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a chemical supply company currently has in stock \(100 \mathrm{lb}\) of a certain chemical, which it sells to customers in 5 -lb lots. Let \(x=\) the number of lots ordered by a randomly chosen customer. The probability distribution of \(x\) is as follows: $$ \begin{array}{lllll} x & 1 & 2 & 3 & 4 \\ p(x) & .2 & .4 & .3 & .1 \end{array} $$ a. Calculate the mean value of \(x\). b. Calculate the variance and standard deviation of \(x\).

Short Answer

Expert verified
The mean of the distribution is 2.3, the variance is 0.81, and the standard deviation is 0.9.

Step by step solution

01

Calculate the mean

To calculate the mean (often called the expected value) of a distribution, multiply each outcome by its probability, and then sum these products. Here, the mean \(\mu\) can be calculated as: \(\mu = 1*.2 + 2*.4 + 3*.3 + 4*.1 = 0.2 + 0.8 + 0.9 + 0.4 = 2.3\)
02

Calculate the variance

The variance is calculated by summing the square of the difference between each outcome and the mean, multiplied by the corresponding probability, or, equivalently, subtracting the square of the mean from the expectation of the square of the variable. Here, the variance \(\sigma^2\) of the distribution is given by: \(\sigma^2 = E[(X-\mu)^2] = (1^2)*.2 + (2^2)*.4 + (3^2)*.3 + (4^2)*.1 - (2.3)^2= 0.2 + 1.6 + 2.7 + 1.6 - 5.29 = 0.81\)
03

Calculate the standard deviation

The standard deviation is simply the square root of the variance. Here, the standard deviation \(\sigma\) is given by: \(\sigma = \sqrt{0.81} = 0.9\)

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Key Concepts

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

Mean Value Calculation
Understanding the mean value of a probability distribution is foundational in statistics, as it represents the average outcome one would expect over a large number of trials. To calculate the mean value, also referred to as the expected value, you multiply each possible outcome of the random variable by its probability. Afterward, you add up all these products to get the mean.

For example, in a scenario where a chemical company sells a product in lots and has a probability distribution for the lots ordered, calculating the mean helps to predict average sales. If a customer can order 1, 2, 3, or 4 lots with respective probabilities of 0.2, 0.4, 0.3, and 0.1, the mean value \(\mu\) is computed as follows:
\[\mu = 1 \times 0.2 + 2 \times 0.4 + 3 \times 0.3 + 4 \times 0.1 = 2.3\]
This mean value provides an insight into the most probable average order quantity, which is a valuable figure for inventory and sales forecasting.
Variance Calculation
Variance is a measure of dispersion in a probability distribution, indicating how much the outcomes deviate from the mean. It gives an idea of the spread or risk associated with the distribution. To calculate the variance, you need to take the squared difference between each outcome and the mean, multiply it by the probability of that outcome, then sum these values. Finally, subtract the square of the mean from this sum.

In practice, knowing the variance helps to understand the risk of deviation from expected sales. For example, with our chemichal company, with mean \(\mu = 2.3\), the variance \(\sigma^2\) is calculated as:
\[\sigma^2 = \sum (x_i - \mu)^2 \times p(x_i) = (1-2.3)^2 \times 0.2 + (2-2.3)^2 \times 0.4 + (3-2.3)^2 \times 0.3 + (4-2.3)^2 \times 0.1\]\[\sigma^2 = 0.2 + 1.6 + 2.7 + 1.6 - 5.29 = 0.81\]

The result is a variance of 0.81. If the variance is high, it indicates a high degree of variability in the number of lots ordered, suggesting that the company should prepare for a greater fluctuation in customer orders.
Standard Deviation Calculation
Standard deviation is another critical statistical measure and is the square root of the variance. It represents the average distance of the data from the mean and is helpful because it has the same units as the original data, making interpretations more intuitive. Standard deviation is used to understand the spread and risk in a more concrete way, summarizing the extent to which individual outcomes differ from the mean.

For our example with the chemical company, the standard deviation \(\sigma\) can be easily derived from the variance:\[\sigma = \sqrt{\sigma^2} = \sqrt{0.81} = 0.9\]
With a standard deviation of 0.9, we now have a more tangible sense of the typical variation in the number of lots ordered. This is essential for the company to grasp the degree to which actual orders might diverge from the average and to plan accordingly.

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Most popular questions from this chapter

Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of Boards 1 and 2 for inspection. a. List the 10 different possible outcomes. b. Suppose that Boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(x\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(x\).

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Suppose that only \(20 \%\) of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are ultimately replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the \(20 \%\) figure? Explain.

Suppose that the \(\mathrm{pH}\) of soil samples taken from a certain geographic region is normally distributed with a mean \(\mathrm{pH}\) of \(6.00\) and a standard deviation of \(0.10 .\) If the \(\mathrm{pH}\) of a randomly selected soil sample from this region is determined, answer the following questions about it: a. What is the probability that the resulting \(\mathrm{pH}\) is between \(5.90\) and \(6.15 ?\) b. What is the probability that the resulting \(\mathrm{pH}\) exceeds \(6.10 ?\) c. What is the probability that the resulting \(\mathrm{pH}\) is at most 5.95? d. What value will be exceeded by only \(5 \%\) of all such pH values?

Determine the value \(z^{*}\) that a. Separates the largest \(3 \%\) of all \(z\) values from the others b. Separates the largest \(1 \%\) of all \(z\) values from the others c. Separates the smallest \(4 \%\) of all \(z\) values from the others d. Separates the smallest \(10 \%\) of all \(z\) values from the others

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