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Chapter 6: Problem 36

Suppose that for a given computer salesperson, the probability distribution of \(x=\) the number of systems sold in 1 month is given by the following table: $$ \begin{array}{lcccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ p(x) & 0.05 & 0.10 & 0.12 & 0.30 & 0.30 & 0.11 & 0.01 & 0.01 \end{array} $$ a. Find the mean value of \(x\) (the mean number of systems sold). b. Find the variance and standard deviation of \(x\). How would you interpret these values? c. What is the probability that the number of systems sold is within 1 standard deviation of its mean value? d. What is the probability that the number of systems sold is more than 2 standard deviations from the mean?

Short Answer

Expert verified
Mean number of systems sold = 4.3, Variance = 1.61, Standard deviation = 1.27. The probability that number of systems sold is within 1 standard deviation of the mean is 0.72. The probability that the number of systems sold deviates by more than 2 standard deviations from the mean is 0.07.

Step by step solution

01

Calculate Mean Value

To calculate the mean or average of this probability distribution, multiply each outcome by its probability and sum those products: \[ \mu = \sum x*p(x) = 1*0.05 + 2*0.10 + 3*0.12 + 4*0.30 + 5*0.30 + 6*0.11 + 7*0.01 + 8*0.01 = 4.3 \] systems.
02

Calculate Variance

The variance is calculated by subtracting the mean from each outcome, squaring the result (so all differences are positive), multiplying by the outcomes' probability, and summing those results. This is represented by the formula: \[ \sigma^2 = \sum (x - \mu)^2 * p(x) \] After plugging in the given values and calculating, the variance comes out to be 1.61.
03

Calculate Standard Deviation

The standard deviation is just the square root of the variance. So \[ \sigma = \sqrt{1.61} = 1.27 \]
04

Calculate Probability within 1 STD Deviation

We need to find the probability that sales fall within 1 standard deviation of the mean. This means sales between \[ \mu - \sigma = 4.3 - 1.27 = 3.03 \] and \[ \mu + \sigma = 4.3 + 1.27 = 5.57 \]. Looking at the given distribution, this corresponds to sales of 3, 4, and 5 systems. Adding up those probabilities: \[ P(3 \leq x \leq 5) = p(3) + p(4) + p(5) = 0.12 + 0.30 + 0.30 = 0.72 \]
05

Calculate Probability More Than 2 STD Deviations

Proceed similarly to the previous step, but look for probabilities outside of 2 standard deviations from the mean: \[ \mu - 2\sigma = 4.3 - 2*1.27 = 1.76 \] and \[ \mu + 2\sigma = 4.3 + 2*1.27 = 6.84 \]. By the table, sales less than 2 or greater than 6 lie more than 2 standard deviations away from the mean. So \[ P(x < 2 \text{ or } x > 6) = p(1) + p(7) + p(8) = 0.05 + 0.01 + 0.01 = 0.07 \]

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Key Concepts

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

Mean Value Computation
Understanding the mean value of a probability distribution is essential for grasping the central tendency of a set of data. In the context of the exercise, the mean denotes the average number of systems a salesperson is likely to sell in a month. To compute the mean value, we multiply each possible outcome by its probability and add the results together. This calculation tells us that on average, the salesperson sells approximately 4.3 systems per month.

An important aspect of mean value computation is recognizing that it provides a single summary figure representing the entire dataset. However, it does not explain the variability or spread of the data, which is why we also look at variance and standard deviation.
Variance and Standard Deviation
Variance and standard deviation are measures of spread, which indicate how much the data varies from the mean value. Variance is the average squared difference of each value from the mean, reflecting the dispersion of the dataset. The variance in our example is 1.61, suggesting that there is a moderate spread around the mean number of systems sold.

On the other hand, standard deviation is the square root of variance and provides a direct measure of spread in the same units as the data, making it more interpretable. With a standard deviation of 1.27, we understand that the number of systems sold typically deviates from the average by about one to two systems.

Interpreting Variance and Standard Deviation

A lower variance and standard deviation indicate that the numbers of systems sold are closely clustered around the mean, while higher values suggest that sales numbers vary more widely. Hence, these metrics give us insight into the consistency and reliability of the salesperson's performance.
Probability Within Standard Deviation
Knowing the probability of an event occurring within a certain number of standard deviations from the mean is a significant concept in probability distribution analysis. For instance, calculating the probability that the sales fall within one standard deviation ('STD Dev') from the mean tells us how often we can expect the typical results to occur. In our exercise, we found a 72% chance that the salesperson sells between 3 to 5 systems, since this range encompasses one standard deviation from the mean value.

This probability is crucial for setting realistic expectations and understanding the likelihood of different sales outcomes. When we state that there's a high probability within one standard deviation of the mean, we're essentially saying that most months, the sales numbers will not be far off from the average, which in this case, was computed as 4.3 systems. To improve comprehension of these probabilities, we visually consider the distribution and observe where the bulk of outcomes fall relative to the average sales.

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Most popular questions from this chapter

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research [1984]: \(1169-1174\) ) suggests the uniform distribution on the interval from 7.5 to 20 as a model for \(x=\) depth (in centimeters) of the bioturbation layer in sediment for a certain region. a. Draw the density curve for \(x\). b. What is the height of the density curve? c. What is the probability that \(x\) is at most \(12 ?\) d. What is the probability that \(x\) is between 10 and 15 ? Between 12 and 17 ? Why are these two probabilities equal?

A business has six customer service telephone lines. Let \(x\) denote the number of lines in use at any given time. Suppose that the probability distribution of \(x\) is as follows: $$ \begin{array}{lccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04 \end{array} $$ Write each of the following events in terms of \(x,\) and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use

Seventy percent of the bicycles sold by a certain store are mountain bikes. Among 100 randomly selected bike purchases, what is the approximate probability that a. At most 75 are mountain bikes? (Hint: See Example 6.33 ) b. Between 60 and 75 (inclusive) are mountain bikes? c. More than 80 are mountain bikes? d. At most 30 are not mountain bikes?

A pizza shop sells pizzas in four different sizes. The 1,000 most recent orders for a single pizza resulted in the following proportions for the various sizes: $$ \begin{array}{lcccc} \text { Size } & 12 \text { in. } & 14 \text { in. } & 16 \text { in. } & 18 \text { in. } \\ \text { Proportion } & 0.20 & 0.25 & 0.50 & 0.05 \end{array} $$ With \(x=\) the size of a pizza in a single-pizza order, the given table is an approximation to the population distribution of \(x\). a. Write a few sentences describing what you would expect to see for pizza sizes over a long sequence of single-pizza orders. b. What is the approximate value of \(P(x<16)\) ? c. What is the approximate value of \(P(x \leq 16) ?\)

An appliance dealer sells three different models of freezers having \(13.5,15.9,\) and 19.1 cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & 0.2 & 0.5 & 0.3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). (Hint: See Example 6.15\()\) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.
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