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Chapter 5: Problem 48

Let \(X_{1}, X_{2}, \ldots, X_{100}\) denote the actual net weights of 100 randomly selected 50 -lb bags of fertilizer. a. If the expected weight of each bag is 50 and the variance is 1 , calculate \(P(49.9 \leq \bar{X} \leq 50.1\) ) (approximately) using the CLT. b. If the expected weight is \(49.8 \mathrm{lb}\) rather than \(50 \mathrm{lb}\) so that on average bags are underfilled, calculate \(P(49.9 \leq \bar{X} \leq 50.1)\).

Short Answer

Expert verified
a) 0.6826 b) 0.1587

Step by step solution

01

Understand the Problem

We are given a sample of 100 bag weights and need to find the probability that the sample mean \( \bar{X} \) lies between 49.9 and 50.1. We have a known mean and variance, and we'll approximate using the Central Limit Theorem (CLT).
02

Define Parameters for Part (a)

For part (a), the expected weight is 50 lb and variance is 1. We have sample size \( n = 100 \).- Population mean \( \mu = 50 \)- Population variance \( \sigma^2 = 1 \)- Standard deviation \( \sigma = 1 \)- Sample standard deviation: \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt{100}} = 0.1 \)
03

Apply the CLT for Part (a)

The Central Limit Theorem suggests \( \bar{X} \sim N(\mu, \sigma_{\bar{X}}^2) \). Therefore, \( \bar{X} \) is approximately normal with mean 50 and standard deviation 0.1.
04

Calculate the Probability for Part (a)

For \( P(49.9 \leq \bar{X} \leq 50.1) \), we convert to Z-scores:- \( Z = \frac{49.9 - 50}{0.1} = -1 \)- \( Z = \frac{50.1 - 50}{0.1} = 1 \)Using the standard normal table, \( P(-1 \leq Z \leq 1) = 0.6826 \).
05

Define Parameters for Part (b)

For part (b), the expected weight changes to 49.8 lb.- Population mean \( \mu = 49.8 \)- Population variance \( \sigma^2 = 1 \)- Standard deviation \( \sigma = 1 \)- Sample standard deviation: \( \sigma_{\bar{X}} = \frac{1}{\sqrt{100}} = 0.1 \) (same as part a).
06

Apply the CLT for Part (b)

\( \bar{X} \sim N(49.8, 0.1^2) \) so \( \bar{X} \) is approximately normal with new mean 49.8 and standard deviation 0.1.
07

Calculate the Probability for Part (b)

For \( P(49.9 \leq \bar{X} \leq 50.1) \), we need new Z-scores:- \( Z = \frac{49.9 - 49.8}{0.1} = 1 \)- \( Z = \frac{50.1 - 49.8}{0.1} = 3 \)Using the standard normal table, \( P(1 \leq Z \leq 3) \approx 0.1587 \).

