Question

Suppose that \(25 \%\) of the fire alarms in a large city are false alarms. Let \(x\) denote the number of false alarms in a random sample of 100 alarms. Give approximations to the following probabilities: a. \(P(20 \leq x \leq 30)\) b. \(P(20

Short answer

The probabilities for the respective ranges are obtained by summing up the appropriate binomial distribution values. The standard deviation gives us the variability in the number of false alarms, and probabilities related to deviations from the mean can be calculated accordingly.

Step-by-step solution

Comprehend the Information Given

From the problem we understand that 25% of the alarms are false in a large city. This translates to 0.25 or a proportion of 1/4. With 'x' denoting the number of false alarms in a random selection of 100 alarms, our objective is to calculate probabilities of 'x' falling within different ranges.

Define Parameters

For the Binomial Distribution corresponding to 'x', the function b(x; n, p) = (nCx) (p^x) (1-p)^(n-x) where 'n' is the number of trials, 'p' is the probability of success (here, an alarm being a false alarm), and 'x' is a particular number of successes will help us to address parts a, b, and c of this problem. Here, n=100, p=0.25.

Calculate for given intervals

a. To calculate \(P(20 \leq x \leq 30)\), we sum up the binomial probabilities for all 'x' from 20 to 30 i.e., b(x=20) + b(x=21) + ... + b(x=30) \n b. To calculate \(P(20<x<30)\), we sum up the binomial probabilities for all 'x' from 21 to 29 i.e., b(x=21) + b(x=22) + ... + b(x=29) \n c. To calculate \(P(35 \leq x)\), we sum up the binomial probabilities for all 'x' from 35 to 100 i.e., b(x=35) + b(x=36) + ... + b(x=100)

Calculating Standard Deviation

For a binomial distribution, mean is given by np and standard deviation is given by sqrt(np(1-p)). Here, mean(μ) = np = 100 * 0.25 = 25 and standard deviation(σ) = sqrt(100*0.25*0.75) = 4.33 (approximately)

Calculate Deviation from Mean

d. To find the probability that 'x' deviates more than 2 standard deviations from its mean, we find \(P(x \leq μ - 2σ)\) + \(P(x \geq μ + 2σ)\) = \(P(x \leq 16.34)\) + \(P(x \geq 33.66)\). This can be calculated by summing up the corresponding binomial probabilities.

Key concepts

Binomial Distribution

When we talk about binomial distribution, we're referring to a specific probability distribution that arises in situations where there are a fixed number of independent trials, and each trial has only two possible outcomes – typically referred to as 'success' and 'failure'. The classic binomial distribution scenario includes a fixed probability of 'success' across each trial.For instance, if we were to flip a coin 10 times, and define a 'head' as a success, we would be looking at a binomial distribution with 10 trials and the probability 'p' for success (getting a head) being 0.5. Similarly, in the given exercise, with each alarm being either a true or a false alarm, and with the probability of a false alarm given as 25%, we have a real-world example of a binomial distribution with 'n' being 100 trials (alarm instances) and 'p' being 0.25 (the probability of a false alarm).Crucially, such a distribution helps us predict the likelihood of various possible numbers of 'successes' (in this case, false alarms) within our set number of trials.

Binomial Probability Calculation

Calculating binomial probabilities involves determining the likelihood of a certain number of successes across a series of trials. The formula for binomial probability, \( b(x; n, p) = \binom{n}{x} p^x (1-p)^{n-x} \), lends itself to this process, with 'n' being the total number of trials, 'x' the desired number of successes, and 'p' the probability of success on any given trial.

Example Calculation

To apply this to the exercise at hand, if we want to calculate the probability of having exactly 20 false alarms out of 100, we'd plug those numbers into our formula as \( b(20; 100, 0.25) \).It's important to remember that when looking for probabilities over a range of successes, we would sum up the individual probabilities. In the exercise, for instance, to calculate the probability of having between 20 and 30 false alarms inclusive, we would sum the probabilities from \( b(20; 100, 0.25) \) through to \( b(30; 100, 0.25) \).This accumulation of probabilities allows us to estimate the likelihood of a specific range of outcomes.

Standard Deviation

The standard deviation is a measure of how spread out numbers are in a dataset, or, in the context of probability distributions, how much variation there is from the average or mean. In the case of a binomial distribution, which is determined by discrete outcomes, the standard deviation provides insight into the variability of the count of successes (or 'successes') we can expect over a number of trials.The formula for the standard deviation of a binomial distribution is \( \sigma = \sqrt{np(1-p)} \),where 'σ' stands for standard deviation, 'n' is the number of trials, and 'p' is the probability of success. It gives us a sense of 'typical' deviations from the expected value. For example, in our exercise, we calculated the standard deviation to be approximately 4.33, offering us an understanding that a number of false alarms that are significantly higher or lower than the mean (by more than 4.33) would be relatively unusual.

Probability Distribution

A probability distribution is a mathematical description of the likelihood of different outcomes in an experiment or process. It tells us how probable it is that a random variable takes on a given value or set of values. Probability distributions can be described by probability mass functions in the case of discrete variables, like the number of false alarms, or by probability density functions for continuous variables.In the broader context, the binomial distribution is a type of probability distribution. The distribution shows us not just one or two likely outcomes, but the entire range of possible outcomes and their corresponding probabilities. It encapsulates the essence of random behavior subject to specific constraints (in this instance, a fixed number of trials and a binary outcome with fixed probability). Understanding the probability distribution enables students to grasp wider patterns in the data, anticipate probabilities for ranges of outcomes, and evaluate the significance of the results as compared to random chance.