At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

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At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

hat 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?

Mathematics

1 answer:

Zina [12.3K] · Answer 1 · 2026-03-05T08:35:47+03:00

Answer:

(a1) The chance that the temperature rises by less than 20°C is 0.667.

(a2) The probability of the temperature increase falling between 20°C and 22°C is 0.133.

(b) The likelihood that the temperature increase could be hazardous at any moment is 0.467.

(c) The anticipated value of the temperature rise is 17.5°C.

Step-by-step explanation:

Let X denote the temperature increase.

The random variable X is distributed uniformly over the interval [10°C, 25°C].

The probability density function for X is shown here:

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

(a1)

The probability of the temperature increase being under 20°C can be calculated as follows:

$P(X$

Consequently, the chance that the temperature increase will be below 20°C is 0.667.

(a2)

The probability of the temperature rise being in the range from 20°C to 22°C is computed as follows:

$P(20$

This leads to the probability of the temperature increase being between 20°C and 22°C being 0.133.

(b)

To find the probability that the increase in temperature could be dangerous, we calculate:

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

This results in a probability of 0.467 that the temperature rise is potentially dangerous at any time.

(c)

The expected value of the uniform random variable X is determined as follows:

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

The expected value for the temperature increase computes to 17.5°C.