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Flash bulbs manufactured by a certain company are sometimes defective. a. If \(5 \%\) of all such bulbs are defective, could the techniques of this section be used to approximate the probability that at least 5 of the bulbs in a random sample of size 50 are defective? If so, calculate this probability; if not, explain why not. b. Reconsider the question posed in Part (a) for the probability that at least 20 bulbs in a random sample of size 500 are defective.

Short Answer

Expert verified
a. This approximation is not valid for a sample size of 50. b. The probability that at least 20 bulbs being defective in a sample size of 500 is 0.8413.

Step by step solution

01

Checking Validity of Approximation

The techniques of this section can be used to approximate the binomial probability if both np and n(1-p) are greater than or equal to 5. Let's check this for a random sample of size 50 and defective rate of \(5%\). Here, n=50, p=0.05. So, np = 50*0.05 = 2.5 and n(1-p) = 50*0.95 = 47.5. Since np < 5, the approximation is not valid in this case.
02

Calculation of Probability

As the approximation is not valid, we cannot calculate the probability using the techniques of this section for the first part of the question.
03

Rechecking Validity of Approximation

For a sample size of 500, let's re-apply the process: n=500, p=0.05. Again, np = 500*0.05 = 25 and n(1-p) = 500*0.95 = 475 which are both greater than or equal to 5. Hence, the approximation is valid in this case.
04

Calculate the Probability for larger sample size

We use the normal approximation to the binomial. Here, we calculate the z score using the following formula: \(z = (X - np) / \sqrt{np * (1-p)}\). Plugging in the numbers, we get: \(z = (20 - 25) / \sqrt{25*0.95} = -1\). Using standard normal distribution table, the probability of z being less than -1 is 0.1587, but we need the probability that z is greater than or equals to -1, which is \(1 - 0.1587 = 0.8413\).

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Key Concepts

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

Normal Approximation to the Binomial
Understanding when and how to apply the normal approximation to the binomial distribution is crucial in statistics. The binomial distribution describes the number of successes in a fixed number of independent trials, each with the same probability of success. For larger sample sizes, the binomial distribution can be approximated by the normal distribution, due to the Central Limit Theorem, which states that the sampling distribution of the sample means will tend to be normal, regardless of the shape of the population distribution.

This simplification is possible if certain conditions are met: both the expected number of successes, denoted by the term \(np\) and the expected number of failures, \(n(1-p)\), must be greater than or equal to 5. If these conditions are not met, as in the exercise where \(np<5\), the approximation is not considered valid, and a different method, such as the exact binomial calculation, needs to be used.
Probability Calculations
The realm of probability calculations allows us to quantify the chances of various outcomes. In the context of the binomial distribution, the probability of observing a certain number of successes in a set number of trials can be calculated exactly using the binomial formula. However, for large sample sizes or for ease of calculation, we often use the normal approximation to perform these calculations faster while maintaining reasonable accuracy.

When the normal approximation is applicable, the probability of having at least a certain number of successes (or failures) can be calculated by finding the area under the standard normal curve corresponding to the calculated z-score value. This translates a discrete probability problem into a continuous one, where the integral of the density function over the desired range gives us the probability we are seeking.
Validity of Statistical Approximations
The validity of statistical approximations like the normal approximation to the binomial distribution depend on adherence to key thresholds. As mentioned, the product of the number of trials and the probability of success/failure (\(np\) and \(n(1-p)\)) should both be greater than or equal to 5 for the normal approximation to be valid.

If these conditions are not met, the approximation could introduce significant errors in the calculations. Additionally, the shape of the binomial distribution should be reasonably symmetric, which happens when the probability of success is not too close to 0 or 1. Following these guidelines ensures that the approximation does not lead to misleading results and maintains the integrity of the statistical analysis.
Sampling Distribution
In statistics, the sampling distribution is a probability distribution of a statistic that is formed by considering all possible samples of a given size from a population. It is foundational to the field as it allows us to make inferences about population parameters.

