Answer:
1) biased sample
2) random sampling
Step-by-step explanation:
Since the results are based on feedback of students in only one class, it's not a good representative of the entire student body.
Also, since it's a math class there will be a bias.
Better method would be randomly selecting students from the entire school students list and take their feedback
A Gantt chart is a bar chart with the bars going left to right instead of going up and down.
Answer:
1. Just plot the points (the first number is x, the second is y)
2. Q3
3. Q1: (5,3) Q2: (-3,5) Q3: (-4,-2) Q4: (1,-4)
a. In order for 
to be a PDF, needs to be non-negative (it is) and the integral over its support must evaluate to 1. is symmetric about 
, i.e. even, so
and so is indeed a PDF.
b. Let 
be a random variable with as its PDF. Then
![\mu=E[X]=\displaystyle\int_{-\infty}^\infty xf(x)\,\mathrm dx=\frac\lambda2\int_{-\infty}^\infty xe^{-\lambda|x|}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cmu%3DE%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac%5Clambda2%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xe%5E%7B-%5Clambda%7Cx%7C%7D%5C%2C%5Cmathrm%20dx)
The integrand is odd, so the integral vanishes and the mean is 
.
The variance of is
![\sigma^2=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=%5Csigma%5E2%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
The second moment is
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f(x)\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2f%28x%29%5C%2C%5Cmathrm%20dx)
This integrand is even, so
![E[X^2]=\displaystyle2\int_0^\infty x^2f(x)\,\mathrm dx=\lambda\int_0^\infty x^2e^{-\lambda x}\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle2%5Cint_0%5E%5Cinfty%20x%5E2f%28x%29%5C%2C%5Cmathrm%20dx%3D%5Clambda%5Cint_0%5E%5Cinfty%20x%5E2e%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx)
Integrate by parts, taking
so that
![\displaystyle E[X^2]=\lambda\left(-\frac{x^2e^{-\lambda x}}\lambda\bigg|_0^\infty+2\int_0^\infty\frac{xe^{-\lambda x}}\lambda\,\mathrm dx\right)=2\int_0^\infty xe^{-\lambda x}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20E%5BX%5E2%5D%3D%5Clambda%5Cleft%28-%5Cfrac%7Bx%5E2e%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5Cbigg%7C_0%5E%5Cinfty%2B2%5Cint_0%5E%5Cinfty%5Cfrac%7Bxe%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5C%2C%5Cmathrm%20dx%5Cright%29%3D2%5Cint_0%5E%5Cinfty%20xe%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx)
Integrate by parts again, this time with
![\displaystyle E[X^2]=2\left(-\frac{xe^{-\lambda x}}\lambda\bigg|_0^\infty+\int_0^\infty\frac{e^{-\lambda x}}\lambda\,\mathrm dx\right)=\frac2\lambda\int_0^\infty e^{-\lambda x}\,\mathrm dx=-\frac2{\lambda^2}e^{-\lambda x}\bigg|_0^\infty=\frac2{\lambda^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20E%5BX%5E2%5D%3D2%5Cleft%28-%5Cfrac%7Bxe%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5Cbigg%7C_0%5E%5Cinfty%2B%5Cint_0%5E%5Cinfty%5Cfrac%7Be%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5C%2C%5Cmathrm%20dx%5Cright%29%3D%5Cfrac2%5Clambda%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx%3D-%5Cfrac2%7B%5Clambda%5E2%7De%5E%7B-%5Clambda%20x%7D%5Cbigg%7C_0%5E%5Cinfty%3D%5Cfrac2%7B%5Clambda%5E2%7D)
and so the variance is 
.
It's pretty simple once you understand how to do it. First we have to find the change in the time. Since it started at 10:00 and went to 15:00 (military time is useful in these problems) the time has changed a total of 5 hours. And since this temperature increases 2.25 degrees each hour, we would multiply 2.25 by 5 which gets us 11.25. Then all we have to do next is add 11.25 and -3 which gets us 8.25 degrees as the answer.
