/*! 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 19 The probability model below desc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The probability model below describes the number of repair calls that an appliance repair shop may receive during an hour. $$\begin{array}{l|c|c|c|c}\text { Repair Calls } & 0 & 1 & 2 & 3 \\\\\hline \text { Probability } & 0.1 & 0.3 & 0.4 & 0.2\end{array}$$ a) How many calls should the shop expect per hour? b) What is the standard deviation?

Short Answer

Expert verified
a) The shop should expect 1.7 calls per hour. b) The standard deviation is approximately 0.9.

Step by step solution

01

Understand the Problem

We have a probability distribution that models the number of repair calls per hour. We need to find the expected number of calls and the standard deviation.
02

Calculate Expected Value

The expected value is calculated using the formula: \( E(X) = \sum x_i \cdot P(x_i) \), where \( x_i \) is the number of repair calls and \( P(x_i) \) is their respective probability. \[ E(X) = 0 \times 0.1 + 1 \times 0.3 + 2 \times 0.4 + 3 \times 0.2 \]\[ E(X) = 0 + 0.3 + 0.8 + 0.6 = 1.7 \]Thus, the shop should expect 1.7 calls per hour.
03

Calculate Variance

Variance is calculated using the formula: \( \text{Var}(X) = \sum (x_i - E(X))^2 \cdot P(x_i) \). We substitute \( E(X) = 1.7 \) into this formula:\[ \text{Var}(X) = (0 - 1.7)^2 \times 0.1 + (1 - 1.7)^2 \times 0.3 + (2 - 1.7)^2 \times 0.4 + (3 - 1.7)^2 \times 0.2 \]\[ \text{Var}(X) = 2.89 \times 0.1 + 0.49 \times 0.3 + 0.09 \times 0.4 + 1.69 \times 0.2 \]\[ \text{Var}(X) = 0.289 + 0.147 + 0.036 + 0.338 = 0.81 \]
04

Calculate Standard Deviation

The standard deviation is the square root of the variance. \[ \sigma = \sqrt{0.81} \approx 0.9 \]Thus, the standard deviation of repair calls is approximately 0.9.

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.

Expected Value
The concept of Expected Value is like the heart of probability distributions. It helps us understand what average outcome we can anticipate from a random process. Imagine you're running an appliance repair shop and want to know how many calls to expect each hour. The Expected Value gives you this estimate. We calculate the Expected Value by taking each possible outcome, multiplying it by its probability, and then adding all those results together. In our repair call case:
  • 0 calls with a probability of 0.1
  • 1 call with a probability of 0.3
  • 2 calls with a probability of 0.4
  • 3 calls with a probability of 0.2
For the calculation:\[ E(X) = 0 \times 0.1 + 1 \times 0.3 + 2 \times 0.4 + 3 \times 0.2 = 1.7 \]So, you would expect to get, on average, 1.7 calls per hour at the shop. This isn't a whole number because it's an average; think of it as a long-term trend rather than a precise count per hour.
Variance
Variance is like a measure of how spread out the repair calls are around the Expected Value. While the Expected Value tells you what you expect to see, the Variance will tell you how much these results can fluctuate from the average. To find the Variance, we calculate the squared difference between each possibility and the Expected Value, multiply by the probability of that outcome, then sum all these numbers. In our example:\[ \text{Var}(X) = (0 - 1.7)^2 \times 0.1 + (1 - 1.7)^2 \times 0.3 + (2 - 1.7)^2 \times 0.4 + (3 - 1.7)^2 \times 0.2 \]Which simplifies to: \[ \text{Var}(X) = 0.289 + 0.147 + 0.036 + 0.338 = 0.81 \]This result of 0.81 shows how much the number of calls is likely to vary from the expected 1.7 calls in any given hour. It gives us a sense of reliability around our Expected Value.
Standard Deviation
The Standard Deviation gives us another view of the spread of our data around the mean, but in a more intuitive way. Where Variance uses squared units, the Standard Deviation returns to the original units by taking the square root of Variance. In plain terms, it tells us the average distance each repair call figure might fall from the mean.For our appliance repair shop example, the Standard Deviation is calculated by:\[ \sigma = \sqrt{0.81} \approx 0.9 \]This value of approximately 0.9 means, on average, each repair call hour can realistically stray from our expected 1.7 calls by about 0.9 calls. It gives the shop a better idea of how much a given hour’s calls might differ from the expected amount. This variability understanding helps in planning resources and managing staff efficiently.

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

An insurance company estimates that it should make an annual profit of \(\$ 150\) on each homeowner's policy written, with a standard deviation of \(\$ 6000\) a) Why is the standard deviation so large? b) If it writes only two of these policies, what are the mean and standard deviation of the annual profit? c) If it writes 10,000 of these policies, what are the mean and standard deviation of the annual profit? d) Is the company likely to be profitable? Explain. e) What assumptions underlie your analysis? Can you think of circumstances under which those assumptions might be violated? Explain.

In the \(4 \times 100\) medley relay event, four swimmers swim 100 yards, each using a different stroke. \(A\) college team preparing for the conference championship looks at the times their swimmers have posted and creates a model based on the following assumptions: \(\bullet\) The swimmers' performances are independent. \(\bullet\) Each swimmer's times follow a Normal model. \(\bullet\) The means and standard deviations of the times (in seconds) are as shown: $$\begin{array}{l|l|c}\text { Swimmer } & \text { Mean } & \text { SD } \\\\\hline 1 \text { (backstroke) } & 50.72 & 0.24 \\\2 \text { (breaststroke) } & 55.51 & 0.22 \\\3 \text { (butterfy) } & 49.43 & 0.25 \\\4 \text { (freestyle) } & 44.91 & 0.21\end{array}$$ a) What are the mean and standard deviation for the relay team's total time in this event? b) The team's best time so far this season was 3: 19.48 (That's 199.48 seconds.) Do you think the team is likely to swim faster than this at the conference championship? Explain.

In a litter of seven kittens, three are female. You pick two kittens at random. a) Create a probability model for the number of male kittens you get. b) What's the expected number of males? c) What's the standard deviation?

A grocery supplier believes that in a dozen eggs, the mean number of broken ones is 0.6 with a standard deviation of 0.5 eggs. You buy 3 dozen eggs without checking them. a) How many broken eggs do you expect to get? b) What's the standard deviation? c) What assumptions did you have to make about the eggs in order to answer this question?

The American Veterinary Association claims that the annual cost of medical care for dogs averages \(\$ 100,\) with a standard deviation of \(\$ 30,\) and for cats averages \(\$ 120\) with a standard deviation of \(\$ 35\) a) What's the expected difference in the cost of medical care for dogs and cats? b) What's the standard deviation of that difference? c) If the costs can be described by Normal models, what's the probability that medical expenses are higher for someone's dog than for her cat? d) What concerns do you have?

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.