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Chapter 5: Problem 10

Consider a binomial distribution with \(n=10\) trials and the probability of success on a single trial \(p=0.05\) (a) Is the distribution skewed left, skewed right, or symmetric? (b) Compute the expected number of successes in 10 trials. (c) Given the low probability of success \(p\) on a single trial, would you expect \(P(r \leq 1)\) to be very high or very low? Explain. (d) Given the low probability of success \(p\) on a single trial, would you expect \(P(r \geq 8)\) to be very high or very low? Explain.

Short Answer

Expert verified
(a) Skewed right. (b) 0.5 successes. (c) Very high. (d) Very low.

Step by step solution

01

Determine Distribution Shape

A binomial distribution is considered symmetric when the probability of success \( p = 0.5 \). It is skewed right when \( p < 0.5 \) and skewed left when \( p > 0.5 \). Since \( p = 0.05 \), the distribution is skewed right.
02

Calculate Expected Number of Successes

The expected number of successes for a binomial distribution is given by the formula \( E = n \times p \). Substitute \( n = 10 \) and \( p = 0.05 \): \[ E = 10 \times 0.05 = 0.5 \]. So, the expected number of successes is 0.5.
03

Evaluate Probability \( P(r \leq 1) \)

With a low probability of success \( p = 0.05 \), the chances of obtaining 1 or fewer successes out of 10 trials is high. Therefore, \( P(r \leq 1) \) is expected to be very high.
04

Evaluate Probability \( P(r \geq 8) \)

Since \( p = 0.05 \) is very small, it is highly unlikely to obtain 8 or more successes in 10 trials. Thus, \( P(r \geq 8) \) is expected to be very low.

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Key Concepts

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

Probability in Binomial Distribution
In a binomial distribution context, probability refers to the chance of achieving a certain number of successes in a series of trials. With parameters like the number of trials, denoted as \( n \), and the probability of success on each trial, represented by \( p \), you can determine the likelihood of different outcomes. The binomial probability formula helps in calculating the probability of getting exactly \( r \) successes in \( n \) trials:
  • \( P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \)
In this exercise, the probability of success for a single trial is 0.05, while the number of trials is 10. Because the probability \( p \) is less than 0.5, most outcomes will lean towards fewer successes, making the probability of \( r \geq 8 \) very low and \( r \leq 1 \) quite high.
Understanding probability in a binomial framework helps to predict outcomes over a set of repeated experiments.
Expected Value in Binomial Distribution
Expected value in a binomial distribution gives a long-term average number of successes you can expect to observe. It represents an `average` outcome if the experiment (with \( n \) trials) were to be repeated many times. The formula to calculate expected value \( E \) in a binomial distribution is simple:
  • \( E = n \times p \)
Using this formula, let's substitute \( n = 10 \) and \( p = 0.05 \) to find the expected number of successes:
  • \( E = 10 \times 0.05 = 0.5 \)
This means you can expect, on average, less than one success in 10 trials. This is useful in estimating future outcomes based on current or past data and probability distribution.
Skewness in Binomial Distribution
Skewness indicates the degree of asymmetry of a distribution around its mean. When examining binomial distributions, the shape depends significantly on the probability \( p \) of success.
  • A symmetric distribution occurs when \( p = 0.5 \).
  • A right-skew occurs when \( p < 0.5 \).
  • A left-skew occurs when \( p > 0.5 \).
In this exercise, since \( p = 0.05 \), we have a right-skewed distribution. This indicates a longer tail on the right side of the distribution, suggesting more probability mass towards lower numbers of successes. Understanding skewness helps in visualizing data distribution, which is crucial for statistical analysis and interpretation.

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Most popular questions from this chapter

Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of passing \(77 \%\) of the students in Western Civilization each term. Let \(n=1\) \(2,3, \ldots\) represent the number of times a student takes western civilization until the first passing grade is received. (Assume the trials are independent.) (a) Write out a formula for the probability distribution of the random variable \(n\) (b) What is the probability that Susan passes on the first try \((n=1) ?\) (c) What is the probability that Susan first passes on the second try \((n=2) ?\) (d) What is the probability that Susan needs three or more tries to pass western civilization? (e) What is the expected number of attempts at western civilization Susan must make to have her (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

USA Today reported that for all airlines, the number of lost bags was May: 6.02 per 1000 passengers December: 12.78 per 1000 passengers Note: A passenger could lose more than one bag. (a) Let \(r=\) number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable \(r\) What is \(\lambda\) to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round \(\lambda\) to the nearest whole number.)

Henry Petroski is a professor of civil engineering at Duke University. In his book To Engineer Is Human: The Role of Failure in Successful Design, Professor Petroski says that up to \(95 \%\) of all structural failures, including those of bridges, airplanes, and other commonplace products of technology, are believed to be the result of crack growth. In most cases, the cracks grow slowly. It is only when the cracks reach intolerable proportions and still go undetected that catastrophe can occur. In a cement retaining wall, occasional hairline cracks are normal and nothing to worry about. If these cracks are spread out and not too close together, the wall is considered safe. However, if a number of cracks group together in a small region, there may be real trouble. Suppose a given cement retaining wall is considered safe if hairline cracks are evenly spread out and occur on the average of 4.2 cracks per 30 -foot section of wall. (a) Explain why a Poisson probability distribution would be a good choice for the random variable \(r=\) number of hairline cracks for a given length of retaining wall. (b) In a 50 -foot section of safe wall, what is the probability of three (evenly spread-out) hairline cracks? What is the probability of three or more (evenly spread-out) hairline cracks? (c) Answer part (b) for a 20 -foot section of wall. (d) Answer part (b) for a 2 -foot section of wall. Round \(\lambda\) to the nearest tenth. (e) Consider your answers to parts (b), (c), and (d). If you had three hairline cracks evenly spread out over a 50 -foot section of wall, should this be cause for concern? The probability is low. Could this mean that you are lucky to have so few cracks? On a 20-foot section of wall [part (c)], the probability of three cracks is higher. Does this mean that this distribution of cracks is closer to what we should expect? For part (d), the probability is very small. Could this mean you are not so lucky and have something to worry about? Explain your answers.

Consider a binomial experiment with \(n=7\) trials where the probability of success on a single trial is \(p=0.60\) (a) Find \(P(r=7)\) (b) Find \(P(r \leq 6)\) by using the complement rule.

USA Today reported that the U.S. (annual) birthrate is about 16 per 1000 people, and the death rate is about 8 per 1000 people. (a) Explain why the Poisson probability distribution would be a good choice for the random variable \(r=\) number of births (or deaths) for a community of a given population size. (b) In a community of 1000 people, what is the (annual) probability of 10 births? What is the probability of 10 deaths? What is the probability of 16 births? 16 deaths? (c) Repeat part (b) for a community of 1500 people. You will need to use a calculator to compute \(P(10 \text { births) and } P(16\text { births). }\) (d) Repeat part (b) for a community of 750 people.
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