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How many different simple random samples of size 7 can be obtained from a population whose size is 100?

Short Answer

Expert verified
There are 16,007,560,800 different simple random samples of size 7 from a population of 100.

Step by step solution

01

- Understand the Concept

To solve this problem, it's important to understand what a simple random sample is. A simple random sample means each subset of the population has an equal probability of being chosen.
02

- Use the Combination Formula

The number of different simple random samples can be calculated using the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \(n\) is the population size and \(k\) is the sample size. Here, \(n = 100\) and \(k = 7\).
03

- Calculate Factorials

Calculate the factorial for each term in the combination formula. \(100!\) is the factorial of 100, \(7!\) is the factorial of 7, and \(93!\) is the factorial of 93.
04

- Apply the Combination Formula

Substitute the factorial values into the combination formula: \[ C(100, 7) = \frac{100!}{7!(100-7)!} = \frac{100!}{7! \cdot 93!} \]
05

- Simplify the Result

Use a calculator or software to simplify and find the final value of the combination formula. This will provide the number of different simple random samples.
06

- Conclude the Solution

The number of different simple random samples of size 7 that can be obtained from a population of 100 is the result obtained in Step 5.

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Key Concepts

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

Combination Formula
To determine the number of different simple random samples, you need the combination formula. This formula is essential for calculating how many ways you can choose a sample of items from a larger set without regard to the order. The combination formula is represented as: i) \( C(n, k) = \frac{n!}{k!(n-k)!} \) Where: n) \(n\) is the population size k) \(k\) is the sample size ii) For our case, you have a population size of 100, and you need a sample size of 7. Hence, \(C(100, 7)\) is what you are solving. When using the formula, it is also important to substitute the values correctly. This helps in getting the accurate count of different possible samples.
Factorials
Factorials play a crucial role in computing combinations. A factorial, denoted by an exclamation mark (!), is the product of all positive integers up to a certain number. For example: i) The factorial of 5, written as \(5!\), is calculated as: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \] ii) To solve \(C(100, 7)\), you will be calculating three factorials: \(100!\), \(7!\), and \(93!\). \(100!\) is the multiplication of all numbers from 100 down to 1, which is a massive number. Factorials grow rapidly with larger numbers and are best computed using a calculator or specific software. Remember, understanding factorials makes it easier to handle problems involving combinations.
Sample Size Selection
The size of the sample chosen represents the number of items you want to pick out from a larger population. For simple random sampling: i) Each item in the population has an equal chance of being selected. ii) In our example, the sample size (\(k\)) is 7, and the population size (\(n\)) is 100. It is important to decide on the sample size carefully to ensure that the sample accurately represents the population. The sample size used influences the precision of your results. Simple random sampling can be depicted as drawing 7 items from a bag containing 100 items, where each selection is independent of the others.

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Most popular questions from this chapter

For a parallel structure of identical components, the system can succeed if at least one of the components succeeds. Assume that components fail independently of each other and that each component has a 0.15 probability of failure. (a) Would it be unusual to observe one component fail? Two components? (b) What is the probability that a parallel structure with 2 identical components will succeed? (c) How many components would be needed in the structure so that the probability the system will succeed is greater than \(0.9999 ?\)

A man has six shirts and four ties. Assuming that they all match, how many different shirt-and-tie combinations can he wear?

In finance, a derivative is a financial asset whose value is determined (derived) from a bundle of various assets, such as mortgages. Suppose a randomly selected mortgage has a probability of 0.01 of default. (a) What is the probability a randomly selected mortgage will not default (that is, pay off)? (b) What is the probability a bundle of five randomly selected mortgages will not default assuming the likelihood any one mortgage being paid off is independent of the others? Note: A derivative might be an investment in which all five mortgages do not default. (c) What is the probability the derivative becomes worthless? That is, at least one of the mortgages defaults? (d) In part (b), we made the assumption that the likelihood of default is independent. Do you believe this is a reasonable assumption? Explain.

The following data represent political party by age from a random sample of registered Iowa Voters $$ \begin{array}{lccccc} & \mathbf{1 7 - 2 9} & \mathbf{3 0 - 4 4} & \mathbf{4 5 - 6 4} & \mathbf{6 5 +} & \text { Total } \\ \hline \text { Republican } & 224 & 340 & 1075 & 561 & \mathbf{2 2 0 0} \\ \hline \text { Democrat } & 184 & 384 & 773 & 459 & \mathbf{1 8 0 0} \\ \hline \text { Total } & \mathbf{4 0 8} & \mathbf{7 2 4} & \mathbf{1 8 4 8} & \mathbf{1 0 2 0} & \mathbf{4 0 0 0} \\ \hline \end{array} $$ (a) Are the events "Republican" and "30-44" independent? Justify your answer. (b) Are the events "Democrat" and "65+" independent? Justify your answer. (c) Are the events "17-29" and "45-64" mutually exclusive? Justify your answer. (d) Are the events "Republican" and "45-64" mutually exclusive? Justify your answer.

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 4th 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?

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