Let's say you want to compute the probability
![\mathbb P(a\le X\le b)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28a%5Cle%20X%5Cle%20b%29)
where
![X](https://tex.z-dn.net/?f=X)
converges in distribution to
![Y](https://tex.z-dn.net/?f=Y)
, and
![Y](https://tex.z-dn.net/?f=Y)
follows a normal distribution. The normal approximation (without the continuity correction) basically involves choosing
![Y](https://tex.z-dn.net/?f=Y)
such that its mean and variance are the same as those for
![X](https://tex.z-dn.net/?f=X)
.
Example: If
![X](https://tex.z-dn.net/?f=X)
is binomially distributed with
![n=100](https://tex.z-dn.net/?f=n%3D100)
and
![p=0.1](https://tex.z-dn.net/?f=p%3D0.1)
, then
![X](https://tex.z-dn.net/?f=X)
has mean
![np=10](https://tex.z-dn.net/?f=np%3D10)
and variance
![np(1-p)=9](https://tex.z-dn.net/?f=np%281-p%29%3D9)
. So you can approximate a probability in terms of
![X](https://tex.z-dn.net/?f=X)
with a probability in terms of
![Y](https://tex.z-dn.net/?f=Y)
:
![\mathbb P(a\le X\le b)\approx\mathbb P(a\le Y\le b)=\mathbb P\left(\dfrac{a-10}3\le\dfrac{Y-10}3\le\dfrac{b-10}3\right)=\mathbb P(a^*\le Z\le b^*)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28a%5Cle%20X%5Cle%20b%29%5Capprox%5Cmathbb%20P%28a%5Cle%20Y%5Cle%20b%29%3D%5Cmathbb%20P%5Cleft%28%5Cdfrac%7Ba-10%7D3%5Cle%5Cdfrac%7BY-10%7D3%5Cle%5Cdfrac%7Bb-10%7D3%5Cright%29%3D%5Cmathbb%20P%28a%5E%2A%5Cle%20Z%5Cle%20b%5E%2A%29)
where
![Z](https://tex.z-dn.net/?f=Z)
follows the standard normal distribution.
-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7, in Order from least to greatest
Fudgin [204]
-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7
Answer:
An equation is ![x+52\°+90\°=180\°](https://tex.z-dn.net/?f=x%2B52%5C%C2%B0%2B90%5C%C2%B0%3D180%5C%C2%B0)
Step-by-step explanation:
Let
be the measure of the unknown acute angle.
According to angle sum property of a triangle, sum of all the angles of a triangle is equal to 180°.
One of the acute angles of a right triangle is equal to 52°.
Also, the triangle is a right angled triangle, one of the angle must be equal to 90°.
![x+52\°+90\°=180\°\\x+142\°=180\°\\x=38\°](https://tex.z-dn.net/?f=x%2B52%5C%C2%B0%2B90%5C%C2%B0%3D180%5C%C2%B0%5C%5Cx%2B142%5C%C2%B0%3D180%5C%C2%B0%5C%5Cx%3D38%5C%C2%B0)
So, an equation is ![x+52\°+90\°=180\°](https://tex.z-dn.net/?f=x%2B52%5C%C2%B0%2B90%5C%C2%B0%3D180%5C%C2%B0)
Answer: ($3.532, $4.166)
Step-by-step explanation:
Given : Significance level : ![\alpha: 1-0.90=0.10](https://tex.z-dn.net/?f=%5Calpha%3A%201-0.90%3D0.10)
Critical value : ![z_{\alpha/2}=1.645](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3D1.645)
Sample size : n= 157
Sample mean : ![\overline{x}=\$\ 3.849](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D%5C%24%5C%203.849)
Standard deviation : ![\sigma= \$\ 2.421](https://tex.z-dn.net/?f=%5Csigma%3D%20%5C%24%5C%202.421)
The confidence interval for population mean is given by :_
![\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%5Cpm%20z_%7B%5Calpha%2F2%7D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![\text{i.e. }\$\ 3.849\pm (1.64)\dfrac{2.421}{\sqrt{157}}\\\\\approx\$\ 3.849\pm0.317\\\\=(\$\ 3.849-0.317,\$\ 3.849+0.317)=(\$\ 3.532,\$\ 4.166)](https://tex.z-dn.net/?f=%5Ctext%7Bi.e.%20%7D%5C%24%5C%203.849%5Cpm%20%281.64%29%5Cdfrac%7B2.421%7D%7B%5Csqrt%7B157%7D%7D%5C%5C%5C%5C%5Capprox%5C%24%5C%203.849%5Cpm0.317%5C%5C%5C%5C%3D%28%5C%24%5C%203.849-0.317%2C%5C%24%5C%203.849%2B0.317%29%3D%28%5C%24%5C%203.532%2C%5C%24%5C%204.166%29)
Hence, the 0% confidence interval for the average tip given at this restaurant = ($3.532, $4.166)
Answer:
The right response is Option c ($185,870,742).
Step-by-step explanation:
Given:
n = 30
r = 5.4%
or,
= 0.054
Periodic payment will be:
![R = \frac{360000000}{30}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7B360000000%7D%7B30%7D)
($)
Now,
The present value will be:
= ![R+R(\frac{1-(1+r)^{-n+1}}{r} )](https://tex.z-dn.net/?f=R%2BR%28%5Cfrac%7B1-%281%2Br%29%5E%7B-n%2B1%7D%7D%7Br%7D%20%29)
By substituting the values, we get
= ![12000000+12000000(\frac{1-(1+0.054)^{-30 + 1}}{0.054} )](https://tex.z-dn.net/?f=12000000%2B12000000%28%5Cfrac%7B1-%281%2B0.054%29%5E%7B-30%20%2B%201%7D%7D%7B0.054%7D%20%29)
= ![12000000+12000000\times 14.4892](https://tex.z-dn.net/?f=12000000%2B12000000%5Ctimes%2014.4892)
=
($)