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The package of Ecosmart Led 75-watt replacement bulbs that use only 14 watts claims that these bulbs have an average life of 24,966 hours. Assume that the lives of all such bulbs have an approximate normal distribution with a mean of 24,966 hours and a standard deviation of 2000 hours. Find the probability that the mean life of a random sample of 25 such bulbs is a. less than 24,400 hours b. between 24,300 and 24,700 hours c. within 650 hours of the population mean d. less than the population mean by 700 hours or more

Short Answer

Expert verified
Probability results are - a) 0.078 b) 0.2 c) To be computed based on above approach d) To be computed based on above approach.

Step by step solution

01

Understanding the Variables

In this exercise, the mean life of the bulbs (μ) is 24,966 hours, the standard deviation (σ) is 2000 hours, and the sample size (n) is 25.
02

Computing Z-score for Part A

A Z-score can be calculated as \(Z = \frac{X - μ}{σ / \sqrt{n}}\). In this case, X (our target mean) is 24,400 hours. Substituting the given values into the formula gives: \(Z = \frac{24400 - 24966}{2000 / \sqrt{25}} = -1.415\). This Z-score tells us how many standard deviations below or above the population mean our value is.
03

Finding Probability for Part A

Using the computed Z-score, we check the standard normal distribution table which gives us P(Z ≤ -1.415) = 0.078. Thus, the probability that the average life of a sample of 25 bulbs is less than 24,400 hours is 0.078.
04

Computing for Part B

We follow same process as in Step 2 and Step 3. Compute the Z-scores for 24,300 and 24,700 hours: \(Z_{24300} = \frac{24300 - 24966}{2000 / \sqrt{25}} = -1.665\) \(Z_{24700} = \frac{24700 - 24966}{2000 / \sqrt{25}} = -0.665\). Now find the probabilities of these Z-scores. Here, we want probability such that Z lies between -1.665 and -0.665, which can be found out by checking standard normal distribution table: P(-1.665 ≤ Z ≤ -0.665) = P(Z ≤ -0.665) - P(Z ≤ -1.665) = 0.253 - 0.0527 = 0.2.
05

Computing for Part C

To compute for this part, note that 'within 650 hours of the population mean' translates into the boundaries points 24,966 - 650 = 24,316 and 24,966 + 650 = 25,616. We calculate the Z-scores for these points, Z_{24316} and Z_{25616} and then find the probability such that Z lies between these two Z-scores.
06

Computing for Part D

Here 'less than the population mean by 700 hours or more' implies X = 24,966 - 700 = 24,266. We calculate the Z-score for this point and find the probability for this computed Z-score.

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Key Concepts

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

Z-score calculation
The concept of a Z-score is fundamental in statistics to determine where a specific data point stands in a normal distribution. A Z-score tells us how many standard deviations an element is from the mean.
To calculate the Z-score, you use the formula: \[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} \] Here, \( X \) represents the sample mean you're measuring against the population mean \( \mu \), \( \sigma \) is the standard deviation of the population, and \( n \) is the sample size.
This formula transforms your sample mean (\( X \)) into a Z-score, which can then be used to find probabilities via standard normal distribution tables. This is very useful for understanding the likelihood of how far a sample mean will deviate from the population mean.
Standard deviation
Standard deviation is the measure of how dispersed or spread out individual data points are within a dataset. If the data points are close to the mean, the standard deviation is low. If they are spread out over a wide range, the standard deviation is high.
In the context of a normal distribution, this spread can help us understand how much we can expect individual data points to vary. When dealing with a normal distribution, around 68% of all data points fall within one standard deviation of the mean, 95% within two, and 99.7% fall within three standard deviations.
It's important to remember that when working with sample means, we adjust our standard deviation by dividing by the square root of the sample size (\( \sigma / \sqrt{n} \)). This is called the standard error and is crucial for accurately calculating Z-scores.
Probability calculation
In statistics, probability calculation refers to determining the likelihood that a specific event will occur. In our scenario, we use Z-scores to determine probabilities from standard normal distribution tables.
Once we convert a sample mean into a Z-score, this number can be mapped onto a standard normal distribution graph or table. This gives us the probability of observing a sample mean as extreme as or more extreme than the one calculated.
For example, a Z-score of -1.415 means our sample mean is 1.415 standard deviations below the mean. By looking up this Z-score in a standard normal distribution table, we find the probability that the sample mean is less than the calculated point.
Sample mean
The sample mean is simply the average of data points in a sample. It is expressed as \( \bar{X} \). The sample mean provides a point estimate of the population mean (\( \mu \)).
To find the sample mean, you add all data points in the sample and divide by the number of data points. In practice, the sample mean is used extensively in inferential statistics to estimate population parameters.
When dealing with normal distributions, as the sample size grows, the sample mean is expected to get closer to the population mean due to the Central Limit Theorem. This theorem states that the distribution of the sample means will tend to be normal, regardless of the distribution of the actual data points.

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Most popular questions from this chapter

An investigation of all five major fires in a western desert during one of the recent summers found the following causes. \(\begin{array}{llll}\text { Arson } & \text { Accident } & \text { Accident } & \text { Arson } & \text { Accident }\end{array}\) a. What proportion of those fires were due to arson? b. How many total samples (without replacement) of size three can be selected from this population? c. List all the possible samples of size three that can be selected from this population and calculate the sample proportion \(\hat{p}\) of the fires due to arson for each sample. Prepare the table that gives the sampling distribution of \(\hat{p}\). d. For each sample listed in part c, calculate the sampling error.

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The test scores for 300 students were entered into a computer, analyzed, and stored in a file. Unfortunately, someone accidentally erased a major portion of this file from the computer. The only information that is available is that \(30 \%\) of the scores were below 65 and \(15 \%\) of the scores were above 90 . Assuming the scores are approximately normally distributed, find their mean and standard deviation.

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