2x1,000,000
9x100,000
3x10,000
7x1,000
0x100
8x10
2x1
Answer:
Probability of a sample that contains exactly two defective parts is .0037 or .37%
Step-by-step explanation:
As we know if P is the probability of achieving k results in n trials then probability formula is P = 
In this formula n = number of trials
k = number of success
(n-k) = number of failures
p = probability of success in one trial
q = (1-p) = probability of failure in one trial
In this sum n = 5
k = 2
number failures (n-k) = (5-2) = 3
p = 2% which can be written as .02
q = 98% Which can be written as .98
Now putting these values in the formula
P = 
P = 
= 5×4×3×2×1/3×2×1×2×1
= 5×2 =10
P = 10×(.02)²×(.98)³
= .0037 or .37%
Answer:
10
Step-by-step explanation:
Here, we want to find the value of ;
h circle h(10)
This mean we are to put h in h, the find the function of 10
we have this as;
6-(6-x)
= 6-6 + x = x
Now, replace the x by 10; so we have h circle h(10)
I'm going to assume the joint density function is

a. In order for
to be a proper probability density function, the integral over its support must be 1.

b. You get the marginal density
by integrating the joint density over all possible values of
:

c. We have

d. We have

and by definition of conditional probability,


e. We can find the expectation of
using the marginal distribution found earlier.
![E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_0%5E1xf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac67%5Cint_0%5E1%282x%5E2%2Bx%29%5C%2C%5Cmathrm%20dx%3D%5Cboxed%7B%5Cfrac57%7D)
f. This part is cut off, but if you're supposed to find the expectation of
, there are several ways to do so.
- Compute the marginal density of
, then directly compute the expected value.

![\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac87)
- Compute the conditional density of
given
, then use the law of total expectation.

The law of total expectation says
![E[Y]=E[E[Y\mid X]]](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5BE%5BY%5Cmid%20X%5D%5D)
We have
![E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}](https://tex.z-dn.net/?f=E%5BY%5Cmid%20X%3Dx%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_%7BY%5Cmid%20X%7D%28y%5Cmid%20x%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac%7B6x%2B4%7D%7B6x%2B3%7D%3D1%2B%5Cfrac1%7B6x%2B3%7D)
![\implies E[Y\mid X]=1+\dfrac1{6X+3}](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5Cmid%20X%5D%3D1%2B%5Cdfrac1%7B6X%2B3%7D)
This random variable is undefined only when
which is outside the support of
, so we have
![E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5Cleft%5B1%2B%5Cdfrac1%7B6X%2B3%7D%5Cright%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cleft%281%2B%5Cfrac1%7B6x%2B3%7D%5Cright%29f_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac87)
Answer:
148
Step-by-step explanation:
AP 200, 196, 192, ...
a14=?
---------
a1= 200
d= 196-200=192-196= -4
a14= a1+13d
a14= 200+13*(-4)= 200- 52= 148