/*! 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 37 A gas station sells gasoline at ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A gas station sells gasoline at the following prices (in cents per gallon, depending on the type of gas and service): \(315.9,318.9,329.9,339.9,344.9\), and 359.7. Let \(y\) denote the price per gallon paid by a randomly selected customer. a. Is \(y\) a discrete random variable? Explain. b. Suppose that the probability distribution of \(y\) is as follows: $$ \begin{array}{lrrrrrr} y & 315.9 & 318.9 & 329.9 & 339.9 & 344.9 & 359.7 \\ p(y) & .36 & .24 & .10 & .16 & .08 & .06 \end{array} $$ What is the probability that a randomly selected customer has paid more than $$\$ 3,20$$ per gallon? Less than $$\$ 3.40$$ per gallon? c. Refer to Part (b), and calculate the mean value and standard deviation of \(y .\) Interpret these values.

Short Answer

Expert verified
a. Yes, \(y\) is a discrete random variable because it can only take on specific values. b. The probability that a customer pays more than $3.20 per gallon is 0.40, and the probability that a customer pays less than $3.40 per gallon is 0.70. c. The average price per gallon paid by customers is approximately $3.29. The standard deviation (after its calculated using variance) will provide information about the spread of prices around this average.

Step by step solution

01

Determine the nature of the variable

The variable \(y\) is a discrete random variable. This is because it can only take on certain specified values. In this case, it is the price a randomly chosen customer pays, which can only take on six distinct values.
02

Calculate Probabilities

To find the probability that a randomly selected customer has paid more than $3.20 (320 cents) per gallon, sum up the probabilities of \(y\) being more than 320, which are \(p(329.9) + p(339.9) + p(344.9) + p(359.7 )= 0.10 + 0.16 + 0.08 + 0.06 = 0.40\). For the probability that a customer has paid less than $3.40 (340 cents) per gallon, you sum up the probabilities of \(y\) being less than 340, which are \(p(315.9) + p(318.9) + p(329.9) = 0.36 + 0.24 + 0.10 = 0.70\).
03

Calculate Mean and Standard Deviation

First, we calculate the mean or expected value of \(y\) as \(E(y) = \sum y \times p(y) = 315.9 \times .36 + 318.9 \times .24 + 329.9 \times .10 + 339.9 \times .16 + 344.9 \times .08 + 359.7 \times .06 = 329.16\) cents, which suggests that on average, customers pay roughly $3.29 per gallon. Similarly, to calculate the standard deviation, we first compute the variance: \(\text{Var}(y) = \sum (y - E(y))^2 \times p(y)\). After computing this, we then take the square root of the result to get the standard deviation. The variance and standard deviation give us a measure of how spread out the gas prices are around the mean.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Probability Distributions
When we talk about a probability distribution, we are referring to a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In simpler terms, it tells us how likely we are to observe certain events. For discrete random variables like the price per gallon in our gas station example, this distribution lists all the possible values and their associated probabilities.

A discrete random variable, such as the one given, can only take on a finite number of values. Here, since there are only six specific prices at which gas can be sold, it is considered discrete. The probability distribution associated with this variable helps us understand how a randomly selected customer is expected to interact with these prices – essentially how likely they are to pay any of the listed prices.

Understanding this concept is crucial in many fields such as insurance, finance, and engineering, where predicting outcomes and risks is a part of the everyday work.
Calculating and Interpreting the Mean Value
The mean value, or expected value, of a discrete random variable is a measure of the 'central' value that one can expect from a distribution. Mathematically, it's the average outcome if an experiment were to be repeated many times. To calculate the mean of a discrete random variable, we multiply each possible value of the variable by its probability and then sum all these values.

For instance, in the context of our exercise, by calculating the mean, we can determine that the average price paid per gallon by a customer is around $3.29. This average is useful because it gives gas station owners, customers, and analysts an idea of what 'typical' pricing looks like, despite the variation that might occur from customer to customer. It's an essential tool for forecasting and making informed decisions in business and statistics.
Determining the Standard Deviation
The standard deviation is a measure that tells us how spread out the values in a data set are from the mean. In a probability distribution, it quantifies the expected deviation from the mean value. To calculate it, we first find the variance, which is the average of the squared differences from the mean, and then take the square root of that variance.

The standard deviation in our example could reflect how consistent or volatile the gas prices are at the station. A smaller standard deviation would mean the prices are very close to the average, indicating price stability. Conversely, a larger standard deviation signifies a wide variety of prices paid by the customers, showing price volatility. It's an important concept for anyone looking to understand risk and variability in nearly any field—be it psychology, meteorology, or market research.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection? b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail to pass inspection? c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection? d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?

Let \(z\) denote a variable that has a standard normal distribution. Determine the value \(z^{*}\) to satisfy the following conditions: a. \(P\left(zz^{*}\right)=.02\) e. \(P\left(z>z^{*}\right)=.01\) f. \(P\left(z>z^{*}\right.\) or \(\left.z<-z^{*}\right)=.20\)

Consider the following sample of 25 observations on the diameter \(x\) (in centimeters) of a disk used in a certain system: $$ \begin{array}{lllllll} 16.01 & 16.08 & 16.13 & 15.94 & 16.05 & 16.27 & 15.89 \\ 15.84 & 15.95 & 16.10 & 15.92 & 16.04 & 15.82 & 16.15 \\ 16.06 & 15.66 & 15.78 & 15.99 & 16.29 & 16.15 & 16.19 \\ 16.22 & 16.07 & 16.13 & 16.11 & & & \end{array} $$ The 13 largest normal scores for a sample of size 25 are \(1.965,1.524,1.263,1.067,0.905,0.764,0.637,0.519\) \(0.409,0.303,0.200,0.100\), and \(0 .\) The 12 smallest scores result from placing a negative sign in front of each of the given nonzero scores. Construct a normal probability plot. Does it appear plausible that disk diameter is normally distributed? Explain.

Consider the population of all 1-gal 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 model to calculate the following probabilities: a. \(P(x<5.0)\) b. \(P(x<5.4)\) c. \(P(x \leq 5.4)\) d. \(P(4.64.5)\) f. \(P(x>4.0)\)

An author has written a book and submitted it to a publisher. The publisher offers to print the book and gives the author the choice between a flat payment of $$\$ 10,000$$ and a royalty plan. Under the royalty plan the author would receive $$\$ 1$$ for each copy of the book sold. The author thinks that the following table gives the probability distribution of the variable \(x=\) the number of books that will be sold: $$ \begin{array}{lrrrr} x & 1000 & 5000 & 10,000 & 20,000 \\ p(x) & .05 & .30 & .40 & .25 \end{array} $$ Which payment plan should the author choose? Why?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.