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Chapter 5: Problem 37

Twenty-three percent of automobiles are not covered by insurance (CNN, February 23 2006 ). On a particular weekend, 35 automobiles are involved in traffic accidents. a. What is the expected number of these automobiles that are not covered by insurance? b. What are the variance and standard deviation?

Short Answer

Expert verified
The expected number is 8.05, variance is 6.1955, and standard deviation is 2.49.

Step by step solution

01

Understand the Given Data

The problem states that 23% of automobiles are not insured, which means the probability \( p = 0.23 \). We have a sample size of 35 automobiles, or \( n = 35 \).
02

Find the Expected Number of Uninsured Automobiles

The expected number \( E(X) \) of uninsured cars is found using the formula for expected value in a binomial distribution: \( E(X) = n \times p \). Here, \( n = 35 \) and \( p = 0.23 \). So, \( E(X) = 35 \times 0.23 = 8.05 \).
03

Calculate the Variance

The variance \( \text{Var}(X) \) for a binomial distribution is calculated using \( \text{Var}(X) = n \times p \times (1-p) \). Substitute \( n = 35 \), \( p = 0.23 \) into the equation to get \( \text{Var}(X) = 35 \times 0.23 \times (0.77) = 6.1955 \).
04

Determine the Standard Deviation

The standard deviation \( \sigma \) is the square root of the variance: \( \sigma = \sqrt{\text{Var}(X)} \). Thus, \( \sigma = \sqrt{6.1955} \approx 2.49 \).

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Key Concepts

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

Expected Value
In statistics, the expected value of a random variable in a binomial distribution provides insight into the average outcome we could anticipate in repeated experiments. Here, we talk about the expected number of uninsured automobiles out of a sample of 35.

To calculate the expected value, use the formula for a binomial distribution:
  • Expected Value, denoted as \( E(X) \), is given by \( E(X) = n \times p \).
  • \( n \) is the sample size, and \( p \) is the probability of a success 鈥� i.e., an automobile being uninsured in this case.
Here, \( n = 35 \) and \( p = 0.23 \), which computes to \( E(X) = 35 \times 0.23 = 8.05 \).

This means, on average, we expect about 8 uninsured cars in a scenario with 35 cars.
Variance
Variance measures the variability of a random variable. In a binomial distribution, variance tells us how much the number of uninsured cars is likely to differ from the expected value.

For a binomial distribution, the variance \( \text{Var}(X) \) can be found using:
  • \( \text{Var}(X) = n \times p \times (1-p) \),
  • where \( 1-p \) represents the probability of failure, or a car being insured.
In our context, substitute \( n = 35 \) and \( p = 0.23 \):
  • \( \text{Var}(X) = 35 \times 0.23 \times 0.77 = 6.1955 \).
This variance value implies that the number of uninsured cars deviates from its expectation (8.05) by an average of about 6.20 units.
Standard Deviation
The standard deviation is a critical statistical measure that reflects the spread or dispersion of data around the expected value in a binomial distribution. It is simply the square root of variance, which offers an intuitive measure of variability.

Here is how it's calculated:
  • \( \sigma = \sqrt{\text{Var}(X)} \)
  • where \( \sigma \) symbolizes the standard deviation.
Using our variance from the previous section, \( \sigma = \sqrt{6.1955} \approx 2.49 \).

Therefore, the standard deviation tells us that in our sample of 35 cars, the number of uninsured ones is spread around the average of 8.05 by approximately 2.49 units. This helps quantify the expected fluctuation in our data, giving a more complete statistical picture.

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Most popular questions from this chapter

The Barron \(s\) Big Money Poll asked 131 investment managers across the United States about their short-term investment outlook (Barron \(s\), October 28,2002 ). Their responses showed that \(4 \%\) were very bullish, \(39 \%\) were bullish, \(29 \%\) were neutral, \(21 \%\) were bearish, and \(7 \%\) were very bearish. Let \(x\) be the random variable reflecting the level of optimism about the market. Set \(x=5\) for very bullish down through \(x=1\) for very bearish. a. Develop a probability distribution for the level of optimism of investment managers. b. Compute the expected value for the level of optimism. c. Compute the variance and standard deviation for the level of optimism. d. Comment on what your results imply about the level of optimism and its variability.

The National Basketball Association (NBA) records a variety of statistics for each team. Two of these statistics are the percentage of field goals made by the team and the percentage of three-point shots made by the team. For a portion of the 2004 season, the shooting records of the 29 teams in the NBA showed that the probability of scoring two points by making a field goal was \(.44,\) and the probability of scoring three points by making a threepoint shot was .34 (NBA website, January 3,2004 ). a. What is the expected value of a two-point shot for these teams? b. What is the expected value of a three-point shot for these teams? c. If the probability of making a two-point shot is greater than the probability of making a three-point shot, why do coaches allow some players to shoot the three-point shot if they have the opportunity? Use expected value to explain your answer.

A regional director responsible for business development in the state of Pennsylvania is concerned about the number of small business failures. If the mean number of small busi- ness failures per month is \(10,\) what is the probability that exactly four small businesses will fail during a given month? Assume that the probability of a failure is the same for any two months and that the occurrence or nonoccurrence of a failure in any month is independent of failures in any other month.

When a new machine is functioning properly, only \(3 \%\) of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. a. Describe the conditions under which this situation would be a binomial experiment. b. Draw a tree diagram similar to Figure 5.3 showing this problem as a two- trial experiment. c. How many experimental outcomes result in exactly one defect being found? d. Compute the probabilities associated with finding no defects, exactly one defect, and two defects.

In a survey conducted by the Gallup Organization, respondents were asked, "What is your favorite sport to watch?" Football and basketball ranked number one and two in terms of preference (Gallup website, January 3,2004 ). Assume that in a group of 10 individuals, 7 prefer football and 3 prefer basketball. A random sample of 3 of these individuals is selected. a. What is the probability that exactly 2 prefer football? b. What is the probability that the majority (either 2 or 3) prefer football?
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