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Given the discrete uniform population $$ I(x)=\left\\{\begin{array}{ll}\frac{1}{3}, & x=2,4,6 \\\0, & \text { }\end{array}\right.$$ find the probability that a random sample of size 54 selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4 . Assume the means to be measured to the nearest tenth.

Short Answer

Expert verified
The probability of having a sample mean between 4.1 and 4.4 given a sample of 54 with replacement is 0.086.

Step by step solution

01

Understand the Distribution

The information given says that there is a discrete uniform population. This means that the values are evenly distributed and only take on a specific few values. In this case, it can only be 2, 4, or 6, each with a probability of 1/3.
02

Calculate the mean

Calculate the mean (expected value) of the population. Since each value has the same probability, this is simply the average of the numbers. \(E(X) = (2+4+6)/3 = 4\). This is the mean of the distribution.
03

Identify the Standard Deviation

Now, calculate the standard deviation of the population. The variance of a discrete uniform distribution is given by \(\frac{(b-a+1)^{2}-1}{12}\). In this case, a = 2, b = 6: \(\sigma^{2} = Var(X) = \frac{(6-2+1)^{2}-1}{12} = \frac{9}{12} = 0.75 \). Hence, the standard deviation \(\sigma = \sqrt{0.75} = 0.866\).
04

Standardize the Desired Range

Now we want to find the probability that a sample of 54 will yield a mean greater than 4.1 but less than 4.4. To do this, we convert these values into z-scores so that we have standard normal variables. The z-value is given by \((X - \mu)/(\sigma/\sqrt{n})\). In this case, n = 54. For x=4.1, \(z_{1} = (4.1 - 4) / (0.866/\sqrt{54}) = 1.33 \) and for x=4.4, \(z_{2} = (4.4 - 4) / (0.866/\sqrt{54}) = 2.51\).
05

Find Probability

Finally, look these up in the z-table or use a standard normal distribution calculator to find the probabilities for these z-scores. The z-table gives the cumulative probability up to the z-value, so to find the probability between two z-scores, subtract the lower from the upper. The z-score of 1.33 has a cumulative probability of 0.908, and a z-score of 2.51 has a cumulative probability of 0.994. Therefore, the probability of having a sample mean between 4.1 and 4.4 given a sample of 54 with replacement is 0.994 - 0.908 = 0.086.

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Key Concepts

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

Discrete Uniform Distribution
In probability and statistics, a discrete uniform distribution refers to a situation where a finite number of outcomes are equally likely to occur. This means every possible outcome has the same probability of happening.

For example, consider a fair six-sided die. When rolled, each side (number) has a probability of 1/6 of showing up. In the exercise, the distribution includes outcomes of 2, 4, and 6, each appearing with a probability of 1/3.

The main characteristics of a discrete uniform distribution are:
  • Equally likely outcomes
  • A finite number of possible outcomes
  • The sum of probabilities for all possible outcomes equals 1
Understanding this distribution helps to compute probabilities and statistics like the mean and variance, as it's evenly spread across its defined range.
Standard Deviation
The standard deviation is a statistical measure that represents the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

For the discrete uniform distribution in our example, we calculate the variance using the formula for a discrete uniform distribution, \[\sigma^{2} = \frac{(b-a+1)^{2}-1}{12}\]

where \(a\) and \(b\) are the minimum and maximum values of the distribution range, respectively. In this problem, the variance is 0.75.

To find the standard deviation \(\sigma\), simply take the square root of the variance, yielding \(\sigma = \sqrt{0.75} = 0.866\).

Standard deviation is crucial when assessing the distribution of data around its mean, particularly when calculating z-scores and probabilities.
Z-score
A z-score indicates how many standard deviations an element is from the mean. It is a measure of how "off" a value is from what is expected based on the mean and standard deviation.

To calculate z-scores for the problem's sample means, we use the formula:\[z = \frac{X - \mu}{\sigma / \sqrt{n}}\]

where:
  • \(X\) is the value of the sample mean
  • \(\mu\) is the mean of the distribution
  • \(\sigma\) is the standard deviation of the distribution
  • \(n\) is the sample size
By calculating z-scores, we translate the values into standard normal form. This helps in determining the probability of a sample mean falling within a certain range. In this exercise, for \(x = 4.1\) and \(x = 4.4\), the z-scores are 1.33 and 2.51, respectively.
Sample Mean
The sample mean represents the average of a sample, which is a subset of the entire population. It's calculated by summing all observed values in the sample set and dividing by the number of observations.

The sample mean is a crucial concept in statistics. It commonly serves as a point of estimate for the population mean. It's particularly useful in large datasets because it allows for predictions and analyses based on the subset's data.

In our problem, we are interested in the probability of the sample mean lying between two specific values: 4.1 and 4.4. When calculating probabilities and z-scores, the sample mean acts as a cornerstone, enabling the conversion between various statistical measures and the standard normal distribution. This is crucial in finding the relevant probabilities within a specified range.

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Most popular questions from this chapter

The heights of 1000 students are approximately normally distributed with a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. If 200 random samples of size 25 are drawn from this population and the means recorded to the nearest tenth of a centimeter, determine (a) the mean and standard deviation of the sampling distribution of \(\bar{X}\); (b) the number of sample means that fall between 172.5 and 175.8 centimeters inclusive; (c) the number of sample means falling below 172.0 centimeters.

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The chemical benzene is highly toxic to humans. However, it is used in the manufacture of many medicine dyes, leather, and many coverings. In any production process involving benzene, the water in the output of the process must not exceed 7950 parts per million (ppm) of benzene because of government regulations. For a particular process of concern the water sample was collected by a manufacturer 25 times randomly and the sample average \(x\) was \(7960 \mathrm{ppm}\). It is known from historical data that the standard deviation \(a\) is \(100 \mathrm{ppm}\) (a) What is the probability that the sample average in this experiment would exceed the government limit if the population mean is equal to the limit? Use the central limit theorem. (b) Is an observed \(\bar{x}=7960\) in this experiment firm evidence that the population mean for the process exceeds the government limit? Answer your question by computing $$P\left(X^{-}>_{-} 7960 \mid p=7950\right).$$ Assume that the distribution of benzene concentration is normal.

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