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Chapter 6: Problem 19

Suppose the sediment density \((\mathrm{g} / \mathrm{cm})\) of a randomly selected specimen from a region is normally distributed with mean \(2.65\) and standard deviation \(.85\) (suggested in "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants,"Water Res., 1984: 1169-1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most \(3.00\) ? Between \(2.65\) and \(3.00\) ? b. How large a sample size would be required to ensure that the first probability in part (a) is at least .99?

Short Answer

Expert verified
a1. Probability is 0.9803; a2. Probability is 0.4803. b. Sample size of 58 required.

Step by step solution

01

Understanding the Sampling Distribution

The sample mean of a sample size of 25, \overline{X}, is normally distributed with the same mean as the population mean \(\mu = 2.65\) and a standard deviation given by the standard error \(\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = 0.85\) and \(n = 25\), so \(\sigma_{\overline{X}} = \frac{0.85}{\sqrt{25}} = 0.17\).
02

Calculate Probability for Sample Mean Below 3.00

We need to find \(P(\overline{X} \leq 3.00)\). First, calculate the z-score: \(z = \frac{3.00 - 2.65}{0.17} = 2.06\). Using standard normal distribution tables or a calculator, find \(P(Z \leq 2.06)\). This probability is approximately \(0.9803\).
03

Calculate Probability for Sample Mean Between 2.65 and 3.00

Find \(P(2.65 \leq \overline{X} \leq 3.00)\). We already know \(P(\overline{X} \leq 3.00) = 0.9803\). The probability \(P(\overline{X} \leq 2.65)\) corresponds to the z-score of 0, which is \(0.5\). So \(P(2.65 \leq \overline{X} \leq 3.00) = 0.9803 - 0.5 = 0.4803\).
04

Determine Required Sample Size for Probability of at Least 0.99

We want \(P(\overline{X} \leq 3.00) \geq 0.99\). Start with the z-score equation \(z = \frac{3.00 - 2.65}{\sigma_{\overline{X}}}\). From standard normal tables, \(P(Z \leq z) = 0.99\) corresponds to a z-score of approximately 2.33. Thus, \(2.33 = \frac{3.00 - 2.65}{\frac{0.85}{\sqrt{n}}}\). Solve for \(n\): \(n = \left(\frac{0.85 \times 2.33}{3.00 - 2.65}\right)^2 = 57.43\). Therefore, at least \(58\) specimens are required to ensure the probability is at least 0.99.

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Key Concepts

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

Sampling Distribution
In statistics, the concept of the sampling distribution is crucial for understanding how we draw inferences about populations. Imagine taking several random samples from a population and calculating the mean of each sample. The distribution of these sample means is called the sampling distribution of the sample mean. For a sample size of 25, the sampling distribution is normally distributed, even if the original population distribution is not, due to the Central Limit Theorem.
The sampling distribution's mean is the same as the population mean, which in this case is 2.65 grams per cubic centimeter. However, the variability or spread of the sample means is determined by the standard error, which is also a key concept here. Understanding this helps us assess probabilities about where the sample mean may lie.
Standard Error
The standard error (SE) measures the dispersion of sample means around the population mean. It quantifies how much your sample mean is expected to vary if you were to take multiple samples. The formula to calculate the standard error of the mean is \(\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}\). Here, \(\sigma\) is the population standard deviation, which is 0.85, and \(n\) is the sample size, which is 25.
By plugging the values into the formula, we get \(\sigma_{\overline{X}} = \frac{0.85}{\sqrt{25}} = 0.17\). This tells us that the variability of sample means around the population mean is reduced in the sample distribution compared to the population. This smaller variability makes the sample mean a more accurate estimate of the population mean.
Z-Score
A Z-score tells you how many standard deviations away a particular value is from the mean of a distribution. In normal distribution problems, we use the Z-score to find the probability associated with a particular value of the sample mean. The Z-score formula is:
  • \( z = \frac{\overline{X} - \mu}{\sigma_{\overline{X}}} \)
For our problem, we need to find the probability that the sample mean is at most 3.00. Plugging into the formula, \( z = \frac{3.00 - 2.65}{0.17} = 2.06 \).
A Z-score of 2.06 indicates that the sample mean of 3.00 is 2.06 standard deviations above the population mean. This aids in calculating the probabilities when using the standard normal distribution table or calculator.
Probability Calculation
Probability calculations on sampling distributions use Z-scores to determine how likely it is for the sample mean to fall within specified ranges. To find the probability that the sample mean is at most 3.00, use the Z-score calculated in the previous section, which is 2.06. From the standard normal distribution table, the probability \( P(Z \leq 2.06) \) is approximately 0.9803, meaning there's a 98.03% chance the sample mean is 3.00 or less.
To find the probability that the sample mean is between 2.65 and 3.00, calculate \( P(2.65 \leq \overline{X} \leq 3.00) \). Since \( P(\overline{X} \leq 2.65) \) corresponds to a Z-score of 0 and a probability of 0.5, the desired probability is \( 0.9803 - 0.5 = 0.4803 \).
This illustrates the use of normal distribution properties to assess the behavior of sample means under specific conditions.

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Most popular questions from this chapter

A large university has 500 single employees who are covered by its dental plan. Suppose the number of claims filed during the next year by such an employee is a Poisson rv with mean value \(2.3\). Assuming that the number of claims filed by any such employee is independent of the number filed by any other employee, what is the approximate probability that the total number of claims filed is at least 1200 ?

The lifetime of a type of battery is normally distributed with mean value \(10 \mathrm{~h}\) and standard deviation \(1 \mathrm{~h}\). There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \(5 \%\) of all packages?

The inside diameter of a randomly selected piston ring is a random variable with mean value \(12 \mathrm{~cm}\) and standard deviation \(.04 \mathrm{~cm}\). a. If \(X\) is the sample mean diameter for a random sample of \(n=16\) rings, where is the sampling distribution of \(X\) centered, and what is the standard deviation of the \(\bar{X}\) distribution? b. Answer the questions posed in part (a) for a sample size of \(n=64\) rings. c. For which of the two random samples, the one of part (a) or the one of part (b), is \(X\) more likely to be within \(.01 \mathrm{~cm}\) of \(12 \mathrm{~cm}\) ? Explain your reasoning.

Show that \(F_{p, v_{1}, v_{2}}=1 / F_{1-p, v_{2}, v_{1}}\).

In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let \(X=\) the number of trees planted in sandy soil that survive 1 year and \(Y=\) the number of trees planted in clay soil that survive I year. If the probability that a tree planted in sandy soil will survive l year is. 7 and the probability of 1 -year survival in clay soil is \(.6\), compute \(P(-5 \leq X-Y \leq 5)\) (use an approximation, but do not bother with the continuity correction).
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