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Suppose \(x=\) the number of courses a randomly selected student at a certain university is taking. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{lccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & 0.02 & 0.03 & 0.09 & 0.25 & 0.40 & 0.16 & 0.05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? More than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3 < x < 6)\). Explain in words why these two probabilities are different.

Short Answer

Expert verified
a. \(P(x=4) = 0.25\), b. \(P(x \leq 4) = 0.39\), c. The probability that the selected student is taking at most five courses is 0.79. d. The probability of selecting a student taking at least five courses is 0.61 and more than five courses is 0.21. e. \(P(3 \leq x \leq 6)\) = 0.90, \(P(3 < x < 6)\) = 0.65. The probabilities are different because the first includes both endpoints whereas the latter excludes them.

Step by step solution

01

Calculate Individual Probability

For part a, referring to the probability distribution table, \(P(x=4)\) can be directly read off the table as 0.25.
02

Calculate Cumulative Probability

For part b, \(P(x \leq 4)\) is calculated by summing the probabilities of \(x=1,\,2,\,3,\,4\). This results in (0.02 + 0.03 + 0.09 + 0.25) = 0.39.
03

At most or At least

For parts c and d, the question asks for the cumulative probability for the student taking at most 5 courses or at least 5 courses. For c, \(P(x \leq 5)\) is calculated by summing the probabilities \(x=1,\,2,\,3,\,4,\,5,\) resulting in (0.02 + 0.03 + 0.09 + 0.25 + 0.40) = 0.79. For d, \(P(x \geq 5)\) is calculated by summing the probabilities \(x=5,\,6,\,7,\) resulting in (0.40 + 0.16 + 0.05) = 0.61, and \(P(x > 5)\) by summing \(x=6,\,7,\) resulting in (0.16 + 0.05) = 0.21.
04

Calculate Ranged Probability

For part e, \(P(3 \leq x \leq 6)\) is calculated by summing the probabilities \(x=3,\,4,\,5,\,6,\) resulting in (0.09 + 0.25 + 0.40 + 0.16) = 0.90. Meanwhile, \(P(3 < x < 6)\) is calculated by summing \(x=4,\,5,\) resulting in ( 0.25 + 0.40) = 0.65.
05

Explanation

The explanation for why \(P(3 \leq x \leq 6)\) and \(P(3 < x < 6)\) are different is because in \(P(3 \leq x \leq 6)\), both boundaries 3 and 6 are included. Whereas in \(P(3 < x < 6)\), the boundaries are excluded, resulting in a smaller range and hence smaller cumulative probability.

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Key Concepts

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

Cumulative Probability
Cumulative probability is a core concept in statistics that relates to the probability of a random variable being less than or equal to a certain value. It is effectively the mechanism that allows the calculation of the probability that an event will happen within an inclusive range.

For example, if you want to know the probability that a student takes 4 or fewer courses, you look at all the probabilities of taking 1, 2, 3, or 4 courses and sum them up. This is what was done in part b of the problem, where the cumulative probability, denoted as \(P(x \leq 4)\), was calculated by adding the individual probabilities up to x=4. Improving understanding of this concept may involve visualizing cumulative probability as a running total: as you move along the possible values, the cumulative probability increases until it reaches 1, or 100 percent certainty that a value in the distribution has been achieved.

For a more relatable explanation, imagine filling a glass with water. Each pour represents adding the probability of one more course, and the cumulative probability would be the level of water in the glass after several pours. Ultimately, the glass will be full — analogous to reaching a cumulative probability of 1, indicating certainty.
Probability Calculation
Probability calculation involves determining the likelihood of a specific outcome in a random event. When you're given a probability distribution, like in the original exercise, each value that a random variable can take has an associated probability. To find the probability for a specific value of the variable, simply refer to that probability in the distribution, as we see in part a for \(P(x=4)\).

To ensure students fully grasp how to perform these calculations, it's important to stress the relationship between the values and their probabilities. You might liken this to looking up a contact in your phonebook where the name is the 'value' and the phone number is the 'probability'. Always remind the student that the probabilities for all possible outcomes must sum up to 1 since one of the outcomes must occur. Moreover, explaining this concept also requires emphasizing the exclusive nature of each probability when considered separately, unlike cumulative probabilities, which are inclusive and additive. This precision is key in understanding and solving more complex probability questions.
Ranged Probability
Ranged probability encompasses the likelihood of a random variable falling within a range of values. When working with a probability distribution, you may encounter questions asking for the probability that a variable is between two values, inclusive or exclusive.

In the textbook problem, for example, understanding the difference between \(P(3 \leq x \leq 6)\) and \(P(3 < x < 6)\) is crucial. In the former, the probability accounts for the students taking 3, 4, 5, or 6 courses — both boundaries are included, thus 'inclusive'. In the latter, it only includes students taking 4 or 5 courses — the boundaries are excluded, thus 'exclusive'.

To illustrate, if a range of pages in a book is selected as possible reading for tonight, inclusivity would mean pages 20 to 30 are all options, while exclusivity means you'll start reading at page 21 and stop before reaching page 30. Helping students differentiate these subtleties in probability calculation, and providing ample examples, will enhance their comprehension of ranged probability, improving their problem-solving capabilities in statistics.

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Most popular questions from this chapter

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

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Consider the population of all one-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture Use the normal distribution to calculate the following probabilities. (Hint: See Example 6.21\()\) a. \(P(x < 5.0)\) b. \(P(x < 5.4)\) c. \(P(x \leq 5.4)\) d. \(P(4.6 < x < 5.2)\) e. \(P(x > 4.5)\) f. \(P(x > 4.0)\)

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than 0 . What is \(p(4)\) ? b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

A chemical supply company currently has in stock 100 pounds of a certain chemical, which it sells to customers in 5 -pound lots. Let \(x=\) the number of lots ordered by a randomly chosen customer. The probability distribution of \(x\) is as follows: $$ \begin{array}{lcccc} x & 1 & 2 & 3 & 4 \\ p(x) & 0.2 & 0.4 & 0.3 & 0.1 \end{array} $$ a. Calculate and interpret the mean value of \(x\). b. Calculate and interpret the variance and standard deviation of \(x\).

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