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91Ó°ÊÓ

Dartmouth Distribution Warehouse makes deliveries of a large number of products to its customers. It is known that \(85 \%\) of all the orders it receives from its customers are delivered on time. Let \(\hat{p}\) be the proportion of orders in a random sample of 100 that are delivered on time. Find the probability that the value of \(\hat{p}\) will be \(\mathbf{a}_{\boldsymbol{x}}\) between \(.81\) and \(.88\) b. less than \(.87\)

Short Answer

Expert verified
The results will depend on the specific values found in a standard normal table or from software calculations, and therefore cannot be pre-determined. The answers to this problem will be in the form \(P(0.81 < \hat{p} < 0.88) = X\) and \(P(\hat{p} < 0.87) = Y\) where X and Y represent probabilities.

Step by step solution

01

Identify Key Parameters

For this task, we need to fill in key parameters. The population proportion is \(p = 0.85\) and the sample size is \(n = 100\). This allows us to begin modeling the distribution of \(\hat{p}\).
02

Calculate the Distribution Mean and Standard Deviation

The expected value of \(\hat{p}\) is \(p = 0.85\), and the standard deviation of \(\hat{p}\) can be calculated by the formula \(\sqrt{(p(1-p)/n)} = \sqrt{(0.85 * 0.15 / 100)}\).
03

Standardize the Desired Proportions

Now we will determine the z-scores associated with the proportions \(.81\) and \(.88\) using the formula \(z = (\hat{p} - p) / \sigma\) where \(\sigma\) is the standard deviation calculated in Step 2. The z-score for \(.81\) will be \(z_1 = (0.81 - 0.85) / \sigma\) and for \(.88\) it will be \(z_2 = (0.88 - 0.85) / \sigma\). These z-scores provide information about how many standard deviations away from the mean these proportions lie.
04

Calculate the Probability Between Specific Bounds

Use a standard normal table, software, or calculator to find the probabilites of obtaining a z score between \(z_1\) and \(z_2\), which corresponds to the probability of the sample proportion being between \(.81\) and \(.88\). This is denoted by \(P(0.81 < \hat{p} < 0.88)\).
05

Calculate the Probability for Proportion Less Than a Value

Similar to Step 4, we'll use the z-score we found for \(.87\) and find the proportion of scores to the left (less than) this score, conveying the probability that the sample proportion is less than \(.87\). This is denoted by \(P(\hat{p} < 0.87)\).

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Key Concepts

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

Understanding Population Proportion
When discussing sampling distributions, the term "population proportion" is key. It refers to the fraction of the total population that exhibits a certain trait. In the given exercise, 85% of all orders from Dartmouth Distribution Warehouse are delivered on time. Therefore, the population proportion is represented as \( p = 0.85 \).
This value is foundational as it lays the groundwork for predicting the behavior of sample proportions. By knowing the population proportion, students can expect that any random sample taken from this population will, on average, also reflect this proportion.
Population proportions are essential for making predictions about different scenarios, especially when it's impractical to assess an entire population.
Hence, knowing \( p \) allows one to determine probabilities for samples, which is an invaluable skill when dealing with inferential statistics.
Significance of Sample Size
Sample size, noted as \( n \), is the number of observations or data points collected from the population. In the exercise, the sample size is stated as 100, meaning 100 orders are examined for punctuality.
Sample size plays a critical role in the accuracy of the sample proportion \( \hat{p} \). A larger sample size typically yields more reliable and stable estimates of the population proportion.
Calculating probabilities using sample distributions requires knowing the sample size, as it impacts the standard deviation and the z-scores.
Using large sample sizes reduces sampling error, leading to a tighter confidence interval around the population proportion. This precision is essential when making future predictions or decisions based on sample data.
Utilizing the Z-Score
Z-scores are a statistical measure that helps in understanding how far a data point is from the population mean, expressed in terms of standard deviations. In the exercise, we calculate z-scores for specific sample proportions (e.g., 0.81 and 0.88) to find probabilities.
The formula for the z-score is given by:
  • \( z = \frac{\hat{p} - p}{\sigma} \)
This formula illustrates how much a sample proportion \( \hat{p} \) deviates from the population proportion \( p \), factoring in the standard deviation \( \sigma \).
Z-scores are instrumental for evaluating the position of a sample proportion within the distribution and are often used with standard normal distribution tables to make probability statements.
For students, grasping z-scores enables them to make statistical inferences accurately and apply these insights in various practical fields like quality control or market analysis.
Importance of Standard Deviation
Standard deviation is a crucial concept in statistics that measures the extent of variability or dispersion in a data set. In this context, the standard deviation of the sample proportion \( \sigma \) captures how much the proportion \( \hat{p} \) is expected to fluctuate from the population proportion \( p \).
The standard deviation of the sample proportion is calculated using the formula:
  • \( \sigma = \sqrt{\frac{p(1-p)}{n}} \)
This equation helps in determining how spread out the sample proportions are around the expected value \( p \).
Understanding standard deviation is vital for interpreting how tightly packed or spread out the data is, offering insight into the stability and reliability of the sample proportions.
When students master this concept, it boosts their ability to assess risks and uncertainties, particularly in areas like finance and statistical quality management.

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Most popular questions from this chapter

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