Question

A fire-detection device uses three temperature-sensitive cells acting independently of one another in such a manner that any one or more can activate the alarm. Each cell has a probability \(p=.8\) of activating the alarm when the temperature reaches \(135^{\circ} \mathrm{F}\) or higher. Let \(x\) equal the number of cells activating the alarm when the temperature reaches \(135^{\circ} \mathrm{F}\) a. Find the probability distribution of \(x\). b. Find the probability that the alarm will function when the temperature reaches \(135^{\circ} \mathrm{F}\). c. Find the expected value and the variance for the random variable \(x\)

Short answer

Answer: The probability distribution of the random variable x representing the number of cells activating the alarm is: - P(x=0) = 0.008 - P(x=1) = 0.096 - P(x=2) = 0.384 - P(x=3) = 0.512 The probability that the alarm will function is 0.992. The expected value of x is 2.4 and the variance is 0.64.

Step-by-step solution

Determine the probability distribution of x

To find the probability distribution of \(x\), we need to calculate the probability of possible outcomes: no cells activating the alarm (\(x=0\)), one cell activating the alarm (\(x=1\)), two cells activating the alarm (\(x=2\)), and all three cells activating the alarm (\(x=3\)). To do this, we use the binomial probability formula: $$P(x) = \binom{n}{x} p^x (1-p)^{n-x}$$ where \(n=3\) and \(p=0.8\). Therefore, the probability distribution of the random variable \(x\) is as follows: \(x=0\): $$P(x=0) = \binom{3}{0} (0.8)^0 (0.2)^3 = 0.008$$ \(x=1\): $$P(x=1) = \binom{3}{1} (0.8)^1 (0.2)^2 = 0.096$$ \(x=2\): $$P(x=2) = \binom{3}{2} (0.8)^2 (0.2)^1 = 0.384$$ \(x=3\): $$P(x=3) = \binom{3}{3} (0.8)^3 (0.2)^0 = 0.512$$

Determine the probability that the alarm will function

To find the probability that the alarm will function, we need to determine the probability that at least one cell activates the alarm, which is the same as the complementary probability of no cells activating the alarm: $$P(\mathrm{Alarm\, functions}) = 1 - P(x=0) = 1 - 0.008 = 0.992$$

Calculate the expected value and variance for x

To find the expected value and variance for the random variable \(x\), we can use the following formulas: Expected Value: $$E(x) = \sum xP(x)$$ Variance: $$Var(x) = E(x^2) - [E(x)]^2$$ Using the probability distribution from Step 1, we can calculate the expected value: $$E(x) = 0(0.008) + 1(0.096) + 2(0.384) + 3(0.512) = 2.4$$ Now, let's calculate \(E(x^2)\): $$E(x^2) = 0^2(0.008) + 1^2(0.096) + 2^2(0.384) + 3^2(0.512) = 6.4$$ Finally, let's find the variance: $$Var(x) = E(x^2) - [E(x)]^2 = 6.4 - (2.4)^2 = 6.4 - 5.76 = 0.64$$ The expected value of \(x\) is 2.4 and the variance is 0.64.

Key concepts

Binomial Distribution

A binomial distribution is a probability distribution that summarizes the likelihood that a variable will take one of two independent values under a given set of parameters or assumptions. Think of it as flipping a coin multiple times, where each flip is an independent event and can either result in a head or a tail. In our exercise, three temperature-sensitive cells are acting independently to activate an alarm.

In this case, each cell has a probability of 0.8 to activate the alarm when a certain temperature is reached. The scenario of cell activation or not parallels our coin flip example, making it a classic case for using the binomial distribution.

• Each cell represents a trial.
• The probability of success (cell activates) is 0.8.
• The probability of failure (cell does not activate) is 0.2.

The key to solving such problems is the Binomial Probability formula: \(P(x) = \binom{n}{x} p^x (1-p)^{n-x}\). This formula helps us find the probability of a specific number of successes among a certain number of trials, considering our parameters of success and failure.

Expected Value

The expected value is a fundamental concept in probability that refers to the average outcome of a random event if it could be repeated many times. It's what you would "expect" to get on average.

For the given exercise with the fire-detection cells, the expected value represents the average number of cells that would activate if the temperature reached the critical level and the scenario played out numerous times. To calculate expected value, you multiply each possible outcome by the probability of its occurrence and sum them up.

• Use the formula: \(E(x) = \sum xP(x)\)
• Compute for each possible value of x: none, one, two, or three cells activating.

So, when calculated for our distribution, the expected value turns out to be 2.4.
This indicates that, on average, 2.4 cells are expected to activate, which makes sense considering the high probability (0.8) of each activation.

Variance

Variance is another vital concept in statistics and probability. It measures how much the outcomes of a random variable differ from the expected value. In simpler terms, variance tells us the degree of spread in a distribution.

Understanding variance helps determine the reliability of the expected value; a lower variance suggests that the outcomes are consistently close to the expected value.

• Calculate variance using: \(Var(x) = E(x^2) - [E(x)]^2 \)
• Determine \(E(x^2) \) by summing the squares of each outcome, weighted by their probabilities.

In our exercise, we find that the variance is 0.64. This low value indicates that the number of cells activating is fairly consistent around the expected value of 2.4.
The variance provides insight into the variability and reliability of the expected number of active alarm cells.