Answer:
negative association
Step-by-step explanation:
This diagram is a scatter plot which represents a set of data. Data on a scatter plot can fit one of three descriptions: positive (increasing), negative (decreasing) or no association (points do not form any kind of line). Given this data and the line of best fit, or the line that pass through the majority of the points, it is a decreasing line. Since the line goes downhill, it is a negative association.
Answer:
10) The first 2 on the left are functions and the one on the top right is a function, but the bottom right is not.
11) Domain = (2, 8, -1, 0) Range = (5, -2, 4, 5)
Answer:
11628
Step-by-step explanation:
Thanks!
Answer:
- 5/48, 3/16, .5, .75, 13
- 1/5, .35, 12/25, .5, 4/5
- -3/4, -7/10, 3/40, 8/10
- -.65, -3/8, 5/16, 2/4
Step-by-step explanation:
- 5/48 = 1.0291666666 | 3/16 = .1875 the rest is obvious
- 1/5 = .2 | 12/25 = .48 | 4/5 = .8
- -3/4 = -.75 | -7/10 = -.7 | 3/40 = .075 | 8/10 = .8
- -3/8 = -.375 | 5/16 = .3125 | 2/4 = .5
a) Add up all the probabilities 
where 
:
b) The expected value is
![E[X]=\displaystyle\sum_xx\,f(x)=3.16](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Csum_xx%5C%2Cf%28x%29%3D3.16)
Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.
c) The variance is
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
where
![E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Csum_xx%5E2%5C%2Cf%28x%29%3D11.58)
So the variance is
![V[X]=11.58-3.16^2\approx1.59](https://tex.z-dn.net/?f=V%5BX%5D%3D11.58-3.16%5E2%5Capprox1.59)
The standard deviation is the square root of the variance:
![\sqrt{V[X]}\approx1.26](https://tex.z-dn.net/?f=%5Csqrt%7BV%5BX%5D%7D%5Capprox1.26)
d) Since 
is a linear combination of 
, computing the expectation and variance of Y is easy:
![E[Y]=E[7X+3]=7E[X]+3=25.12](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5B7X%2B3%5D%3D7E%5BX%5D%2B3%3D25.12)
![V[Y]=V[7X+3]=7^2V[X]\approx78.13](https://tex.z-dn.net/?f=V%5BY%5D%3DV%5B7X%2B3%5D%3D7%5E2V%5BX%5D%5Capprox78.13)
e) The covariance of X and Y is
![\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%5BX%2CY%5D%3DE%5B%28X-E%5BX%5D%29%28Y-E%5BY%5D%29%5D%3DE%5BXY%5D-E%5BX%5DE%5BY%5D)
We have
so
![E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54](https://tex.z-dn.net/?f=E%5BXY%5D%3DE%5B7X%5E2%2B3X%5D%3D7E%5BX%5E2%5D%2B3E%5BX%5D%3D90.54)
Then the covariance is
![\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%5BX%2CY%5D%3D90.54-3.16%5Ccdot25.12%5Capprox11.16)
f) Dividing the covariance by the variance of X gives
![\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%7BCov%7D%5BX%2CY%5D%7D%7BV%5BX%5D%7D%5Capprox%5Cdfrac%7B11.16%7D%7B1.59%7D%5Capprox0.9638)
