This distribution has expectation
![E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\int_1^\infty\frac3{x^3}\,\mathrm dx=\frac32](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_1%5E%5Cinfty%5Cfrac3%7Bx%5E3%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac32)
a. The probability that 
falls below the average/expectation is
b. Denote by 
the largest of the three claims 
. Then the density of this maximum order statistic is
where 
is the distribution function for . This is given by
So we have
and the expectation is
![E[X_{(3)}]=\displaystyle\int_{-\infty}^\infty xf_{X_{(3)}}(x)\,\mathrm dx=\int_1^\infty\frac9{x^3}\left(1-\frac1{x^3}\right)^2\,\mathrm dx=\frac{81}{40}=\boxed{2.025}](https://tex.z-dn.net/?f=E%5BX_%7B%283%29%7D%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf_%7BX_%7B%283%29%7D%7D%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_1%5E%5Cinfty%5Cfrac9%7Bx%5E3%7D%5Cleft%281-%5Cfrac1%7Bx%5E3%7D%5Cright%29%5E2%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7B81%7D%7B40%7D%3D%5Cboxed%7B2.025%7D)
c. Denote by 
the smallest of the three claims. has density
so the expectation is
![E[X_{(1)}]=\displaystyle\int_{-\infty}^\infty xf_{X_{(1)}}(x)\,\mathrm dx=\int_1^\infty\frac9{x^9}\,\mathrm dx=\frac98=\boxed{1.125}](https://tex.z-dn.net/?f=E%5BX_%7B%281%29%7D%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf_%7BX_%7B%281%29%7D%7D%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_1%5E%5Cinfty%5Cfrac9%7Bx%5E9%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac98%3D%5Cboxed%7B1.125%7D)
Answer:
45 Houses
Step-by-step explanation:
5 houses for every 9 blocks means you need to multiply 5 and 9.
Answer:
3.2
Step-by-step explanation:
it is closer to 3.2 than 3.3
Answer:
The answer could be 0 or 1 out of 4, because choosing a card at first and not putting it back, then picking a second card at random, it would likely be one in zero attempts to pick 3 again but otherwise that it could be 1 out of 4 because it could be possible picking 3 at the first try.
