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Chapter 6: Problem 27

Assume a binomial probability distribution has \(p=.60\) and \(n=200\) a. What are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. What is the probability of 100 to 110 successes? d. What is the probability of 130 or more successes? e. What is the advantage of using the normal probability distribution to approximate the binomial probabilities? Use part (d) to explain the advantage.

Short Answer

Expert verified
a. Mean = 120, SD 鈮� 6.93. b. Yes, normal approximation is valid. c. Probability 鈮� 0.0837. d. Probability 鈮� 0.0853. e. Normal approximation simplifies calculations for large n.

Step by step solution

01

Calculate the Mean

For a binomial distribution, the mean is calculated using the formula \( \mu = n \cdot p \). For this problem, \( n = 200 \) and \( p = 0.60 \). So, the mean \( \mu \) is: \[ \mu = 200 \times 0.60 = 120 \]
02

Calculate the Standard Deviation

The standard deviation for a binomial distribution is calculated as \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \).Using the given values: \[ \sigma = \sqrt{200 \times 0.60 \times 0.40} = \sqrt{48} \approx 6.93 \]
03

Check Conditions for Normal Approximation

The rule of thumb for the normal approximation of the binomial distribution is that both \( n \cdot p \) and \( n \cdot (1 - p) \) should be greater than 5. Calculate: - \( n \cdot p = 200 \times 0.60 = 120 \)- \( n \cdot (1-p) = 200 \times 0.40 = 80 \)Both values are greater than 5, so the binomial distribution can be approximated by a normal distribution.
04

Probability of 100 to 110 Successes using Normal Approximation

First, use the continuity correction by adjusting the intervals to 99.5 to 110.5. Then, convert these to the z-scores: For 99.5 (lower bound):\[ z = \frac{99.5 - 120}{6.93} \approx -2.95 \]For 110.5 (upper bound):\[ z = \frac{110.5 - 120}{6.93} \approx -1.37 \]Using a standard normal distribution table or calculator, find:- The probability of z less than -1.37: \( P(Z < -1.37) \approx 0.0853 \)- The probability of z less than -2.95: \( P(Z < -2.95) \approx 0.0016 \)The probability of success between 100 to 110 is:\( 0.0853 - 0.0016 = 0.0837 \).
05

Probability of 130 or More Successes using Normal Approximation

Use the continuity correction for 129.5 as the lower bound. Calculate the z-score:\[ z = \frac{129.5 - 120}{6.93} \approx 1.37 \]Find the probability using standard normal distribution:- \( P(Z > 1.37) = 1 - P(Z < 1.37) \)- \( P(Z < 1.37) \approx 0.9147 \)So, \( P(Z > 1.37) \approx 1 - 0.9147 = 0.0853 \).
06

Advantage of Normal Approximation

The process of finding binomial probabilities using the formulas for each value can be very tedious, especially for large n and a range of values as in part (d). Using the normal approximation simplifies this calculation by converting it into a simpler form involving z-scores and probability tables, which is faster and more practical for large samples.

