Question

At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water that is discharged back into the ocean. The amount that the water temperature is raised has a uniform distribution over the interval from 10 to 25 degree C. (a) What is the probability that the temperature increase will be (1) less than 20 degrees C? (2) between 20 and 22C?(b) Suppose that a temperature increase of more than 18 degrees C is considered to be potentially harmful to the environment. What is the probability, at any point of time, that the temperature increase is potentially dangerous? (c) what is the expected value of the temperature increase?

Answer 1

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 X = temperature increase.

The random variable X follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of X is:

$f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.$

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

$P(X<20)=\int\limits^{20}_{10}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{20}_{10}{dx}\,\\=\frac{1}{15}[x]^{20}_{10}=\frac{1}{15}[20-10]=\frac{10}{15}\\=0.667$

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:

$P(20<X<22)=\int\limits^{22}_{20}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{22}_{20}{dx}\,\\=\frac{1}{15}[x]^{22}_{20}=\frac{1}{15}[22-20]=\frac{2}{15}\\=0.133$

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:

$P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467$

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 X as follows:

$E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5$

Thus, the expected value of the temperature increase is 17.5°C.