/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 An insurance company offers its ... [FREE SOLUTION] | 91Ó°ÊÓ

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An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let \(X=\) the number of months between successive payments. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<1 \\ .30 & 1 \leq x<3 \\ .40 & 3 \leq x<4 \\ .45 & 4 \leq x<6 \\ .60 & 6 \leq x<12 \\ 1 & 12 \leq x\end{cases} $$ a. What is the pmf of \(X\) ? b. Using just the cdf, compute \(P(3 \leq X \leq 6)\) and \(P(4 \leq X)\).

Short Answer

Expert verified
a. PMF: 1: 0.30, 3: 0.10, 4: 0.05, 6: 0.15, 12: 0.40; b. \(P(3 ≤ X ≤ 6) = 0.30\), \(P(4 ≤ X) = 0.60\).

Step by step solution

01

Understand the CDF Range and Values

The cumulative distribution function (CDF) assigns probabilities to different intervals of the random variable values. Here, the specified intervals for certain CDF values of \(X\) are given. This will help us determine the probability mass function (PMF) later.
02

Calculate the PMF of X

The probability mass function (PMF) is computed by finding the probability difference at each interval point. Identify points where the CDF value changes and note:- For \(1 \leq x < 3\), \(F(x) - F(x-1) = 0.30 - 0 = 0.30\)- For \(3 \leq x < 4\), the jump is \(0.40 - 0.30 = 0.10\)- For \(4 \leq x < 6\), the jump is \(0.45 - 0.40 = 0.05\)- For \(6 \leq x < 12\), the jump is \(0.60 - 0.45 = 0.15\)- For \(x \geq 12\), the value is \(1 - 0.60 = 0.40\)Thus, PMF values are \(P(X=1)=0.30\), \(P(X=3)=0.10\), \(P(X=4)=0.05\), \(P(X=6)=0.15\), \(P(X=12)=0.40\).
03

Compute P(3 ≤ X ≤ 6)

To compute \(P(3 \leq X \leq 6)\), use the cumulative probabilities:\[P(3 \leq X \leq 6) = F(6) - F(2.999) = 0.60 - 0.30 = 0.30\] Since 6 and 3 are included in \(F(x)\)'s segment, the probabilities add up to 0.30.
04

Compute P(4 ≤ X)

For \(P(4 \leq X)\), subtract \(F(3.999)\) from 1 (as \( x\geq 4\) would encompass all further probabilities):\[P(4 \leq X) = 1 - F(3.999) = 1 - 0.40 = 0.60\] This computation shows the cumulative probability beyond 4.

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Key Concepts

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

Probability Mass Function
A probability mass function (PMF) is an essential concept when dealing with discrete random variables. It tells us the probability that a discrete random variable is exactly equal to some value. In more straightforward terms, think of PMF as translating the likelihood of each possible outcome of a mathematical experiment into numbers that add up to one.
The PMF of our variable, say \(X\), can be formalized as \(P(X = x)\), where \(x\) represents the possible values our random variable can take.
In the provided exercise, we have observed such a translation through different intervals. For instance, \(P(X = 1)\) is 0.30, implying that the probability of a policyholder making payments every month is 0.30.
Similarly, other PMF values are calculated based on the changes between cumulative distribution values, showing the jumps from one interval to the next.
Cumulative Distribution Function
The cumulative distribution function (CDF) is pivotal in understanding how probabilities accumulate over a range of values. For a random variable \(X\), the CDF, denoted as \(F(x)\), represents the probability that \(X\) will take a value less than or equal to \(x\).
This helps us when we need to calculate probabilities over intervals rather than single points. In the given exercise, the CDF is defined piecewise, showing how probabilities build up as \(X\) reaches different thresholds.
To compute probabilities between two points, we rely on the difference between their CDF values. For example, \(P(3 \leq X \leq 6)\) was found by subtracting the cumulative probability at X equals just less than 3 from that at 6, yielding a 0.30 cumulative probability for that segment.
Random Variable
A random variable is a fundamental concept in probability and statistics, representing outcomes of a mathematical experiment numerically. It serves as a function that assigns numerical values to each event in a sample space, allowing us to analyze more intricate probability scenarios.
In the context of the exercise, \(X\) is our random variable, reflecting the intervals between payment options for insurance policyholders. Different intervals indicate varied payment structures directly linked to the probabilities defined by the PMF and CDF.
By understanding \(X\) as a function that can take on specific values based on a given distribution, students can correlate the probabilities of these payment intervals in real-world scenarios, aiding their grasp of randomness and probability calculations in practical applications.

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Most popular questions from this chapter

The \(n\) candidates for a job have been ranked 1,2 , \(3, \ldots, n\). Let \(X=\) the rank of a randomly selected candidate, so that \(X\) has pmf $$ p(x)= \begin{cases}1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise }\end{cases} $$ (this is called the discrete uniform distribution). Compute \(E(X)\) and \(V(X)\) using the shortcut formula.

Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the probability that any particular couple or individual arrives late is .4 (a couple will travel together in the same vehicle, so either both people will be on time or else both will arrive late). Assume that different couples and individuals are on time or late independently of one another. Let \(X=\) the number of people who arrive late for the seminar. a. Determine the probability mass function of \(X\). b. Obtain the cumulative distribution function of \(X\), and use it to calculate \(P(2 \leq X \leq 6)\).

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is $$ \begin{array}{l|cccc} y & 0 & 1 & 2 & 3 \\ \hline p(y) & .60 & .25 & .10 & .05 \end{array} $$ a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

An appliance dealer sells three different models of upright freezers having \(13.5,15.9\), and \(19.1\) cubic feet of storage space, respectively. Let \(X=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(X\) has pmf $$ \begin{array}{l|rrr} x & 13.5 & 15.9 & 19.1 \\ \hline p(x) & .2 & .5 & .3 \end{array} $$ a. Compute \(E(X), E\left(X^{2}\right)\), and \(V(X)\). b. If the price of a freezer having capacity \(X\) cubic feet is \(25 X-8.5\), what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price \(25 X-8.5\) paid by the next customer? d. Suppose that although the rated capacity of a freezer is \(X\), the actual capacity is \(h(X)=X-.01 X^{2}\). What is the expected actual capacity of the freezer purchased by the next customer?

Individuals \(A\) and \(B\) begin to play a sequence of chess games. Let \(S=\\{\) A wins a game \(\\}\), and suppose that outcomes of successive games are independent with \(P(S)=p\) and \(P(F)=1-p\) (they never draw). They will play until one of them wins ten games. Let \(X=\) the number of games played (with possible values \(10,11, \ldots, 19\) ). a. For \(x=10,11, \ldots, 19\), obtain an expression for \(p(x)=\) \(P(X=x)\). b. If a draw is possible, with \(p=P(S), q=P(F), 1-p-\) \(q=P(\) draw \()\), what are the possible values of \(X\) ? What is \(P(20 \leq X)\) ? [Hint: \(P(20 \leq X)=1-P(X<20)\).]

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