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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

Studies have found that women diagnosed with cancer in one breast also sometimes have cancer in the other breast that was not initially detected by mammogram or physical examination ("MRI Evaluation of the Contralateral Breast in Women with Recently Diagnosed Breast Cancer," The New England Journal of Medicine [2007]: 1295-1303). To determine if magnetic resonance imaging (MRI) could detect missed tumors in the other breast, 969 women diagnosed with cancer in one breast had an MRI exam. The MRI detected tumors in the other breast in 30 of these women. a. Use \(p=\frac{30}{969}=0.031\) as an estimate of the probability that a woman diagnosed with cancer in one breast has an undetected tumor in the other breast. Consider a random sample of 500 women diagnosed with cancer in one breast. Explain why it is reasonable to think that the random variable \(x=\) number in the sample who have an undetected tumor in the other breast has a binomial distribution with \(n=500\) and \(p=0.031\). b. Is it reasonable to use the normal distribution to approximate probabilities for the random variable \(x\) defined in Part (a)? Explain why or why not. c. Approximate the following probabilities: i. \(\quad P(x<10)\) ii. \(\quad P(10 \leq x \leq 25)\) iii. \(P(x>20)\) d. For each of the probabilities calculated in Part (c), write a sentence interpreting the probability.

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