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The times that college students spend studying per week have a distribution that is skewed to the right with a mean of \(8.4\) hours and a standard deviation of \(2.7\) hours. Find the probability that the mean time spent studying per week for a random sample of 45 students would be a. between 8 and 9 hours b. less than 8 hours

Short Answer

Expert verified
The probability that the mean time spent studying per week for a random sample of 45 students would be:a. between 8 and 9 hours is the difference between the probabilities at Z-scores for X=8 and X=9.b. less than 8 hours is the probability at Z-score calculated for X=8.

Step by step solution

01

Identify the given data

In this exercise, following values are given: The mean (\(\mu\)) is 8.4 hours, the standard deviation (\(\sigma\)) is 2.7 hours, and the sample size (\(n\)) is 45 students.
02

Compute Standard Error

Calculate the standard error, which is the standard deviation divided by the square root of the sample size. In this case, the standard error (\(\sigma / \sqrt{n}\)) would be \(2.7 / \sqrt{45}\).
03

Calculate Z-scores

For each part of the problem, compute the relevant Z-score. The Z score is a measure of how many standard deviations an observation or datum is from the mean. Z-score is given by the formula: (a) For time between 8 to 9 hours:Z = (X - \(\mu\)) / ( \(\sigma / \sqrt{n}) )Calculate Z using X=8 and X=9(b) For time less than 8 hours:Z = (X - \(\mu\)) / ( \(\sigma / \sqrt{n}\))Calculate Z using X=8
04

Find the probability

Look up each Z-score calculated in the previous step in a standard normal probability table (or use a calculator with a normal distribution conversion function) to find the respective probabilities. Use the property that the total probability under the normal curve is 1 to find probabilities corresponding to the given scenarios. For instances (a) and (b), find the difference between probabilities corresponding to the calculated Z-scores.

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Key Concepts

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

Z-score calculation
The Z-score is a crucial concept in statistics that helps us understand how far an observation is from the mean, measured in terms of standard deviations. It is a useful tool when we wish to find the probability of a particular observation within a normal distribution. The formula for calculating the Z-score is:\[Z = \frac{X - \mu}{\sigma / \sqrt{n}}\]Where:
  • \(X\) is the observation.
  • \(\mu\) is the mean of the population.
  • \(\sigma\) is the standard deviation of the population.
  • \(n\) is the sample size, which modifies the standard deviation component according to the Central Limit Theorem.
These components allow us to translate an observation into a standardized normal variable, enabling us to determine probabilities efficiently.

For example, in our exercise, we compute Z-scores for times 8 hours and 9 hours to determine how far they lie from the mean of 8.4 hours, using a sample of 45 students. These Z-scores help us pinpoint probabilities on a standard normal distribution.
Standard error
The standard error plays a key role in converting a normal distribution into a standard distribution by taking into account sample variability. It reflects the amount by which the sample mean's estimates of the population mean vary.

Mathematically, the standard error (SE) is calculated by:\[SE = \frac{\sigma}{\sqrt{n}}\]Where:
  • \(\sigma\) is the standard deviation of the population.
  • \(n\) is the sample size.
The larger the sample size, the smaller the standard error, suggesting a tighter estimate around the population mean. The standard error modifies the standard deviation in the Z-score formula, capturing the increased reliability of the sample mean. For our exercise on study times, the standard deviation of 2.7 hours becomes more precise when divided by the square root of the sample size, 45, yielding a smaller SE that is used in the Z-score calculation.
Normal distribution
The normal distribution is often called a bell curve because of its symmetrical shape. It is widespread in nature and statistics due to the Central Limit Theorem, which asserts that the distribution of the sample mean will approach a normal distribution as the sample size increases, even if the original data is skewed. In our exercise, although the original data on study times is skewed, with a large enough sample size (like 45), the mean study time can be assumed to follow a normal distribution.

The characteristics of a normal distribution include:
  • Symmetry about the mean.
  • Mean, median, and mode all located at the peak.
  • Predictable proportions of data within one (68%), two (95%), and three (99.7%) standard deviations from the mean.
Understanding this concept allows us to use the standard normal distribution to find probabilities related to sample means, using calculated Z-scores.
Probability
Probability helps quantify the chance of an event occurring, in this case, the likelihood of specific mean study times. By calculating Z-scores and using the properties of the standard normal distribution, we can find these probabilities.
  • Once we have the Z-scores for 8 and 9 hours, we can look these up in a standard normal distribution table to find the corresponding probabilities.
  • For studying time of less than 8 hours, we use the Z-score calculated for 8 hours to determine this probability directly.
The area under the curve of the standard normal distribution gives us these probabilities, where the total area sums to 1.

This process translates raw data into meaningful insights, informing us about the distribution and likelihood of certain outcomes within the population, incredibly useful for making decisions based on statistical data.

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Most popular questions from this chapter

Briefly explain the meaning of a population distribution and a sampling distribution. Give an example of each.

A machine at Katz Steel Corporation makes 3 -inch-long nails. The probability distribution of the lengths of these nails is normal with a mean of 3 inches and a standard deviation of \(.1\) inch. The quality control inspector takes a sample of 25 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than \(2.95\) inches or greater than \(3.05\) inches, the inspector concludes that the machine needs an adjustment. What is the probubility that based on a sample of 25 nails, the inspector will conclude that the machine needs an adjustment?

A certain elevator has a maximum legal carrying capacity of 6000 pounds. Suppose that the population of all people who ride this elevator have a mean weight of 160 pounds with a standard deviation of 25 pounds. If 35 of these people board the elevator, what is the probability that their combined wcight will exceed 6000 pounds? Assume that the 35 people constitute a random sample from the population.

What condition or conditions must hold true for the sampling distribution of the sample mean to be normal when the sample size is less than 30 ?

Johnson Electronics Corporation makes electric tubes. It is known that the standard deviation of the lives of these tubes is 150 hours. The company's research department takes a sample of 100 such tubes and finds that the mean life of these tubes is 2250 hours. What is the probability that this sample mean is within 25 hours of the mean life of all tubes produced by this company?

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