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Chapter 7: Problem 55

Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection? b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail to pass inspection? c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection? d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?

Short Answer

Expert verified
a) Probability for at most 5 cars failing is calculated using binomial distribution. b) For 5 to 10 cars failing, use the same concept as a) and subtract the probabilities. c) Use the formula for mean and standard deviation for binomial distribution. d) Convert into standard normal variable and use standard distribution table.

Step by step solution

01

Calculate probability for at most 5 cars failing

We use the formula for the cumulative binomial distribution probability, given by \(P(k鈮) = \sum_{i=0}^{k} C(n,i) * p^i * (1-p)^{n-i}\) for k=0 to 5. Here, n=15 (number of trials), p=0.3 (probability of failing) and C(n,i) is the binomial coefficient.
02

Calculate probability for between 5 and 10 cars failing

As above, we use the binomial distribution formula. However, this time we calculate the cumulative distribution from k=0 to 10, then subtract the cumulative distribution from k=0 to 4, thus giving the probability of 5 to 10 inclusive.
03

Calculate the mean and standard deviation

The mean value of a binomial distribution is \(np\) and the standard deviation is \(\sqrt{np(1-p)}\). Substituting the given values n=25 and p=0.3, we first find the mean and standard deviation for the cars failing the inspection. To find the values for those passing, we subtract from the total number of cars.
04

Calculate probability for number of passed cars within 1 standard deviation of the mean

Convert it into a standard normal variable Z=\((X-\mean)/\sd\). In this case, X is the number of cars that pass and is within one standard deviation, meaning X should be between \(\mean - \sd\) and \(\mean + \sd\). Use the standard normal distribution table to find the probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cumulative Probability
Cumulative probability is a concept that helps us determine the likelihood of a random variable falling within a particular range. When dealing with the binomial distribution, cumulative probability measures the probability that the variable is equal to or less than a certain value.

In the context of car inspections, when we're interested in finding the probability that at most five cars fail the inspection out of fifteen, we are calculating cumulative probability. The formula we use is given by: \[P(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}\] Here, \( n \) is the total number of trials (cars), \( p \) is the probability of a car failing inspection, and \( \binom{n}{i} \) represents the binomial coefficient, which counts the ways to choose \( i \) failures from \( n \) trials.

This method is useful because it adds up the probabilities from zero up to the maximum value (in this case, five), making the result easy to interpret and apply.
Mean and Standard Deviation
The mean and standard deviation are key statistical measures that help us understand and describe binomial distributions. The mean gives us an average measure, representing the expected number of outcomes, while the standard deviation measures the variability or spread of these outcomes.

For a binomial distribution, the mean \( \mu \) and standard deviation \( \sigma \) are calculated using the following formulas: - Mean: \( \mu = np \) - Standard Deviation: \( \sigma = \sqrt{np(1-p)} \) where \( n \) is the number of trials and \( p \) is the probability of success (or in context, failing inspection).

In the car inspection problem, if we take \( n = 25 \) and \( p = 0.3 \) for failing, the mean and standard deviation can help us determine how many cars we expect will pass or fail inspection on average. To find these values for cars that pass, we would subtract those who fail from the total number of cars examined. It provides an essential glimpse into the performance and reliability of the inspection process.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. It is often used in statistics to transform normal variables into a standard scale, allowing for the use of Z-scores.

In problems involving binomial distributions, such as the number of cars passing inspections, standard normalization allows us to find probabilities related to the average number of cars passing. To find whether the number of passing cars is within one standard deviation of the mean, the Z-score transformation is utilized by the formula: \[Z = \frac{X - \mu}{\sigma}\] Here, \( X \) is the observed number of cars passing, and \( \mu \) and \( \sigma \) are the mean and standard deviation we calculated earlier.

By using the standard normal distribution table, we can find the probability that the number of cars passing is within one standard deviation from the mean. This method provides a robust outcome, as it quantifies the likelihood of this event occurring amidst normal variability, making it an essential tool in quality control and assessment processes.

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Most popular questions from this chapter

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