also at

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

h_{2} = 401 KJ/K

v_{2} =  1.15\frac{m^{3}}{kg}

a) Mass flow rate is given by:

m' = \frac{Av}{v_{1}}

Now,

m = \frac{0.2\times 40}{0.172} = 46.51 kg/s

b) for the power produced by turbine, P = m'(h_{1} - h_{2})

P = 46.51\times(793 - 401) = 18.231 MW

5 0
2 years ago
2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam
Natasha2012 [34]

Answer:

\eta = 91.7%

Explanation:

Determine the initial velocity

v_1 = \frac{\dot v}{A_1}

    = \frac{0.1}{\pi}{4} 0.08^2

     = 19.89 m/s

final velocity

v_2 =\frac{\dot v}{A_2}

      = \frac{0.1}{\frac{\pi}{4} 0.12^2}

      =8.84 m/s

total mechanical energy is given as

E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}

\dot v = \dot m v                       ( v =v_1 =v_2)

E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}

                = mv\Delta P + \dot m  \frac{v_2^2 -v_1^2}{2}

                 = \dot v \Delta P  + \dot v \rho \frac{v_2^2 -v_1^2}{2}

              = 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}

E_{mech} = 36.34 W

Shaft power

W = \eta_[motar} W_{elec}

    =0.9\times 44 =39.6

mechanical efficiency

\eta{pump} =\frac{ E_{mech}}{W}

=\frac{36.34}{39.6} = 0.917  = 91.7%

8 0
2 years ago
In Female, the twenty-third pair of chromosomes is called in in
zalisa [80]
The twenty-third pair of chromosomes is called the sex chromosomes. Females have two X chromosomes and males have one X and one Y
6 0
2 years ago
A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055
Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}

So the mean conductivity

K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )

T_m=\dfrac{160+273+40+273}{2}

T_m=373 K

K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )

K_m=0.112 \frac{W}{mK}

So heat conduction through cylinder

Q=kA\dfrac{\Delta T}{L}

Q=0.112\times \pi \times 0.15^2\times 120

Q=0.95 W/m

4 0
2 years ago
A spherical hot-air balloon is initially filled with air at 120 kPa and 20°C with an initial diameter of 5 m. Air enters this ba
adoni [48]

Answer:

time is 17.43 min

Explanation:

given data

initial diameter = 5 m

velocity = 3 m/s

final diameter = 17 m

solution

we will apply here change in change in volume equation that is express as

ΔV = \frac{4}{3} \pi * (rf)^3 - \frac{4}{3} \pi * (ri)^3    .............1

here ΔV is change in volume and rf is final radius and ri is initial radius

ΔV = \frac{4}{3} \pi * (8.5)^3 - \frac{4}{3} \pi * (2.5)^3

ΔV = 2507 m³

so

Q = velocity × Area

Q = 3 × π ×(0.5)² = 2.356 m³/s

and

change in time is express as

Δt = \frac{\Delta V}{Q}

Δt = \frac{2507}{2.356}

Δt = 1046 sec

so change in time is 17.43 min

8 0
2 years ago
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