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Key Concepts

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

Understanding the Sample Mean
When dealing with statistics, the **sample mean** is a fundamental concept. It is the average of a set of values from a sample of a population. Finding the sample mean helps us understand the central tendency of the data we're analyzing. For example, if we have 100 bags of fertilizer and each bag's weight is listed in our data, the sample mean is simply the sum of these weights divided by 100. This is important because the sample mean gives us a point estimate of the population mean, helping us make inferences about the larger group without measuring every single item. The sample mean's reliability increases with the size of the sample, thanks to the Central Limit Theorem (CLT). The CLT tells us that as the sample size gets larger, the distribution of the sample mean approaches a normal distribution, even if the population distribution is not normal. This means the sample mean becomes a powerful tool for probability calculations.
Probability Calculation and Its Application
Probability calculations allow us to determine the likelihood of an event happening. In the context of our fertilizer example, we need to calculate the probability that the sample mean falls within a specific range (49.9 to 50.1 pounds). We achieve this by using Z-scores, which are a measure of how many standard deviations a data point is from the mean. Calculating Z-scores requires knowing both the mean and the standard deviation of the data distribution. To calculate the Z-score for our example, we subtract the population mean from our target value and divide by the standard deviation:
  • For 49.9 lb: \(Z = \frac{49.9 - 50}{0.1} = -1\)
  • For 50.1 lb: \(Z = \frac{50.1 - 50}{0.1} = 1\)
Using standard normal distribution tables or calculators, we find the probability that the Z-score is between -1 and 1, which tells us how likely the sample mean falls in our specified range. This method becomes particularly reliable as we apply the Central Limit Theorem.
The Role of Normal Distribution in Statistics
**Normal distribution** is a probability distribution that is symmetric around the mean, showing that data near the mean are more frequent in occurrence. It forms a bell-shaped curve when plotted, where the center represents the average value of the data. In statistical analysis, normal distribution is crucial because many tests and methods assume that the data follows this distribution. When the data behaves normally, we can apply many statistical techniques with more accuracy and reliability. In our example with fertilizer bags, the Central Limit Theorem helps us by showing that the distribution of the sample mean is approximately normal. This is true even if the actual weights of the bags are not perfectly normally distributed. This allows us to use normal distribution tools, such as Z-scores and probability tables, to calculate the likelihood of the sample mean falling within a certain range, making complex probability calculations simpler and more intuitive.

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Most popular questions from this chapter

A stockroom currently has 30 components of a certain type, of which 8 were provided by supplier 1,10 by supplier 2 , and 12 by supplier 3. Six of these are to be randomly selected for a particular assembly. Let \(X=\) the number of supplier 1s components selected, \(Y=\) the number of supplier \(2 \mathrm{~s}\) components selected, and \(p(x, y)\) denote the joint pmf of \(X\) and \(Y\). a. What is \(p(3,2)\) ? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, \(p(3,2)=\) (number of outcomes with \(X=3\) and \(Y=2) /\) (total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.] b. Using the logic of part (a), obtain \(p(x, y)\). (This can be thought of as a multivariate hypergeometric distributionsampling without replacement from a finite population consisting of more than two categories.)

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of \(500 \mathrm{psi}\). a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200 ? b. If the sample size had been 15 rather than 40 , could the probability requested in part (a) be calculated from the given information?

Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\), and \(X_{6}\) denote the numbers of blue, brown, green, orange, red, and yellow M\&M candies, respectively, in a sample of size \(n\). Then these \(X_{i}\) s have a multinomial distribution. According to the M\&M web site, the color proportions are \(p_{1}=.24, p_{2}=.13, p_{3}=.16, p_{4}=\) \(.20, p_{5}=.13\), and \(p_{6}=.14\). a. If \(n=12\), what is the probability that there are exactly two M\&Ms of each color? b. For \(n=20\), what is the probability that there are at most five orange candies? [Hint: Think of an orange candy as a success and any other color as a failure.] c. In a sample of \(20 \mathrm{M} \& \mathrm{Ms}\), what is the probability that the number of candies that are blue, green, or orange is at least 10 ?

a. Show that \(\operatorname{Cov}(X, Y+Z)=\operatorname{Cov}(X, Y)+\operatorname{Cov}(X, Z)\). b. Let \(X_{1}\) and \(X_{2}\) be quantitative and verbal scores on one aptitude exam, and let \(Y_{1}\) and \(Y_{2}\) be corresponding scores on another exam. If \(\operatorname{Cov}\left(X_{1}, Y_{1}\right)=5, \operatorname{Cov}\left(X_{1}, Y_{2}\right)=1\), \(\operatorname{Cov}\left(X_{2}, Y_{1}\right)=2\), and \(\operatorname{Cov}\left(X_{2}, Y_{2}\right)=8\), what is the covariance between the two total scores \(X_{1}+X_{2}\) and \(Y_{1}+Y_{2} ?\)

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of \(1.2\). a. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51 ? b. What is the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 ?
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