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Key Concepts

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

Normal Approximation
The normal approximation is a technique used to estimate the probabilities of a binomial distribution with a normal distribution. This is particularly useful for large sample sizes or when calculating exact binomial probabilities is cumbersome. To decide if normal approximation can be applied, check two main conditions:
  • Both the expected number of successes (\(n \cdot p\)) and failures (\(n \cdot (1-p)\)) should each be greater than 5.
  • The formula for verifying these conditions is quite straightforward: if both \(n \cdot p\) and \(n \cdot (1-p)\) are greater than 5, the normal approximation is appropriate.
In the given exercise, since \(n=200\), \(p=0.60\), we have \(n \cdot p=120\) and \(n \cdot (1-p)=80\). Both values satisfy the condition, which means a normal approximation can be applied, simplifying our calculations greatly.
Probability Distribution
A probability distribution describes how probabilities are assigned to each possible outcome of a statistical experiment. In simpler terms, it's a mathematical function providing the probabilities of occurrence of different possible outcomes in an experiment.
  • In a binomial distribution, which is our main focus in this exercise, we deal with a fixed number of trials, each with two possible outcomes (success or failure).
  • Probability distributions can also be approximated by continuous distributions, such as the normal distribution, especially when there is a large number of trials involved.
  • The normal distribution is particularly useful for approximating binomial distributions when the conditions are met, allowing easier calculation of probabilities over a range of values.
The exercise involves approximating the binomial distribution with a normal distribution to find the probability of achieving a certain number of successes. This is done by using z-scores and applying them to the standard normal distribution.
Standard Deviation
Standard deviation is a measure that indicates the amount of variation or dispersion in a set of values. In the context of a binomial distribution, standard deviation quantifies how much the probability of different outcomes deviates from the mean number of successes.
  • For a binomial distribution, the standard deviation is calculated with the formula \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \), where \(n\) is the number of trials, and \(p\) is the probability of success.
  • In the exercise, the calculated standard deviation is approximately 6.93, indicating this is the average amount by which the number of successes will typically differ from the mean.
  • This measure is essential when using the normal approximation, as it allows conversion of values into z-scores, determining how many standard deviations away we are from the mean.
The standard deviation helps us understand the spread of the binomial distribution, making it a vital component when approximating with the normal distribution.
Mean Calculation
Calculating the mean of a binomial distribution is a straightforward yet crucial step in understanding the distribution's behavior. The mean, represented by \( \mu \), gives us the average number of successes expected in the given number of trials.
  • The formula used is \( \mu = n \cdot p \), where \(n\) is the number of trials, and \(p\) is the probability of success.
  • In our exercise, substituting the values \(n = 200\) and \(p = 0.60\) yields a mean of 120, signifying we expect 120 successes out of 200 trials on average.
  • This mean is a central component in standardizing our normal approximation, serving as the reference point around which standard deviations are calculated.
Knowing the mean is crucial as it allows us to center our normal approximation correctly, ensuring our calculations for probabilities over ranges are accurate and meaningful.

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Most popular questions from this chapter

In an article about the cost of health care, Money magazine reported that a visit to a hospital emergency room for something as simple as a sore throat has a mean cost of \(\$ 328\) (Money, January 2009 ). Assume that the cost for this type of hospital emergency room visit is normally distributed with a standard deviation of \(\$ 92 .\) Answer the following questions about the cost of a hospital emergency room visit for this medical service. a. What is the probability that the cost will be more than \(\$ 500 ?\) b. What is the probability that the cost will be less than \(\$ 250 ?\) c. What is the probability that the cost will be between \(\$ 300\) and \(\$ 400 ?\) d. If the cost to a patient is in the lower \(8 \%\) of charges for this medical service, what was the cost of this patient's emergency room visit?

Given that \(z\) is a standard normal random variable, find \(z\) for each situation. a. The area to the right of \(z\) is .01 b. The area to the right of \(z\) is .025 c. The area to the right of \(z\) is .05 d. The area to the right of \(z\) is .10

In January \(2003,\) the American worker spent an average of 77 hours logged on to the Internet while at work (CNBC, March 15,2003 ). Assume the population mean is 77 hours, the times are normally distributed, and that the standard deviation is 20 hours. a. What is the probability that in January 2003 a randomly selected worker spent fewer than 50 hours logged on to the Internet? b. What percentage of workers spent more than 100 hours in January 2003 logged on to the Internet? c. \(A\) person is classified as a heavy user if he or she is in the upper \(20 \%\) of usage. In January \(2003,\) how many hours did a worker have to be logged on to the Internet to be considered a heavy user?

The U.S. Bureau of Labor Statistics reports that the average annual expenditure on food and drink for all families is \(\$ 5700\) (Money, December 2003 ). Assume that annual expenditure on food and drink is normally distributed and that the standard deviation is \(\$ 1500\) a. What is the range of expenditures of the \(10 \%\) of families with the lowest annual spending on food and drink? b. What percentage of families spend more than \(\$ 7000\) annually on food and drink? c. What is the range of expenditures for the \(5 \%\) of families with the highest annual spending on food and drink?

Collina's Italian Caf茅 in Houston, Texas, advertises that carryout orders take about 25 minutes (Collina's website, February 27,2008 ). Assume that the time required for a carryout order to be ready for customer pickup has an exponential distribution with a mean of 25 minutes. a. What is the probability than a carryout order will be ready within 20 minutes? b. If a customer arrives 30 minutes after placing an order, what is the probability that the order will not be ready? c. \(\quad\) A particular customer lives 15 minutes from Collina's Italian Caf茅. If the customer places a telephone order at 5: 20 P.M., what is the probability that the customer can drive to the caf茅, pick up the order, and return home by 6: 00 p.n.?
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