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A population has mean 12 and standard deviation \(1.5 .\) a. Find the mean and standard deviation of \(x-\) for samples of size 90 . b. Find the probability that the mean of a sample of size 90 will differ from the population mean 12 by at least 0.3 unit, that is, is either less than 11.7 or more than 12.3 .

Short Answer

Expert verified
The sample mean is 12, standard deviation is approximately 0.1586. The probability of sample mean differing from 12 by at least 0.3 is 5.88%.

Step by step solution

01

Understanding the Central Limit Theorem

The Central Limit Theorem states that for a large enough sample size (usually n > 30), the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the original population distribution. This allows us to calculate the mean and standard deviation of the sample means.
02

Identify the Mean of the Sample Mean (x̄)

The mean of the sample means (also called the expected value of \( \bar{x} \)) is equal to the population mean. Hence, the mean of \( \bar{x} \) for samples of size 90 is \[\mu_{\bar{x}} = \mu = 12.\]
03

Identify the Standard Deviation of the Sample Mean (x̄)

The standard deviation of the sample means (also called the standard error) is calculated by dividing the population standard deviation by the square root of the sample size:\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1.5}{\sqrt{90}} \approx 0.15857.\]
04

Calculate the Z-scores for the Sample Mean

To find the probability that the sample mean differs from the population mean by at least 0.3, calculate the Z-scores for \( \bar{x} = 11.7 \) and \( \bar{x} = 12.3 \):For \( \bar{x} = 11.7 \):\[Z = \frac{11.7 - 12}{0.15857} \approx -1.89\]For \( \bar{x} = 12.3 \):\[Z = \frac{12.3 - 12}{0.15857} \approx 1.89\]
05

Calculate the Probability

Using the standard normal distribution table, find the probability of Z being less than -1.89 and greater than 1.89. These probabilities are both approximately \( 0.0294 \). Since the events are mutually exclusive, add these probabilities together:\[P(Z < -1.89) + P(Z > 1.89) = 0.0294 + 0.0294 = 0.0588.\]Thus, the probability that the mean of a sample of size 90 will differ from the population mean by at least 0.3 is \( 0.0588 \) or 5.88%.

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Key Concepts

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

Sampling Distribution
The sampling distribution refers to the probability distribution of a given sample statistic. In simpler terms, it's how a sample mean or other statistics would behave across multiple samples taken from the same population. In our exercise, the sample statistic at play is the sample mean. The Central Limit Theorem (CLT) plays a crucial role here. It tells us that regardless of the population's distribution, the distribution of the sample means will tend to approximate a normal distribution as the sample size increases. This is particularly true for sample sizes over 30.
  • The sampling distribution of the mean will have the same mean as the population mean.
  • The variability or spread is less than that of the population, determined by the standard error.
This allows us to make predictions and calculations about the sample mean, like determining probabilities of variances from the population mean, as illustrated in the exercise.
Standard Deviation
Standard deviation is an essential measure in statistics that tells us how much individual data points on average deviate from the mean. In the context of the sampling distribution, we deal with the 'standard error', which is the standard deviation of the sample mean. It reflects how much the sample mean would vary from one sample to the next.
  • The formula for the standard error of the mean is: \[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\] where \(\sigma_{\bar{x}}\) is the standard error, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
  • In our exercise, it was calculated as \(0.15857\) for a sample of size 90.
A smaller standard deviation in the sampling distribution indicates that the sample means are closely clustered around the population mean, making it a critical concept for probability calculation.
Probability Calculation
Probability calculation in this context involves determining how likely it is for our sample mean to fall within a certain range. Using the Central Limit Theorem, we can calculate these probabilities using standard scores (z-scores) and standard normal distribution tables.
  • We are interested in the probability that the sample mean falls outside of the specified range (11.7 to 12.3 in our exercise).
  • This involves finding the probability for the corresponding z-scores (probabilities are the areas under the z-score curve).
The outcome of the probability calculation in the exercise was 5.88%. This calculation is a powerful tool for understanding and predicting variance in sample means and guides decision making.
Z-scores
Z-scores are statistical measurements that describe a value's relation to the mean of a group of values. They are expressed in terms of standard deviations from the mean. Z-scores are very helpful in determining how far a sample mean lies from the population mean, and hence, they play a crucial role in probability calculations.
  • The formula to calculate the z-score is: \[Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}\] where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, and \(\sigma_{\bar{x}}\) is the standard error.
  • In our context, we calculated z-scores for two sample means (11.7 and 12.3) and obtained values of approximately -1.89 and 1.89.
Understanding z-scores helps in interpreting the spread and distribution of sample means and allows us to leverage the normal distribution for calculating probabilities.

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Most popular questions from this chapter

A population has mean 16 and standard deviation 1.7 . a. Find the mean and standard deviation of \(x-\) for samples of size 80 . b. Find the probability that the mean of a sample of size 80 will be more than 16.4 .

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