/*! 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

91Ó°ÊÓ

Bottling cola A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters \((\mathrm{ml})\). 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 a randomly selected bottle contains less than \(295 \mathrm{ml}\) ? Show your work. (b) What is the probability that the mean contents of six randomly selected bottles are less than \(295 \mathrm{ml}\) ? Show your work.

Short Answer

Expert verified
(a) P(X < 295) ≈ 0.1587; (b) P(mean < 295) ≈ 0.0071.

Step by step solution

01

Understand the Normal Distribution

The problem states that the contents of the bottles follow a normal distribution with a mean \( \mu = 298 \, \text{ml} \) and a standard deviation \( \sigma = 3 \, \text{ml} \). We will use this information to calculate probabilities for normal distributed variables.
02

Calculate Z-score for a Single Bottle (Part a)

To find the probability that a randomly selected bottle contains less than 295 ml, we need to calculate the Z-score. The Z-score formula is:\[ Z = \frac{X - \mu}{\sigma} \]where \( X \) is the value we are interested in (295 ml).Substitute the given values:\[ Z = \frac{295 - 298}{3} = \frac{-3}{3} = -1 \]
03

Use Z-table for Probability (Part a)

Using a Z-table (or normal distribution calculator), find the probability corresponding to a Z-score of -1. This gives us the probability that a randomly selected bottle contains less than 295 ml.The probability corresponding to \( Z = -1 \) is approximately 0.1587.
04

Understand the Sample Mean Distribution (Part b)

When considering a sample mean of multiple bottles, the distribution of the sample mean also follows a normal distribution. It is called the sampling distribution of the sample mean, which has mean \( \mu_{\bar{X}} = \mu = 298 \) and standard deviation \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \) where \( n = 6 \) is the sample size.
05

Calculate Standard Error for the Sample Mean (Part b)

Calculate the standard error for the sample mean using the formula:\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{6}} \]Calculate:\[ \sigma_{\bar{X}} = \frac{3}{2.449} \approx 1.225 \]
06

Calculate Z-score for Sample Mean (Part b)

To find the probability that the mean content of six bottles is less than 295 ml, calculate the Z-score using:\[ Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} \]Substitute the values:\[ Z = \frac{295 - 298}{1.225} \approx \frac{-3}{1.225} = -2.45 \]
07

Use Z-table for Probability (Part b)

Using a Z-table, find the probability corresponding to a Z-score of -2.45. This represents the likelihood that the mean of the six bottles is less than 295 ml. The probability is approximately 0.0071.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-score
The Z-score is a fundamental concept when dealing with the normal distribution. It provides a way to determine how far away a specific data point is from the mean, in terms of standard deviations. This statistical measure helps us understand where a data point falls within the distribution.
The formula for the Z-score is given by:\[ Z = \frac{X - \mu}{\sigma} \]Where:
  • \(X\) is the value of interest.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
In the context of our exercise, calculating the Z-score for a bottle containing 295 ml helps identify how uncommon or far from the average it is. A Z-score of -1, for instance, shows that the bottle is 1 standard deviation below the mean.
Probability Calculation
Once the Z-score is determined, the next step is to find the probability associated with this Z-score. Probabilities in a normal distribution can be found using a Z-table or a normal distribution calculator. These tools provide the probability that a value lies below a specific Z-score.
For example, a Z-score of -1 corresponds to a probability of about 0.1587. This means there is a 15.87% chance that a randomly selected cola bottle contains less than 295 ml. Similarly, for a Z-score of -2.45, the probability is roughly 0.0071, indicating just a 0.71% chance that the average content of six bottles falls below 295 ml.
Understanding how to navigate and interpret Z-tables is crucial for these calculations, allowing predictions about how often certain values occur within your data set.
Sampling Distribution
Sampling distribution is essential when examining the averages from multiple samples. It describes the distribution of a statistic, such as the sample mean, over numerous samples taken from the same population.
In our exercise, when multiple bottles are taken as a group, the mean content of these bottles becomes the focus, not just individual bottles. This collective distribution is also normal provided the underlying population distribution is normal (or close to it).
The mean of the sampling distribution is the same as the population mean, while the standard deviation, also known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size \( n \):\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \]For six bottles, this would adjust the standard deviation, helping understand how the average across these bottles deviates from the mean.
Central Limit Theorem
The Central Limit Theorem (CLT) is a cornerstone of probability theory. It states that the distribution of sample means approaches a normal distribution as the sample size grows, regardless of the original distribution's shape, provided the samples are independent and identically distributed.
This theorem underlies the trustworthiness of cooking modern-day statistics, allowing statisticians to make inferences. Even when the population distribution is not normal, the sample mean distribution will approximate a normal distribution if the sample is sufficiently large (usually \(n \geq 30\)).
In our bottling example, the sample size was just six, but because the original distribution is normal, the sampling distribution is also normal. This makes it possible to apply the concept of normal distribution, making calculations for probabilities straightforward and reliable, even for small sample sizes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of teens A study of the health of teenagers plans to measure the blood cholesterol levels of an SRS of 13 - to 16 -year-olds. The researchers will report the mean \(\bar{x}\) from their sample as an estimate of the mean cholesterol level \(\mu\) in this population. Explain to someone who knows little about statistics what it means to say that \(\bar{x}\) is an unbiased estimator of \(\mu\).

The candy machine Suppose a large candy machine has \(15 \%\) orange candies. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion \(\hat{p}\) of orange candies. (a) What is the mean of the sampling distribution of \(\hat{p}\) ? Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p}\). Check to see if the \(10 \%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Large Counts condition is met. (d) If the sample size were 225 rather than \(25,\) how would this change the sampling distribution of \(\hat{p} ?\)

Scores on the mathematics part of the SAT exam in a recent year were roughly Normal with mean 515 and standard deviation 114 . You choose an SRS of 100 students and average their SAT Math scores. Suppose that you do this many, many times. Which of the following are the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) (a) \(\quad\) Mean \(=515, \mathrm{SD}=114\) (b) \(\quad\) Mean \(=515, \mathrm{SD}=114 / \sqrt{100}\) (c) \(\quad\) Mean \(=515 / 100, \mathrm{SD}=114 / 100\) (d) \(\quad\) Mean \(=515 / 100, \mathrm{SD}=114 / \sqrt{100}\) (e) Cannot be determined without knowing the 100 scores.

The sampling distribution of \(\hat{p}\) is approximately Normal because (a) there are at least 7500 Division I college athletes. (b) \(n p=225\) and \(n(1-p)=525\) are both at least 10 . (c) a random sample was chosen. (d) the athletes' responses are quantitative. (e) the sampling distribution of \(\hat{p}\) always has this shape.

Hispanic workers A factory employs 3000 unionized workers, of whom \(30 \%\) are Hispanic. The 15 -member union executive committee contains 3 Hispanics. What would be the probability of 3 or fewer Hispanics if the executive committee were chosen at random from all the workers?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.