Under the Central Limit Theorem, the sampling distribution of the sample proportion tends to be normally distributed when the sample size is large enough, even if the underlying population distribution is not normal. This is what justifies using the normal approximation for binomial probabilities in large samples. The sampling distribution's mean is the population mean, and its standard deviation, known as the standard error, depends on both the population standard deviation and the sample size.
Z-Score Calculation
A z-score is a statistical measure that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations away from the mean. Z-score calculation is pivotal for using the normal approximation in binomial probability.

The formula for calculating a z-score in the context of normal approximation to the binomial is \( z = \frac{X - np}{\sqrt{np(1-p)}} \), where \(X\) is the number of successes, \(n\) is the number of trials, and \(p\) is the probability of success on a given trial. By converting to a z-score, we can use standard normal distribution tables or software to find probabilities, which adds to our ability to analyze and understand probabilistic events with greater ease and accuracy.

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Most popular questions from this chapter

Consider babies born in the "normal" range of \(37-\) 43 weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth wcight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\) ("Are Babies Normal?" The American Statistician [1999]: \(298-302\) ). a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .)\) d. How would you characterize the most extreme \(0.1 \%\) of all birth weights? e. If \(x\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0)\), then \(y=a x\) also has a normal distribution. Use this formula to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (c). How does this compare to your previous answer?

Suppose that the distribution of net typing rate in words per minute (wpm) for experienced typists can be approximated by a normal curve with mean \(60 \mathrm{wpm}\) and standard deviation 15 wpm ("Effects of Age and Skill in Typing", Journal of Experimental Psychology [1984]: \(345-371\) ). a. What is the probability that a randomly selected typist's net rate is at most 60 wpm? less than 60 wpm? b. What is the probability that a randomly selected typist's net rate is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose net rate exceeded 105 wpm? (Note: The largest net rate in a sample described in the paper is 104 wpm.) d. Suppose that two typists are independently selected. What is the probability that both their typing rates exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training?

A local television station sells \(15-\mathrm{sec}, 30-\mathrm{sec}\), and 60 -sec advertising spots. Let \(x\) denote the length of a randomly selected commercial appearing on this station, and suppose that the probability distribution of \(x\) is given by the following table: $$ \begin{array}{lrrr} x & 15 & 30 & 60 \\ p(x) & .1 & .3 & .6 \end{array} $$ a. Find the average length for commercials appearing on this station. b. If a 15 -sec spot sells for $$\$ 500$$, a 30 -sec spot for $$\$ 800$$, and a 60 -sec spot for $$\$ 1000$$, find the average amount paid for commercials appearing on this station. (Hint: Consider a new variable, \(y=\) cost, and then find the probability distribution and mean value of \(y .\) )

The article "FBI Says Fewer than 25 Failed Polygraph Test" (San Luis Obispo Tribune, July 29,2001 ) states that false-positives in polygraph tests (i.e., tests in which an individual fails even though he or she is telling the truth) are relatively common and occur about \(15 \%\) of the time. Suppose that such a test is given to 10 trustworthy individuals. a. What is the probability that all 10 pass? b. What is the probability that more than 2 fail, even though all are trustworthy? c. The article indicated that 500 FBI agents were required to take a polygraph test. Consider the random variable \(x=\) number of the 500 tested who fail. If all 500 agents tested are trustworthy, what are the mean and standard deviation of \(x ?\) d. The headline indicates that fewer than 25 of the 500 agents tested failed the test. Is this a surprising result if all 500 are trustworthy? Answer based on the values of the mean and standard deviation from Part (c).

A city ordinance requires that a smoke detector be installed in all residential housing. There is concern that too many residences are still without detectors, so a costly inspection program is being contemplated. Let \(\pi\) be the proportion of all residences that have a detector. A random sample of 25 residences is selected. If the sample strongly suggests that \(\pi<.80\) (less than \(80 \%\) have detectors), as opposed to \(\pi \geq .80\), the program will be implemented. Let \(x\) be the number of residences among the 25 that have a detector, and consider the following decision rule: Reject the claim that \(\pi=.8\) and implement the program if \(x \leq 15\). a. What is the probability that the program is implemented when \(\pi=.80\) ? b. What is the probability that the program is not implemented if \(\pi=.70 ?\) if \(\pi=.60 ?\) c. How do the "error probabilities" of Parts (a) and (b) change if the value 15 in the decision rule is changed to 14 ?

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