Compare the probabilities, Civil Engineering

Assignment Help:

Compare the probabilities:

Use the central limit theorem to compute P (¦ X - μ ¦ ≥ k), k = 1, 2,3,4,5 where X and μ are as defined in Example. Compare the probabilities with the probabilities computed in Example.

Solution:

We have np = 2, npq = 1, n = 4 and so

P(¦X - μ¦≥ k)=P (¦x-2¦√1 ≥ k/√1) = P(¦x-2¦≥ k)

+1-P(¦x-2¦< k) ∼ 1-[ Φ (k) - Φ (-k)]

=2(1- Φ(k))

Thus we tabulate the probabilities as below

 

496_Compare the probabilities.png

As expected, since the sample size 4 is not large sufficient, the difference between the exact and the approximate probabilities in Example are considerable. Thus the approximation will be better for larger values of n.

Approximations of equations are good for large n but better approximations are acquired by making some corrections, called the continuity corrections, for representing a discrete distribution through a continuous distribution. For the discrete binomial distribution P (x < X < x + 1) is equal to zero if x is an integer hut while this probability is approximated by the normal probability it is never equal to zero. This necessitates the continuity correction. A reasonable approach to get a better approximations to write P (X = x) as equal to

P (x-1/2

P ( xl ≤ X ≤ x2 ) as P( x1-1/2 < X ≤x2 + 1/2 ). This suggests that a better

approximation to binomial cdf is obtained by using

F(x) ∼ Φ  {(x+1/2 -np)/√(npq))} and

P(x1≤ X≤ x2) ∼ Φ  ((x2+1/2-np)/√(npq)) - Φ ((x1-1/2 -np)/√(npq))

alternatively of previous equations respectively. In fact, a theoretical justification of why this equation gives a better approximation than previous one can also be given. Thus, we do not go into the theoretical details.


Related Discussions:- Compare the probabilities

Survey, relationship between included angles and whole circle bearing

relationship between included angles and whole circle bearing

Determine the major and minor principal effective stresses, The stresses ac...

The stresses acting on the plane of maximum shearing stress by a given point in sand are as follows: total normal stress = 250 kN/m 2 ; pore-water pressure = 88.5 kN/ m 2 ; shearin

Important parts of road pavements, Important parts of road pavements: ...

Important parts of road pavements: Prime Coat : Application of a low viscosity bituminous binder over a granular base to fill up the voids. Tack Coat : Application of b

Estimate the size estimating the rcc structures, PART - A 1.  Can you e...

PART - A 1.  Can you enumerate the benefits of individual wall method of estimation? 2.  Estimate the single room of size  3m x 4m x height 3m  using individual wall method?

Calculate the monthly payment for home mortgage, Your home mortgage is $300...

Your home mortgage is $300,000 for 30 years with a nominal annual rate of 7%. The monthly payment is most nearly: Solution: n = 360 months interest = 7%-annual ÷ 12-months/year

Afflux on account of obstruction to flow, Q. Afflux on account of obstructi...

Q. Afflux on account of obstruction to flow? Even if the river constriction on account of approach banks is not excessive and bridge waterway is wider than that causing critica

Hydraulics, A rectangular channel, 20ft wide is shown. The channel consists...

A rectangular channel, 20ft wide is shown. The channel consists of three reaches of different slopes. The channel has a roughness coefficient of n=0.015 and carries a discharge of

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd