/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Let \(x_{1}, x_{2}, \ldots, x_{1... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x_{1}, x_{2}, \ldots, x_{100}\) denote the actual net weights (in pounds) of 100 randomly selected bags of fertilizer. Suppose that the weight of a randomly selected bag has a distribution with mean \(50 \mathrm{lb}\) and variance \(1 \mathrm{lb}^{2}\). Let \(\bar{x}\) be the sample mean weight \((n=100)\). a. Describe the sampling distribution of \(\bar{x}\). b. What is the probability that the sample mean is between \(49.75 \mathrm{lb}\) and \(50.25 \mathrm{lb}\) ? c. What is the probability that the sample mean is less than \(50 \mathrm{lb}\) ?

Short Answer

Expert verified
a. The sampling distribution will be a normal distribution with mean 50 lb and standard deviation 0.1 lb. b. The probability that the sample mean is between 49.75 lb and 50.25 lb is 0.98758. c. The probability that the sample mean is less than 50 lb is 0.5

Step by step solution

01

Describe the sampling distribution of \( \bar{x} \)

The sampling distribution of \( \bar{x} \) should also follow a normal distribution. According to the central limit theorem, the mean of \(\bar{x}\) is same as the population mean (i.e. 50 lb), and the standard deviation of \(\bar{x}\) (called standard error) can be calculated as the population standard deviation divided by the square root of the number of samples, i.e. \(SE = \sqrt{1}/\sqrt{100} = 0.1\) lb.
02

Compute the probability that the sample mean is between 49.75 lb and 50.25 lb

To compute this probability, we first standardize these values by subtracting the population mean and dividing the result by the standard error. The z-scores for 49.75 and 50.25 are \(Z_{49.75}= (49.75-50)/0.1 = -2.5\) and \(Z_{50.25}= (50.25-50)/0.1 = 2.5\) respectively. The probability can be calculated using Z-table (or a normal distribution calculator), P(-2.5 < Z < 2.5) = 0.98758
03

Compute the probability the sample mean is less than 50 lb

To find this probability, the Z-score for 50 should be calculated first. Since the population mean itself is 50, the Z-score is 0. Thus, the required probability is P(Z < 0), which is equal to 0.5 (as per standard normal distribution).

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that describes the distribution of sample means. When you have a sufficiently large sample size, the theorem states that the distribution of the sample means will be approximately normal, regardless of the population's initial distribution. For the problem at hand, since we are dealing with 100 samples, the CLT implies that the sample mean \(\bar{x}\) will be distributed normally around the population mean of 50 lb. The beauty of the CLT lies in its ability to simplify complex distributions, enabling statisticians to make inferences about the population using the normal distribution.

It's important to note that the 'sufficiently large' sample size typically refers to a sample with 30 or more observations, making our scenario with 100 bags of fertilizer a clear candidate for the application of the CLT.
Standard Error
Standard error (SE) measures the variability or precision of the sample mean as an estimate of the population mean. It's calculated by taking the standard deviation of the population and dividing it by the square root of the sample size. In the problem given, we compute the standard error for the mean weight of fertilizer bags as \(SE = \sqrt{1}/\sqrt{100} = 0.1\) lb.

Understanding standard error is crucial because it helps determine how much the sample mean will vary from one sample to another. A smaller standard error indicates that the sample mean is a more precise estimate of the population mean. It's this principle that allows us to create confidence intervals around the sample mean.
Normal Distribution
The normal distribution, commonly referred to as the bell curve, describes how the values of a variable are distributed. It is symmetrical, with most of the observations clustering around the central peak and probabilities for values further away from the mean tapering off equally in both directions.

In the context of our sample means, the CLT ensures that they will be distributed normally around the true population mean (\(\mu = 50\) lb). This predictable pattern is what allows us to calculate probabilities, such as the chance that the sample mean falls within a certain range.
Z-score
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. For a given value, its Z-score represents how many standard deviations it is above or below the mean. In the exercise, we computed Z-scores to find how far in standard deviations \(49.75\) lb and \(50.25\) lb are from the mean of \(50\) lb.

Calculating Z-scores is fundamental when dealing with normal distributions because this standardization allows us to use the standard normal distribution to find probabilities. For instance, knowing that a Z-score of \( -2.5\) corresponds to a value well below the mean informs us about the rarity or likelihood of that value occurring by chance.
Probability
Probability is the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, with 0 indicating impossibility and 1 indicating certainty. In the exercise, we used the concept of probability to calculate the chances of the sample mean weight being within a specific range (between \(49.75\) lb and \(50.25\) lb) and also for it being less than the population mean of \(50\) lb.

Through the normal distribution and its Z-scores, we're able to find that there's a 98.758% chance for the sample mean to fall within the specified range, and a 50% chance for it to be less than \(50\) lb. These probabilities help us understand and quantify the variability inherent in any sampling process.

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Most popular questions from this chapter

A random sample is to be selected from a population that has a proportion of successes \(\pi=.65 .\) Determine the mean and standard deviation of the sampling distribution of \(p\) for each of the following sample sizes: a. \(n=10\) b. \(n=20\) c. \(n=30\) d. \(n=50\) e. \(n=100\) f. \(n=200\)

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5 . a. What are the mean and standard deviation of the \(\bar{x}\) sampling distribution? Describe the shape of the \(\bar{x}\) sampling distribution. b. What is the approximate probability that \(\bar{x}\) will be within \(0.5\) of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

A random sample is selected from a population with mean \(\mu=100\) and standard deviation \(\sigma=10 .\) Determine the mean and standard deviation of the \(\bar{x}\) sampling distribution for each of the following sample sizes: a. \(n=9\) b. \(n=15\) c. \(n=36\) d. \(n=50\) e. \(n=100\) f. \(n=400\)

Water permeability of concrete can be measured by letting water flow across the surface and determining the amount lost (in inches per hour). Suppose that the permeability index \(x\) for a randomly selected concrete specimen of a particular type is normally distributed with mean value 1000 and standard deviation 150 . a. How likely is it that a single randomly selected specimen will have a permeability index between 850 and \(1300 ?\) b. If the permeability index is to be determined for each specimen in a random sample of size 10 , how likely is it that the sample average permeability index will be between 950 and \(1100 ?\) between 850 and 1300 ?

Explain the difference between \(\sigma\) and \(\sigma_{\bar{x}}\) and between \(\mu\) and \(\mu_{-x}\)

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