Answer:
(a1) The probability that temperature increase will be less than 20°C is 0.667.
(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.
(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.
(c) The expected value of the temperature increase is 17.5°C.
Step-by-step explanation:
Let <em>X</em> = temperature increase.
The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].
The probability density function of <em>X</em> is:
![f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.](https://tex.z-dn.net/?f=f%28X%29%3D%5Cleft%20%5C%7B%20%7B%7B%5Cfrac%7B1%7D%7B25-10%7D%3D%5Cfrac%7B1%7D%7B15%7D%3B%5C%20x%5Cin%20%5B10%2C%2025%5D%7D%20%5Catop%20%7B0%3B%5C%20otherwise%7D%7D%20%5Cright.)
(a1)
Compute the probability that temperature increase will be less than 20°C as follows:
![P(X](https://tex.z-dn.net/?f=P%28X%3C20%29%3D%5Cint%5Climits%5E%7B20%7D_%7B10%7D%7B%5Cfrac%7B1%7D%7B15%7D%7D%5C%2C%20dx%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Cint%5Climits%5E%7B20%7D_%7B10%7D%7Bdx%7D%5C%2C%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Bx%5D%5E%7B20%7D_%7B10%7D%3D%5Cfrac%7B1%7D%7B15%7D%5B20-10%5D%3D%5Cfrac%7B10%7D%7B15%7D%5C%5C%3D0.667)
Thus, the probability that temperature increase will be less than 20°C is 0.667.
(a2)
Compute the probability that temperature increase will be between 20°C and 22°C as follows:
![P(20](https://tex.z-dn.net/?f=P%2820%3CX%3C22%29%3D%5Cint%5Climits%5E%7B22%7D_%7B20%7D%7B%5Cfrac%7B1%7D%7B15%7D%7D%5C%2C%20dx%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Cint%5Climits%5E%7B22%7D_%7B20%7D%7Bdx%7D%5C%2C%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Bx%5D%5E%7B22%7D_%7B20%7D%3D%5Cfrac%7B1%7D%7B15%7D%5B22-20%5D%3D%5Cfrac%7B2%7D%7B15%7D%5C%5C%3D0.133)
Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.
(b)
Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:
![P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467](https://tex.z-dn.net/?f=P%28X%3E18%29%3D%5Cint%5Climits%5E%7B25%7D_%7B18%7D%7B%5Cfrac%7B1%7D%7B15%7D%7D%5C%2C%20dx%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Cint%5Climits%5E%7B25%7D_%7B18%7D%7Bdx%7D%5C%2C%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Bx%5D%5E%7B25%7D_%7B18%7D%3D%5Cfrac%7B1%7D%7B15%7D%5B25-18%5D%3D%5Cfrac%7B7%7D%7B15%7D%5C%5C%3D0.467)
Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.
(c)
Compute the expected value of the uniform random variable <em>X</em> as follows:
![E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5](https://tex.z-dn.net/?f=E%28X%29%3D%5Cfrac%7B1%7D%7B2%7D%5B10%2B25%5D%3D%5Cfrac%7B35%7D%7B2%7D%3D17.5)
Thus, the expected value of the temperature increase is 17.5°C.