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Chapter 17: Problem 16

The length of time required by students to complete a paper 2 HL exam is a random variable with a pdf \(f(y)=\left\\{\begin{array}{ll}\frac{1}{2}\left(c y+\frac{y^{2}}{3}\right) & 0 \leqslant y \leqslant 2 \\ 0 & \text { otherwise }\end{array}\right.\) a) Find \(c\) b) Find the probability that a randomly selected student will finish in less than 1 hour. c) In a randomly selected group of 10 students, what is the probability that 3 students will finish the exam in less than 1 hour? d) Given that Casper, who happened to be randomly selected, needs at least 1 hour to finish the exam, what is the probability that he will require at least 90 minutes?

Short Answer

Expert verified
a) c = \frac{1}{3}; b) P(Y < 1) = \frac{5}{36}.

Step by step solution

01

Find c using the Property of PDF

First, remember that the total probability over all possible values of a probability density function (pdf) must equal 1. This gives us the equation \(\int_{0}^{2} \frac{1}{2}\left(c y + \frac{y^{2}}{3}\right) \, dy = 1\). Evaluating the integral:\[\int_{0}^{2} \left(c y + \frac{y^{2}}{3}\right) \, dy = \left[ \frac{cy^2}{2} + \frac{y^3}{9} \right]_{0}^{2}\]Calculating, we find:\(\frac{2c}{2} + \frac{8}{9} = c + \frac{8}{9}\). Therefore, the equation is:\[ \frac{c}{2}(c + \frac{8}{9}) = 1\]. Solving for \(c\) gives \(c = \frac{1}{3}\).
02

Find Probability of Finishing in Less Than 1 Hour

To find the probability that a student completes the exam in less than 1 hour, we calculate:\[ P(Y < 1) = \int_{0}^{1} \frac{1}{2}\left(\frac{1}{3}y + \frac{y^{2}}{3}\right) \, dy \]. Solving this integral:\[\int_{0}^{1} \left(\frac{1}{6}y + \frac{y^{2}}{6}\right) \, dy = \left[ \frac{1}{12}y^2 + \frac{y^3}{18} \right]_{0}^{1} = \left( \frac{1}{12} + \frac{1}{18} \right) = \frac{5}{36}.\]

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Key Concepts

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

Random Variables
Random variables are an essential concept in probability and statistics. In simple terms, a random variable is a variable that can take on different values, each with an associated probability. They are often denoted by capital letters such as \(X\) or \(Y\). Each value a random variable can assume is determined by some outcome of a random phenomenon.
  • Random variables can be classified as either discrete or continuous.
  • A discrete random variable has specific, separate values (e.g., rolling a die), while a continuous random variable can take on any value within a range (e.g., time spent completing an exam).
The time a student takes to complete an exam is a perfect example of a continuous random variable, as it can range anywhere within a specific time span, such as the 0 to 2 hours assumed in the exercise.
Probability Density Function
A probability density function (pdf) is a function that describes the likelihood of a random variable taking on a particular value. For a continuous random variable, the pdf indicates the probability of the variable falling within a specific range of values rather than taking on any one value.
  • The area under the curve of a pdf over a certain interval represents the probability that the variable falls within that interval.
  • For continuous variables, the total area under the pdf across all possible values must equal 1.
In the exercise, the given pdf \( f(y) = \frac{1}{2}\left(c y + \frac{y^{2}}{3}\right) \) applies from 0 to 2 hours, meaning outside this interval, the probability is zero. By determining the constant \(c\), the function ensures the required properties of a pdf are satisfied.
Integration
Integration is a mathematical process used to find the area under a curve described by a function. In probability, integration aids in determining the probability that a continuous random variable falls between two limits.
  • The integral of a pdf over an interval gives the probability that a random variable will fall within that interval.
  • To solve for \(c\), the function was integrated over its defined interval to ensure the area equals 1.
  • Similarly, to find the probability a student finishes in less than 1 hour, we integrate the pdf from 0 to 1.
Through integration, the exercise calculates specific probabilities, demonstrating how this mathematical tool is vital for evaluating continuous probability distributions.
Distribution
In statistics, a distribution describes how the values of a random variable are spread or distributed. For continuous random variables, the probability distribution is often visualized using a probability density function.
  • A random variable's distribution provides insights into its expected value, variance, and the probabilities of certain outcomes.
  • In the context of the exercise, the distribution of exam completion time is continuous and defined by the given pdf.
  • This distribution can answer various probability questions, such as the likelihood a student will finish quickly or how a group might perform overall.
Understanding a distribution enables prediction and analysis in real-world scenarios, making it an indispensable concept in probability and statistics.

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Most popular questions from this chapter

Houses in a large city are equipped with alarm systems to protect them from burglary. A company claims their system to be \(98 \%\) reliable. That is, it will trigger an alarm in \(98 \%\) of the cases. In a certain neighbourhood, 10 houses equipped with this system experience an attempted burglary. a) Find the probability that all the alarms work properly. b) Find the probability that at least half of the houses trigger an alarm. c) Find the probability that at most 8 alarms will work properly.

\(X \sim N\left(\mu, \sigma^{2}\right) . P(x>19.6)=0.16\) and \(P(x<17.6)=0.012 .\) Find \(\mu\) and \(\sigma\)

Consider the binomial random variable with \(n=6\) and \(p=0.3\) a) Fill in the probabilities below. $$\begin{array}{|l|l|l|l|l|l|l|l|}\hline k & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline P(x \leqslant k) & & & & & & & \\\\\hline\end{array}$$ b) Fill in the table below. Some cells have been filled for you to guide you. (TABLE CAN'T COPY)

A supplier of copper wire looks for flaws before despatching it to customers. It is known that the number of flaws follows a Poisson probability distribution with a mean of 2.3 flaws per metre. a) Determine the probability that there are exactly 2 flaws in 1 metre of the wire. b) Determine the probability that there is at least one flaw in 2 metres of the wire.

The scores on a public schools examination are normally distributed with a mean of 550 and a standard deviation of 100 a) What is the probability that a randomly chosen student from this population scores below 400? b) What is the probability that a student will score between 450 and \(650 ?\) c) What score should you have in order to be in the 90 th percentile? d) Find the IQR of this distribution.
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