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Chapter 5: Problem 51

How many different simple random samples of size 5 can be obtained from a population whose size is \(50 ?\)

Short Answer

Expert verified
There are 2,118,760 different simple random samples of size 5 that can be obtained from a population of size 50.

Step by step solution

01

- Understand the Problem

Determine the number of different simple random samples of size 5 that can be drawn from a population of size 50. This involves selecting 5 items from 50 without regard to the order of selection.
02

- Identify the Formula

The number of ways to choose a sample of size 5 from 50 can be calculated using the combination formula: \ \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \( n = 50 \) and \( k = 5 \).
03

- Plug in the Values

Substitute \( n = 50 \) and \( k = 5 \) into the combination formula: \ \[ \binom{50}{5} = \frac{50!}{5!(50-5)!} = \frac{50!}{5! \times 45!} \]
04

- Simplify the Expression

Recognize that \( 50! \text{ can be expanded and then simplified with terms in } 45! \): \ \[ \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} \]
05

- Calculate the Result

Perform the multiplication and division: \ \[ \frac{50 \times 49 \times 48 \times 47 \times 46}{120} = 2,118,760 \]

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Key Concepts

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

simple random sampling
Simple random sampling is a fundamental concept in statistics. It is used to ensure that every member of a population has an equal chance of being selected in the sample. This method is crucial because it minimizes bias.
To create a simple random sample, you can use:
  • Random number generators
  • Lottery methods
  • Random digit dialing
The exercise involves selecting 5 items out of 50 without considering the order. This ensures a fair selection process, which is a key principle of simple random sampling.
combination formula
The combination formula is used to determine how many ways you can choose a subset of items from a larger set. It is especially useful when the order of selection does not matter.
The formula is:
\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)
Here, \(n\) is the total number of items, and \(k\) is the number of items to choose. In the exercise, \(n = 50\) and \(k = 5\). So, using the formula, we get:
\(\binom{50}{5} = \frac{50!}{5!(50-5)!} = \frac{50!}{5! \times 45!}\)
This formula simplifies complex selections into manageable calculations.
statistical calculations
Statistical calculations often involve breaking down complex problems using mathematical formulas. This helps in simplifying and solving them step-by-step.
For the combination formula, it's important to:
  • Plug in the given values
  • Simplify both the numerator and the denominator
  • Perform the final division to get the result
In our exercise, after substituting \(n = 50\) and \(k = 5\), the calculation becomes:
\[ \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} = \frac{254251200}{120} = 2,118,760\]
Performing these calculations accurately ensures your results are correct.
population sampling
Population sampling is an important step in statistical analysis. It involves selecting a subset (sample) from the entire population to make inferences about the whole group.
In our problem, we are sampling 5 individuals from a population of 50. This helps in conducting studies without having to observe the entire population, which is often impractical.
There are various methods of population sampling like:
  • Simple Random Sampling
  • Stratified Sampling
  • Cluster Sampling
Understanding these methods helps in choosing the best approach for your study. Simple random sampling, like in this exercise, is one of the most basic yet powerful methods to collect unbiased data.

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Most popular questions from this chapter

Forming a Committee Four members from a 50 -person committee are to be selected randomly to serve as chairperson, vice-chairperson, secretary, and treasurer. The first person selected is the chairperson; the second, the vice- chairperson; the third, the secretary; and the fourth, the treasurer. How many different leadership structures are possible?

In the Big Game, an urn contains balls numbered 1 to \(50,\) and a second urn contains balls numbered 1 to \(36 .\) From the first urn, 5 balls are chosen randomly, without replacement. From the second urn, 1 ball is chosen randomly. For a \(\$ 1\) bet, a player chooses one set of five numbers to match the balls selected from the first urn and one number to match the ball selected from the second urn. To win, all six numbers must match; that is, the player must match the first 5 balls selected from the first urn and the single ball selected from the second urn. What is the probability of winning the Big Game with a single ticket?

Suppose you are dealt 5 cards from a standard 52 -card deck. Determine the probability of being dealt three of a kind (such as three aces or three kings) by answering the following questions: (a) How many ways can 5 cards be selected from a 52 card deck? (b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck? (c) The remaining 2 cards must be different from the 3 chosen and different from each other. For example, if we drew three kings, the 4 th card cannot be a king. After selecting the three of a kind, there are 12 different ranks of card remaining in the deck that can be chosen. If we have three kings, then we can choose twos, threes, and so on. Of the 12 ranks remaining, we choose 2 of them and then select one of the 4 cards in each of the two chosen ranks. How many ways can we select the remaining 2 cards? (d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind. That is, what is the probability of selecting three of a kind and two cards that are not like?

Find the value of each combination. $$_{52} C_{1}$$

Find the value of each combination. $$_{12} C_{3}$$
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