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Chapter 3: Problem 68

A certain type of digital camera comes in either a 3megapixel version or a 4-megapixel version. A camera store has received a shipment of 15 of these cameras, of which 6 have 3 -megapixel resolution. Suppose that 5 of these cameras are randomly selected to be stored behind the counter; the other 10 are placed in a storeroom. Let \(X=\) the number of 3 -megapixel cameras among the 5 selected for behindthe-counter storage. a. What kind of a distribution does \(X\) have (name and values of all parameters)? b. Compute \(P(X=2), P(X \leq 2)\), and \(P(X \geq 2)\). c. Calculate the mean value and standard deviation of \(X\).

Short Answer

Expert verified
X follows a hypergeometric distribution: \(P(X=2) \approx 0.42, P(X \leq 2) \approx 0.714, P(X \geq 2) \approx 0.706; \mu = 2, \sigma \approx 0.917\).

Step by step solution

01

Identify the Distribution

The scenario can be described using a hypergeometric distribution. The reason for this is because we are drawing a sample of cameras (without replacement) from a finite population with two types of outcomes (3-megapixel and 4-megapixel). Thus, \(X\) follows a hypergeometric distribution with the following parameters: the population size \(N = 15\), the number of success states in the population (3-megapixel cameras) \(K = 6\), and the sample size \(n = 5\).
02

Calculate P(X=2)

Using the hypergeometric distribution formula, the probability \(P(X=k)\) is calculated as: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] For \(P(X = 2)\), plug in the values \(K = 6\), \(n = 5\), \(k = 2\), and \(N = 15\): \[ P(X = 2) = \frac{\binom{6}{2} \binom{9}{3}}{\binom{15}{5}} = \frac{15 \times 84}{3003} \approx 0.42 \]
03

Calculate P(X 鈮� 2)

To find \(P(X \leq 2)\), sum the probabilities \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\). First, calculate \(P(X = 0)\) and \(P(X = 1)\) using the same formula: \[ P(X = 0) = \frac{\binom{6}{0} \binom{9}{5}}{\binom{15}{5}} = \frac{1 \times 126}{3003} = \frac{126}{3003} \approx 0.042 \] \[ P(X = 1) = \frac{\binom{6}{1} \binom{9}{4}}{\binom{15}{5}} = \frac{6 \times 126}{3003} = \frac{756}{3003} \approx 0.252 \] So, \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.042 + 0.252 + 0.42 = 0.714 \]
04

Calculate P(X 鈮� 2)

To find \(P(X \geq 2)\), use the complement rule: \[ P(X \geq 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.042 - 0.252 = 0.706 \]
05

Calculate Mean and Standard Deviation of X

For the hypergeometric distribution, the mean \(\mu\) and standard deviation \(\sigma\) are given by: \[ \mu = n \frac{K}{N} = 5 \frac{6}{15} = 2 \] and \[ \sigma = \sqrt{n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}} = \sqrt{5 \times \frac{6}{15} \times \frac{9}{15} \times \frac{10}{14}} \approx 0.9165 \]

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Key Concepts

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

Probability Calculation
To calculate probabilities using the hypergeometric distribution, it's important to understand its structure. This distribution applies to scenarios without replacement, meaning each item selected affects the probabilities of subsequent items. In our case, we have a store with 15 cameras, 6 of which are 3-megapixel. We need to find the probability of selecting a specific number of 3-megapixel cameras out of 5 total chosen.

The formula used is: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]where:
  • K is the number of successes in the population (3-megapixel cameras).
  • N is the total population size (total cameras).
  • n is the sample size (cameras chosen).
  • k is the number of successes in the sample (3-megapixel cameras chosen).
This formula allows us to compute probabilities for different values of k, such as \( P(X=2) \), which represents the probability of selecting 2 out of 5 cameras to be 3-megapixel.
Mean and Standard Deviation
The mean and standard deviation in a hypergeometric distribution provide insights into the expected average and variation of results in repeated sampling. For our camera selection problem, these statistics help us know the 'center' and spread of the number of 3-megapixel cameras among any 5 selected.
  • The mean \( \mu \) is calculated as: \[ \mu = n \frac{K}{N} \]This gives the average number of 3-megapixel cameras expected in our sample of 5. In this case, the calculation \( 5 \frac{6}{15} = 2 \) indicates we generally expect two 3-megapixel cameras in a random draw of five.
  • The standard deviation \( \sigma \) is given by: \[ \sigma = \sqrt{n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}} \] This measures the dispersion from the mean. In our example, \( \approx 0.9165 \) tells us how much the number of selected 3-megapixel cameras might vary across different samples.
Understanding these statistics is crucial for predicting and interpreting how our selection might deviate from what's expected.
Probability and Statistics Concepts
Grasping the fundamental concepts of probability and statistics is key to solving problems like the camera selection. The hypergeometric distribution itself is part of a broader set of tools used in probability and statistics to handle finite populations.
Key concepts to understand include:
  • Discrete Probability Distributions: These deal with variables that have discrete values. Hypergeometric distribution is one example, alongside others like binomial and Poisson distributions.
  • Expectation and Variability: The mean gives a long-term average outcome of the random variable, while the standard deviation measures how spread out the outcomes are.
  • Without Replacement: This implies each selection reduces the population size, directly impacting probability calculations, unlike with replacement scenarios.
By applying these concepts, we can make informed predictions and decisions based on statistical evidence, relevant in many real-world situations.

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Most popular questions from this chapter

An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let \(X=\) the number of months between successive payments. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<1 \\ .30 & 1 \leq x<3 \\ .40 & 3 \leq x<4 \\ .45 & 4 \leq x<6 \\ .60 & 6 \leq x<12 \\ 1 & 12 \leq x\end{cases} $$ a. What is the pmf of \(X\) ? b. Using just the cdf, compute \(P(3 \leq X \leq 6)\) and \(P(4 \leq X)\).

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A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is \(P(Y=2)\) ? b. What is \(p(3)\) ? c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

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Three brothers and their wives decide to have children until each family has two female children. What is the pmf of \(X=\) the total number of male children born to the brothers? What is \(E(X)\), and how does it compare to the expected number of male children born to each brother?
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