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A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. a. What is the pmf of the number of granite specimens selected for analysis? b. What is the probability that all specimens of one of the two types of rock are selected for analysis? c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?

Short Answer

Expert verified
a. PMF: Hypergeometric distribution; b. \( P(X=10) + P(X=5) \); c. Sum \( P(X=6)+P(X=7)+P(X=8)+P(X=9) \).

Step by step solution

01

Identify Population and Sample

The geologist has 20 rock specimens in total, 10 basaltic and 10 granite. The assistant selects 15 specimens from these 20 without replacement.
02

Define Random Variables

Define the random variable \( X \) as the number of granite specimens selected from the total 15 specimens.
03

Determine Probability Mass Function (PMF)

The PMF of \( X \) is described by the hypergeometric distribution: \( P(X = k) = \frac{{\binom{10}{k} \binom{10}{15-k}}}{\binom{20}{15}} \), where \( k \) is the number of granite specimens selected.
04

Calculate Each Term of PMF (Part a)

Calculate the PMF values for \( k = 5 \) to \( k = 10 \), as it's not possible to choose more than 10 granites or fewer than 5 given the total number of specimens. Each \( P(X=k) \) corresponds to computing the combinations and dividing by the total combinations.
05

Probability All of One Type Rock Selected (Part b)

All specimens of one type (granite or basaltic) would mean selecting either 10 granite and 5 basaltic or 10 basaltic and 5 granite. The probability is \( P(X=10) + P(X=5) \).
06

Find Mean and Standard Deviation (Part c)

Use the hypergeometric distribution formulas for mean \( \mu = 15 \times \frac{10}{20} = 7.5 \) and standard deviation \( \sigma \approx \sqrt{15 \times \frac{10}{20} \times \frac{10}{20} \times \frac{5}{19}} \approx 1.58 \).
07

Probability Within 1 Standard Deviation (Part c)

The range within 1 standard deviation is \([6, 9]\). Calculate \( P(X=6) + P(X=7) + P(X=8) + P(X=9) \) to find this probability.

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Key Concepts

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

Understanding Probability Mass Function (PMF)
The Probability Mass Function (PMF) is crucial in understanding discrete random variables, especially in cases like the hypergeometric distribution. In this exercise, the PMF pertains to the random variable \( X \), which represents the number of granite specimens selected from the 15 chosen specimens. The PMF provides the probability that \( X \) takes a specific value. Specifically, for the hypergeometric distribution used here, it gives the likelihood of selecting a particular number of granite specimens from the total available.
The PMF in hypergeometric distribution is calculated using the formula:
  • \( P(X = k) = \frac{{\binom{10}{k} \binom{10}{15-k}}}{\binom{20}{15}} \)
where \( \binom{n}{k} \) represents combinations, reflecting the number of ways to choose \( k \) items from \( n \) items.
By understanding this PMF, students can determine the probability for any chosen number of granite specimens \( k \). This allows for precise calculations and insights into the behavior of the random variable within the dataset, as demonstrated in the exercise.
Exploring Standard Deviation in a Hypergeometric Distribution
Standard deviation helps quantify the amount of variation or dispersion of a set of values. In the context of our hypergeometric distribution, it indicates how much the number of granite specimens selected is expected to vary from its mean under repeated sampling conditions.
For the exercise, the standard deviation \( \sigma \) of the number of granite specimens selected is evaluated using a specific formula for hypergeometric distributions:
  • \( \sigma \approx \sqrt{15 \times \frac{10}{20} \times \frac{10}{20} \times \frac{5}{19}} \approx 1.58 \)
This indicates that on average, the number of granite specimens selected will usually deviate by about 1.58 from the mean of 7.5. Understanding standard deviation allows students to estimate the extent of variation around the mean value, providing insights such as the range of granite specimens typically selected. Additionally, it helps in understanding probability calculations, such as finding the probability that the selection is within one standard deviation of the mean.
The Role of a Random Variable
A random variable is a fundamental concept in statistics, representing a numerical outcome of a random process. In the hypergeometric distribution exercise, the random variable \( X \) is defined as the number of granite specimens selected from the total of 15 specimens chosen.
Unlike fixed numbers, a random variable can take different values, each with a certain probability associated with it. In our exercise, these values are discrete, ranging usually from 5 to 10, as outlined by the problem conditions. Understanding \( X \) as a random variable allows students to grasp its probabilistic nature and its relationship with the PMF.
The random variable helps us to convey all possible outcomes of a statistical experiment in a structured way and is instrumental in the formation and use of probability mass functions, as well as calculating mean and standard deviation. Mastering the notion of a random variable is essential for engaging deeply with concepts in probability, statistics, and their practical application to exercises like this.

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Most popular questions from this chapter

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