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Chapter 6: Problem 76

Twenty-five percent of the customers of a grocery store use an express checkout. Consider five randomly selected customers, and let \(x\) denote the number among the five who use the express checkout. a. Calculate \(p(2),\) that is, \(P(x=2)\). b. Calculate \(P(x \leq 1)\). c. Calculate \(P(x \geq 2)\). (Hint: Make use of your answer to Part (b).) d. Calculate \(P(x \neq 2)\).

Short Answer

Expert verified
Based on the given question and solution, the short answer is: a. \(P(x=2) = 0.2637\) b. \(P(x \leq 1) = 0.6328\) c. \(P(x \geq 2) = 0.3672\) d. \(P(x \neq 2) = 0.7363\)

Step by step solution

01

Identify variables of the binomial distribution

: We first need to identify the variables for the binomial distribution formula, which are: - n: Number of trials (in this case, n=5 randomly selected customers). - x: Number of successes we are interested in (number of customers using the express checkout). - p: Probability of a successful trial (25% or 0.25 customers using the express checkout). - q: Probability of an unsuccessful trial (75% or 0.75 customers not using the express checkout).
02

Apply the binomial distribution formula

: To calculate the probability, we apply the binomial distribution formula, which is: \[P(x) = \binom{n}{x}p^xq^{(n-x)}\] a. Calculate the probability P(x=2): For this part, we are looking for the probability that exactly 2 out of 5 customers use the express checkout. Thus, x = 2. \(P(x=2) = \binom{5}{2}(0.25)^2(0.75)^3\)
03

Calculate the binomial coefficient

: The binomial coefficient is calculated using: \[\binom{n}{x} = \frac{n!}{x!(n-x)!}\] In our case \(n=5\) and \(x=2\): \[\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{120}{2*6} = 10\]
04

Calculate the probabilities

: Now, we can plug in the values and calculate the required probabilities: a. Calculate P(x=2): \[P(x=2) = 10*(0.25)^2(0.75)^3 = 0.2637\] Thus, the probability that exactly 2 out of 5 customers use the express checkout is 0.2637. b. Calculate P(x鈮�1): To do this, we have to calculate the probability of P(x=0) and P(x=1) and sum them: \[P(x \leq 1) = P(x = 0) + P(x = 1)\] \[P(x=0) = \binom{5}{0}(0.25)^0(0.75)^5 = 0.2373\] \[P(x=1) = \binom{5}{1}(0.25)^1(0.75)^4 = 0.3955\] \[P(x \leq 1) = 0.2373 + 0.3955 = 0.6328\] Thus, the probability that at most 1 out of 5 customers uses the express checkout is 0.6328. c. Calculate P(x鈮�2): Since P(x鈮�2) is the complement of P(x鈮�1), we can find it by subtracting P(x鈮�1) from 1: \[P(x \geq 2) = 1 - P(x \leq 1) = 1 - 0.6328 = 0.3672\] Thus, the probability that at least 2 out of 5 customers use the express checkout is 0.3672. d. Calculate P(x鈮�2): Since P(x鈮�2) is the complement of P(x=2), we can find it by subtracting P(x=2) from 1: \[P(x \neq 2) = 1 - P(x = 2) = 1 - 0.2637 = 0.7363\] Thus, the probability that a different number than 2 out of 5 customers use the express checkout is 0.7363.

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Key Concepts

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

Binomial Distribution Formula
Understanding the binomial distribution is integral to grasping various probability scenarios, especially when evaluating discrete outcomes. The formula to calculate probabilities in a binomial distribution is denoted as:
\[P(x) = \binom{n}{x}p^xq^{(n-x)}\]
Here's how each component contributes to the formula:
  • \(n\) represents the number of trials or events.
  • \(x\) represents the number of successful outcomes you're seeking.
  • \(p\) is the probability of a single success.
  • \(q\) is the probability of a single failure, which equals \(1-p\).
By applying this formula, you can evaluate situations where there are two outcomes such as pass/fail, yes/no, or success/failure, which makes it a powerful tool in probability theory. In our exercise, we used this formula to model the scenario where customers use or do not use the express checkout at a grocery store.
Probability Calculations
To execute probability calculations using the binomial distribution, follow these logical steps:
  • Identify and substitute the values of \(n\), \(x\), \(p\), and \(q\) into the binomial formula.
  • Calculate the binomial coefficient, which might require factorial computations.
  • Compute each term separately before multiplying them together for precision and ease of understanding.
For example, when we looked for the probability that exactly 2 out of 5 customers use an express checkout, we computed the binomial coefficient for \(n=5\) and \(x=2\), raised the probability of success and failure to their respective powers, and then multiplied them together. This systematic approach reduces the chance of making a mistake and helps in validating the steps when dealing with more complex probability questions.
Binomial Coefficient
The binomial coefficient is a fundamental component of calculating probabilities in binomial distributions. It dictates the number of ways that \(x\) successes can occur in \(n\) trials, regardless of order, and is defined by the formula:
\[\binom{n}{x} = \frac{n!}{x!(n-x)!}\]
Where \(!\) signifies the factorial operation, i.e., the product of an integer and all the positive integers below it. In our exercise, to find the probability of seeing 2 customers using the express checkout out of 5, we computed the binomial coefficient for \(x=2\) and \(n=5\), resulting in:\[\binom{5}{2} = 10\].
This calculation is essential because it provides the number of distinct ways the successes can be arranged, which directly impacts the overall probability result. Overall, mastery of finding the binomial coefficient is vital for accurate probability calculations in binomial scenarios.

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Most popular questions from this chapter

A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) when applying for a building permit. Let \(y\) be the number of forms required of the next applicant. Suppose the probability that \(y\) forms are required is known to be proportional to \(y ;\) that is, \(p(y)=\) \(k y\) for \(y=1, \ldots, 5\) a. What is the value of \(k ?\) (Hint: \(\Sigma p(y)=1 .)\) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required?

The paper referenced in Example 6.21 suggested that a normal distribution with mean 3500 grams and standard deviation 550 grams is a reasonable model for birth weights of babies born in Canada. a. One common medical definition of a large baby is any baby that weighs more than 4000 grams at birth. What is the probability that a randomly selected Canadian baby is a large baby? b. What is the probability that a randomly selected Canadian baby weighs either less than 2000 grams or more than 4000 grams at birth? c. What birth weights describe the \(10 \%\) of Canadian babies with the greatest birth weights? 6.52 The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm ("The Effects of Split Keyboard Geometry on Upper Body Postures, Ergonomics [2009]: 104-111). a. What is the probability that a randomly selected typist's speed is at most 60 wpm? Less than 60 wpm? b. What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? d. Suppose that two typists are independently selected. What is the probability that both their speeds 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? (Hint: See Example \(6.23 .)\)

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1\). Determine each of the following probabilities: a. \(P(z<0.10)\) b. \(P(z<-0.10)\) c. \(P(0.40-1.25)\) g. \(P(z<-1.50\) or \(z>2.50)\)

6.99 A symptom validity test (SVT) is sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for Posttraumatic Stress Disorder: Application of the Binomial Distribution" (Journal of Anxiety Disorders [2008]: 1297-1302) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is 0.7 (compared to 0.96 for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=0.7\) to calculate various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: $$ \text { i. } \quad P(x=42) $$ ii. \(P(x<42)\) $$ \text { iii. } P(x \leq 42) $$ c. Explain why the probabilities calculated in Part (b) are not all equal. d. The authors calculated the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=0.96,\) they found $$ P(x \leq 42)=.000000000013 $$ Explain why the authors calculated this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.
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