Math Problem Statement

The amount of water in a bottle is approximately normally distributed with a mean of 2.80 liters with a standard deviation of 0.025 liter. Complete parts​ (a) through​ (c) below. a. What is the probability that an individual bottle contains less than 2.78 ​liters?

enter your response here

​(Round to three decimal places as​ needed.)

b. If a sample of 4 bottles is​ selected, what is the probability that the sample mean amount contained is less than 2.78 ​liters?

enter your response here

​(Round to three decimal places as​ needed.)

c. If a sample of 25 bottles is​ selected, what is the probability that the sample mean amount contained is less than 2.78 ​liters?

enter your response here

​(Round to three decimal places as​ needed.)

Solution

To solve these questions, we can use properties of the normal distribution. Let’s break it down step by step.

We are given:

  • Mean μ=2.80\mu = 2.80 liters
  • Standard deviation σ=0.025\sigma = 0.025 liters

We'll use the standard normal distribution ZZ, where:

Z=XμσZ = \frac{X - \mu}{\sigma}

where XX is the random variable we are interested in. For part (b) and (c), we use the concept of the sampling distribution of the sample mean, which has a standard error σXˉ=σn\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}.


a. What is the probability that an individual bottle contains less than 2.78 liters?

We calculate the z-score using the formula:

Z=2.782.800.025=0.020.025=0.8Z = \frac{2.78 - 2.80}{0.025} = \frac{-0.02}{0.025} = -0.8

Now, using the standard normal table or a calculator:

P(Z<0.8)=0.2119P(Z < -0.8) = 0.2119

Thus, the probability that an individual bottle contains less than 2.78 liters is 0.212 (rounded to three decimal places).


b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.78 liters?

For a sample of n=4n = 4, the standard error is:

σXˉ=0.0254=0.0252=0.0125\sigma_{\bar{X}} = \frac{0.025}{\sqrt{4}} = \frac{0.025}{2} = 0.0125

Now, calculate the z-score for the sample mean:

Z=2.782.800.0125=0.020.0125=1.6Z = \frac{2.78 - 2.80}{0.0125} = \frac{-0.02}{0.0125} = -1.6

Using the standard normal table or a calculator:

P(Z<1.6)=0.0548P(Z < -1.6) = 0.0548

Thus, the probability that the sample mean is less than 2.78 liters for 4 bottles is 0.055 (rounded to three decimal places).


c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.78 liters?

For a sample of n=25n = 25, the standard error is:

σXˉ=0.02525=0.0255=0.005\sigma_{\bar{X}} = \frac{0.025}{\sqrt{25}} = \frac{0.025}{5} = 0.005

Now, calculate the z-score for the sample mean:

Z=2.782.800.005=0.020.005=4.0Z = \frac{2.78 - 2.80}{0.005} = \frac{-0.02}{0.005} = -4.0

Using the standard normal table or a calculator:

P(Z<4.0)0.00003P(Z < -4.0) \approx 0.00003

Thus, the probability that the sample mean is less than 2.78 liters for 25 bottles is approximately 0.000 (rounded to three decimal places).


Summary of Answers:

  • a. Probability for one bottle P(X<2.78)=0.212P(X < 2.78) = 0.212
  • b. Probability for a sample of 4 bottles P(Xˉ<2.78)=0.055P(\bar{X} < 2.78) = 0.055
  • c. Probability for a sample of 25 bottles P(Xˉ<2.78)0.000P(\bar{X} < 2.78) \approx 0.000

Would you like further clarification on any part of this process?

5 Related Questions:

  1. How is the standard error of the mean calculated for different sample sizes?
  2. What is the difference between the probability for a single observation and for a sample mean?
  3. How would the probabilities change if the standard deviation were larger?
  4. What would happen to the probability if the mean were closer to 2.78 liters?
  5. How can we interpret the result of the probability being near 0 in part c?

Tip:

When dealing with sample means, always remember that increasing the sample size reduces the standard error, making it less likely for the sample mean to deviate significantly from the population mean.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Normal Distribution
Sampling Distribution
Probability
Standard Error

Formulas

Z = (X - μ) / σ
Z = (X - μ) / (σ / √n)

Theorems

Central Limit Theorem
Properties of the Normal Distribution

Suitable Grade Level

College Level Statistics