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A certain type of automobile battery is known to last an average of 1110 days with a standard deviation of 80 days. If 400 of these batteries are selected, use the Normal Probabilities for Means applet to find the following probabilities for the average length of life of the selected batteries: a. The average is between 1100 and 1110 . b. The average is greater than 1120 . c. The average is less than 900 .

Short Answer

Expert verified
Answer: (a) The probability that the average length of life of the selected batteries is between 1100 and 1110 days is approximately 49.38%. (b) The probability that the average length of life of the selected batteries is greater than 1120 days is approximately 0.62%. (c) The probability that the average length of life of the selected batteries is less than 900 days is approximately 0%.

Step by step solution

01

Identify the given data

We are given the following information about the population of batteries: - Mean (\(\mu\)) = 1110 days - Standard Deviation (\(\sigma\)) = 80 days - Sample size (n) = 400 batteries
02

Calculate the standard error for the sample mean

Standard error represents the standard deviation of the sampling distribution for the sample mean. It can be calculated using the following formula: Standard Error (SE) = \(\frac{\sigma}{\sqrt{n}}\) SE = \(\frac{80}{\sqrt{400}}\) = \(\frac{80}{20}\) = 4
03

Use the standard normal distribution table or calculator for finding probabilities

Now, we will find the probabilities of the average length of life of the selected batteries under the given conditions (a, b, and c), using the standard normal distribution table or calculator.
04

Step 3a: Probability for average length between 1100 and 1110 days

First, we need to find the z-scores for 1100 and 1110 by using the formula: z = \(\frac{\text{sample mean} - \mu}{\text{SE}}\) z(1100) = \(\frac{1100 - 1110}{4}\) = -2.5 z(1110) = \(\frac{1110 - 1110}{4}\) = 0 Now, we can find the probability between these z-scores. P(-2.5 < z < 0) = P(z < 0) - P(z < -2.5) Using the standard normal distribution table or calculator: P(-2.5 < z < 0) ≈ 0.5000 - 0.0062 = 0.4938 Therefore, the probability that the average length of life of the selected batteries is between 1100 and 1110 days is approximately 49.38%.
05

Step 3b: Probability for average length greater than 1120 days

Next, we need to find the z-score for 1120: z(1120) = \(\frac{1120 - 1110}{4}\) = 2.5 Now, we can find the probability that z > 2.5. P(z > 2.5) = 1 - P(z < 2.5) Using the standard normal distribution table or calculator: P(z > 2.5) ≈ 1 - 0.9938 = 0.0062 Therefore, the probability that the average length of life of the selected batteries is greater than 1120 days is approximately 0.62%.
06

Step 3c: Probability for average length less than 900 days

Finding the z-score for 900: z(900) = \(\frac{900 - 1110}{4}\) = -52.5 P(z < -52.5) is extremely close to 0, as it is far from the mean in the negative direction. Therefore, the probability that the average length of life of the selected batteries is less than 900 days is approximately 0%.

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Key Concepts

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

Standard Deviation
The standard deviation is a key concept in understanding the normal distribution. It measures how much individual data points in a dataset deviate from the mean. In simple terms, it tells us how spread out the values are. A larger standard deviation means the data points are more spread out from the mean, while a smaller standard deviation indicates they are closer to the mean.

In our scenario with the automobile batteries, the standard deviation of 80 days means that the life spans of the batteries vary by an average of 80 days from the mean of 1110 days. This value provides a sense of consistency or variability in the battery life span. It is essential because it helps to quantify the amount of variation or dispersion in a set of values.
Standard Error
The standard error is a statistical term that provides an estimate of the standard deviation of a sampling distribution. It is crucial when we are dealing with sample data rather than population data.

In the exercise, we use the formula to find the standard error:\[\text{Standard Error} (SE) = \frac{\sigma}{\sqrt{n}}\]where \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size. For our 400 batteries, the standard error is 4 days. This means that the spread or variability of the average battery life is smaller with a larger sample size. Thus, it gives us a clearer and more accurate picture of where the true mean might lie.
Z-score
A z-score is a measure that describes the position of a raw score in terms of its distance from the mean, measured in standard deviation units. Z-scores are a way to standardize different datasets, allowing for direct comparison.

In our battery example, to find out how far a certain sample mean is from the overall mean of 1110 days, we compute the z-score as:\[z = \frac{\text{sample mean} - \mu}{\text{SE}}\]The z-score helps us determine the probability of a particular data point occurring within the context of the normal distribution. For instance, a z-score of 0 means the sample mean is exactly at the population mean, while a z-score of 2.5 suggests it is 2.5 standard errors above the mean. This standardization enables us to utilize standard normal distribution tables to calculate probabilities effectively.
Probability Calculation
Probability calculation within the context of a normal distribution allows us to determine the likelihood of certain outcomes. In our exercise, we are interested in probabilities for different ranges of the average battery life.

To calculate probabilities, we've used z-scores. Once we have a z-score, we look up the probability in a standard normal distribution table or use a calculator. This probability shows us how often we can expect a particular outcome based on our model and assumptions. For example, a z-score for 1100 gives us the probability of a sample mean being out this far from the mean, which we calculated as approximately 49.38%. This indicates there's a high chance of the average falling within this range. Similarly, low probabilities for extremely high or low averages underscore the rarity of such events, as seen with probabilities for averages greater than 1120 or less than 900, calculating to be just 0.62% and virtually 0%, respectively.

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