Problem 37
Question
Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed. Within what limits will 95 percent of the sample means occur?
Step-by-Step Solution
Verified Answer
The 95% confidence interval is from 5953.53 to 6046.47.
1Step 1: Understand the Problem
We are given a normal distribution with a mean (\( \mu = 6000 \) pounds) and standard deviation (\( \sigma = 150 \) pounds). We need to find the interval for the sample means of 40 trucks with a 95% confidence level.
2Step 2: Identify Relevant Statistics Concepts
In this scenario, we use the Central Limit Theorem, which tells us that the sampling distribution of the sample mean will be normally distributed with mean \( \mu = 6000 \) and standard deviation \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \) where \( n = 40 \).
3Step 3: Calculate the Standard Error of the Mean
The standard error (SE) is calculated as follows:\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{150}{\sqrt{40}} \approx 23.71 \]
4Step 4: Find the Z-Score for 95% Confidence
For a 95% confidence interval, we use the z-score corresponding to 95% confidence, which is approximately \( z = 1.96 \). This z-score represents the number of standard deviations from the mean.
5Step 5: Calculate the Confidence Interval
The confidence interval is calculated using the formula:\[ \text{Interval} = \mu \pm z \times \sigma_{\bar{x}} \]Plugging in the values, we get:\[ 6000 \pm 1.96 \times 23.71 \]This simplifies to:\[ 6000 \pm 46.47 \]
6Step 6: Determine the Interval Limits
Calculate the interval limits:- Lower Limit: \( 6000 - 46.47 = 5953.53 \)- Upper Limit: \( 6000 + 46.47 = 6046.47 \)Therefore, the 95% confidence interval is from 5953.53 to 6046.47.
Key Concepts
Central Limit TheoremStandard ErrorNormal Distribution
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It tells us that when we take the mean of a large number of samples from a population, the distribution of these sample means will approximate a normal distribution. This holds true even if the original population distribution is not normal.
Key points of the Central Limit Theorem include:
In the case of Crossett Trucking Company, the sample size was 40—a quantity large enough for the CLT to apply. Hence, we could assume a normal distribution for the sample means.
Key points of the Central Limit Theorem include:
- As the sample size increases, the sampling distribution becomes more like a normal distribution.
- The mean of the sample means is equal to the mean of the population (\( \mu \)).
- The standard deviation of the sampling distribution of the sample mean (\( \sigma_{\bar{x}} \)) is given by the formula \[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the standard deviation of the population and \( n \) is the sample size.
In the case of Crossett Trucking Company, the sample size was 40—a quantity large enough for the CLT to apply. Hence, we could assume a normal distribution for the sample means.
Standard Error
The standard error (SE) is a measure used to quantify the amount of variation or dispersion in the sampling distribution of a statistic, most often the sample mean. The standard error of the mean indicates how much we expect the sample mean to deviate from the actual population mean.
To calculate the standard error, you would use the formula:\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]Where:
Therefore, the standard error is calculated as:\[\sigma_{\bar{x}} = \frac{150}{\sqrt{40}} \approx 23.71 \]This tells us that the individual sample means are expected to vary by about 23.71 pounds from the true population mean. In the context of the problem, this helps determine the 'tightness' of the confidence interval for the sample means.
To calculate the standard error, you would use the formula:\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]Where:
- \( \sigma \) is the population standard deviation.
- \( n \) is the sample size.
Therefore, the standard error is calculated as:\[\sigma_{\bar{x}} = \frac{150}{\sqrt{40}} \approx 23.71 \]This tells us that the individual sample means are expected to vary by about 23.71 pounds from the true population mean. In the context of the problem, this helps determine the 'tightness' of the confidence interval for the sample means.
Normal Distribution
A normal distribution, often known as a "bell curve," is a type of continuous probability distribution for a real-valued random variable. It is symmetric around its mean, showing that data near the mean are more frequent in occurrence.
Characteristics of a normal distribution include:
In our scenario, the problem assumed that the population of delivery truck weights follows a normal distribution. This assumption allowed us to use the properties of the normal distribution, particularly the z-score, to calculate the confidence interval for the sample means, ensuring predictive precision.
Characteristics of a normal distribution include:
- The mean, median, and mode of the distribution are all equal.
- The graph of the normal distribution is bell-shaped and symmetrical about the mean.
- It is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).
- Approximately 68% of the data falls within one standard deviation of the mean; about 95% within two standard deviations; and about 99.7% within three standard deviations (Empirical Rule).
In our scenario, the problem assumed that the population of delivery truck weights follows a normal distribution. This assumption allowed us to use the properties of the normal distribution, particularly the z-score, to calculate the confidence interval for the sample means, ensuring predictive precision.
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