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Chapter 3: Problem 97

Of all customers purchasing automatic garage-door openers, \(75 \%\) purchase a chain-driven model. Let \(X=\) the number among the next 15 purchasers who select the chain-driven model. a. What is the pmf of \(X\) ? b. Compute \(P(X>10)\). c. Compute \(P(6 \leq X \leq 10)\). d. Compute \(\mu\) and \(\sigma^{2}\). e. If the store currently has in stock 10 chaindriven models and 8 shaft- driven models, what is the probability that the requests of these 15 customers can all be met from existing stock?

Short Answer

Expert verified
a. The pmf uses the binomial distribution formula. b. P(X>10)=0.403. c. P(6鈮鈮�10)=0.290. d. 渭=11.25, 蟽虏=2.8125. e. P(X鈮�10)=0.597.

Step by step solution

01

Understanding the Problem

To solve these questions, recognize that we're dealing with a binomial distribution. The probability of purchasing a chain-driven model is 0.75, and we have 15 customers. Thus, we define the random variable \( X \) as the number of customers purchasing a chain-driven model. Each customer purchase is an independent Bernoulli trial.
02

Identify the PMF of X

The probability mass function (pmf) for a binomial distribution is given by \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n = 15 \), \( k \) is the number of successes (chain-driven purchases), and \( p = 0.75 \). Substitute \( n \) and \( p \) into the formula to express the pmf.
03

Compute P(X > 10)

To find \( P(X > 10) \), calculate the sum of probabilities from \( X = 11 \) to \( X = 15 \). Use the binomial probability formula:\[ P(X > 10) = \sum_{k=11}^{15} \binom{15}{k} (0.75)^k (0.25)^{15-k} \]Compute each term and sum them to find the total probability.
04

Compute P(6 鈮� X 鈮� 10)

Calculate \( P(6 \leq X \leq 10) \) by summing probabilities from \( X = 6 \) to \( X = 10 \):\[ P(6 \leq X \leq 10) = \sum_{k=6}^{10} \binom{15}{k} (0.75)^k (0.25)^{15-k} \]Calculate each term and sum them.
05

Compute 渭 and 蟽虏

For a binomial distribution, the mean \( \mu \) is \( np = 15 \times 0.75 = 11.25 \) and the variance \( \sigma^2 \) is \( np(1-p) = 15 \times 0.75 \times 0.25 = 2.8125 \). Calculate these to find the expected value and the variance.
06

Probability of Meeting Stock Requests

The store has 10 chain-driven and 8 shaft-driven models. The purchases can be met if \( X \leq 10 \). Thus, find \( P(X \leq 10) \):\[ P(X \leq 10) = \sum_{k=0}^{10} \binom{15}{k} (0.75)^k (0.25)^{15-k} \]Compute each probability term and sum them to find the probability of all requests being met.

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Key Concepts

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

Probability Mass Function
The probability mass function (PMF) is crucial for understanding the distribution of a discrete random variable. It tells us the probability that a random variable equals a particular value. In our exercise, we're dealing with a binomial random variable, which is defined by two parameters: the number of trials (n) and the probability of success (p) on each trial.

For our problem, the PMF of the binomial distribution is given by the formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where:
  • \(n = 15\) is the number of customers.
  • \(k\) is the number of customers choosing the chain-driven model.
  • \(p = 0.75\) is the success probability for each customer.
By substituting these into the formula, we compute the probability of exactly \(k\) customers purchasing the chain-driven model.
Bernoulli Trials
Bernoulli trials are basic experiments with two possible outcomes: success or failure. In this context, a success is a customer buying a chain-driven garage door, and a failure is buying a different model.

Each purchase decision is an independent Bernoulli trial, meaning the outcome of one does not affect any other. This property is key to using the binomial distribution, which is essentially a sum of multiple Bernoulli trials. Understanding this concept helps in recognizing that the binomial model fits our scenario well, as we know the probability for each trial and the number of trials.
Expected Value and Variance
Understanding the expected value and variance helps us predict overall behavior in a probability scenario. In a binomial distribution:
  • The expected value (mean) \(\mu\) tells us the average outcome we could expect if we repeated the buying process many times.
  • Variance \(\sigma^2\) provides insight into how spread out the values might be around the mean.
For our problem, the formulas are straightforward:
  • The expected value is computed as \(\mu = np = 15 \times 0.75 = 11.25\).
  • The variance is \(\sigma^2 = np(1-p) = 15 \times 0.75 \times 0.25 = 2.8125\).
These values offer us a picture of the average purchase preference and variability among the customers.
Probability Calculation
Calculating probabilities in a binomial model involves summing the appropriate PMF values. Here, we have examples such as:
  • \(P(X > 10)\) which requires calculating probabilities for all outcomes where more than 10 customers choose the chain-driven model. This involves summing the PMFs from \(X=11\) to \(X=15\).
  • \(P(6 \leq X \leq 10)\), where we sum probabilities for outcomes between 6 and 10.
The calculations show us how probable it is for certain numbers of customers to make specific purchase choices. This helps in business planning and inventory management by providing a probabilistic outlook.
Stock Probability
Stock probability addresses whether the available stock can satisfy customer demands. Here, we have 10 chain-driven models in stock and need to calculate the probability that this stock will be sufficient.

We solve this problem by finding \(P(X \leq 10)\). This entails calculating the sum of all probabilities from \(X=0\) to \(X=10\), representing scenarios where 10 or fewer chain models are purchased.

This probability is critical for the store's inventory management as it helps predict if current stock levels can meet customer demand, reducing the possibility of shortages or excess.

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Most popular questions from this chapter

The pmf for \(X=\) the number of major defects on a randomly selected appliance of a certain type is \begin{tabular}{l|ccccc} \(x\) & 0 & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.08\) & \(.15\) & \(.45\) & \(.27\) & \(.05\) \end{tabular} Compute the following: a. \(E(X)\) b. \(V(X)\) directly from the definition c. The standard deviation of \(X\) d. \(V(X)\) using the shortcut formula

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\lambda=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20 ? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

Each time a component is tested, the trial is a success \((S)\) or failure \((F)\). Suppose the component is tested repeatedly until a success occurs on three consecutive trials. Let \(Y\) denote the number of trials necessary to achieve this. List all outcomes corresponding to the five smallest possible values of \(Y\), and state which \(Y\) value is associated with each one.

A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most \(2 .\) a. What is the probability that the batch will be accepted when the actual proportion of defectives is \(.01 ? .05 ? .10\) ? \(.20 ? .25 ?\) b. Let \(p\) denote the actual proportion of defectives in the batch. A graph of \(P\) (batch is accepted) as a function of \(p\), with \(p\) on the horizontal axis and \(P\) (batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for \(0 \leq p \leq 1\). c. Repeat parts (a) and (b) with " 1 " replacing " 2 " in the acceptance sampling plan. d. Repeat parts (a) and (b) with " 15 " replacing " 10 " in the acceptance sampling plan. e. Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

Suppose that \(90 \%\) of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? What assumptions did you make in the course of answering the question posed?
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