/*! 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 55 Bottling cola A bottling company... [FREE SOLUTION] | 91Ó°ÊÓ

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Bottling cola A bottling company uses a flling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliters \((\mathrm{ml}) .\) In fact, the contents vary according to a Normal distribution with mean \(\mu=298 \mathrm{ml}\) and standard deviation \(\sigma=3 \mathrm{ml}\) (a) What is the probability that an individual bottle contains less than 295 \(\mathrm{ml}\) ? Show your work. (b) What is the probability that the mean contents of six randomly selected bottles is less than 295 \(\mathrm{ml}\) ? Show your work.

Short Answer

Expert verified
(a) 15.87% probability for a single bottle. (b) 0.71% probability for the mean of six bottles.

Step by step solution

01

Understanding the Problem

We are given that the contents of the cola bottles are normally distributed with a mean \( \mu = 298 \mathrm{ml} \) and a standard deviation \( \sigma = 3 \mathrm{ml} \). We need to find two probabilities: (a) for a single bottle containing less than 295 ml, and (b) that the mean contents of six randomly selected bottles is less than 295 ml.
02

Calculate Z-Score for Part (a)

To find the probability that an individual bottle contains less than 295 ml, we calculate the Z-score. The formula for the Z-score is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value we are interested in, 295 ml in this case. Substituting the values:\[ Z = \frac{295 - 298}{3} = \frac{-3}{3} = -1 \]
03

Find Probability Using Z-Score for Part (a)

Using the Z-score calculated in Step 2, we can find the probability by looking up the Z-score of \(-1\) in the standard normal distribution table. This gives us a probability of approximately \(0.1587\) (or 15.87%).
04

Calculate Standard Error for Part (b)

For six randomly selected bottles, we need to consider the sampling distribution of the sample mean. The standard error of the mean \( \sigma_{\bar{X}} \) is calculated using \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \), where \( n = 6 \). Substituting the known values gives:\[ \sigma_{\bar{X}} = \frac{3}{\sqrt{6}} \approx 1.2247 \]
05

Calculate Z-Score for Part (b)

Now, we calculate the Z-score for the mean contents of six bottles. \( Z = \frac{X - \mu}{\sigma_{\bar{X}}} \), where \( X = 295 \). Substituting the values:\[ Z = \frac{295 - 298}{1.2247} \approx \frac{-3}{1.2247} \approx -2.4495 \]
06

Find Probability Using Z-Score for Part (b)

Using the Z-score calculated in Step 5, look up the probability for \(-2.4495\) in the standard normal distribution table. This gives a probability of approximately \(0.0071\) (or 0.71%).

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental concept in statistics, especially when dealing with normal distributions. In a normal distribution, we calculate the probability of a particular event happening by determining how frequently an event might occur under specific conditions. For instance, in our cola bottling scenario, finding the probability that a single bottle contains less than 295 ml involves several steps:
  • Identify the mean (\( \mu = 298 \text{ ml}\)).
  • Determine the standard deviation (\( \sigma = 3 \text{ ml}\)).
  • Calculate or find the Z-score using the value of interest (295 ml).
  • Use the Z-score to find the corresponding probability from a standard normal distribution table.
For part (b), calculating the probability that the mean of six bottles is less than 295 ml involves considering a sample (not just a single observation). Thus, the standard error (SE) is crucial, affecting the resulting probability. Overall, probability calculations in the context of normal distributions provide valuable insights into likely outcomes based on known data attributes like mean and standard deviation.
Z-Score
The Z-score is a statistical measure that helps us understand the significance of data points relative to the mean of a data set. It indicates how many standard deviations a data point is from the mean. Calculating the Z-score is a crucial step for converting a value into the standard normal distribution. The formula is:\[ Z = \frac{X - \mu}{\sigma} \]where:
  • \( X \) is the value in question (e.g., 295 ml for one bottle),
  • \( \mu \) is the mean of the distribution, and
  • \( \sigma \) is the standard deviation.
By transforming raw data using Z-scores, we gain access to standardized tables that tell us the probability of observing a value less than the given data point. In part (b) of the exercise, we use an adjusted Z-score calculation because we are considering a sample mean rather than a single observation. This involves using the standard error of the mean for the calculation.
Standard Deviation
Standard deviation (\( \sigma \)) is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates a wider spread of values. In our example, a standard deviation of 3 ml tells us that most bottles are filled close to 298 ml, but there's some variation.
In probability calculations involving a sample, the standard deviation is used to determine the standard error, which then helps calculate the Z-score for the probability of a sample mean. The formula for standard error (\( \sigma_{\bar{X}} \)) is:\[\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\]where:
  • \( \sigma \) is the population standard deviation, and
  • \( n \) is the sample size (e.g., 6 in case of multiple bottles).
Understanding the role of standard deviation is crucial in interpreting the variability of data and calculating the probability of outcomes in normal distributions.

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