/*! 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 23 If the temperature at which a ce... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the temperature at which a certain compound melts is a random variable with mean value \(120^{\circ} \mathrm{C}\) and standard deviation \(2^{\circ} \mathrm{C}\), what are the mean temperature and standard deviation measured in \({ }^{\circ} \mathrm{F}\) ?

Short Answer

Expert verified
Mean is 248°F and standard deviation is 3.6°F.

Step by step solution

01

Convert Mean Temperature from Celsius to Fahrenheit

The formula to convert a temperature from Celsius to Fahrenheit is given by \[ F = \left(\frac{9}{5}\right)C + 32 \] where \( C \) is the temperature in Celsius and \( F \) in Fahrenheit. The mean temperature in Celsius is \( 120^{\circ} \). Substituting this into the formula, the mean temperature in Fahrenheit is \[ F = \left(\frac{9}{5}\right)\times 120 + 32 = 248^{\circ} \].
02

Convert Standard Deviation from Celsius to Fahrenheit

The standard deviation in a linear transformation is affected only by the multiplicative factor in the transformation. Since the formula to convert Celsius to Fahrenheit is \( F = \left(\frac{9}{5}\right)C + 32 \), the standard deviation transformation is given by \( \sigma_F = \left(\frac{9}{5}\right)\sigma_C \). Given \( \sigma_C = 2^{\circ} \), the standard deviation in Fahrenheit is \[ \sigma_F = \left(\frac{9}{5}\right) \times 2 = 3.6^{\circ} \].

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.

Celsius to Fahrenheit conversion
Converting temperatures between Celsius and Fahrenheit is a common task in science and everyday life. The conversion formula is straightforward:
  • The temperature in Fahrenheit is calculated by multiplying the Celsius temperature by \( \frac{9}{5} \)
  • Then add 32 to the result.
So, if you have a temperature, like the mean melt temperature of a compound, at \(120^{\circ} \mathrm{C}\), you would calculate: \[ F = \left(\frac{9}{5}\right)\times 120 + 32 = 248^{\circ} \mathrm{F}.\] This formula derives from aligning the freezing and boiling points of water in both scales. Celsius is based on water freezing at 0 and boiling at 100. Meanwhile, Fahrenheit uses 32 for freezing and 212 for boiling. Thus, the formula bridges the numerical gap with a fractional conversion factor and a constant.
Standard deviation transformation
Standard deviation transformation is crucial when dealing with different measurement units. The standard deviation shows how much values typically vary from the mean. When converting from Celsius to Fahrenheit, it's important to maintain the proportion of variability, while adjusting the scale. The transformation formula for standard deviation when changing units is simpler than for converting individual data points. This is because it only requires considering the multiplicative factor:
  • In the formula \( F = \left(\frac{9}{5}\right)C + 32 \),
  • only the fractional part \( \frac{9}{5} \) affects the standard deviation, not the addition of 32.
Thus, for a standard deviation \( \sigma_{C} = 2^{\circ} \mathrm{C} \), you use:\[ \sigma_{F} = \left(\frac{9}{5}\right) \times 2 = 3.6^{\circ} \mathrm{F}.\] The additive constant in temperature conversions (like +32 in F conversions) doesn't influence variability.
Mean temperature calculation
Calculating mean temperature involves averaging a set of values, giving you a central value for the dataset. It is a vital part of statistical analysis in many fields, helping describe the typical state or amount. Imagine measuring the melting points of many samples of a compound. By averaging them in Celsius, suppose the mean is \(120^{\circ} \mathrm{C}\). This value captures the general melting behavior. However, in cases where you'll use or communicate this information in a Fahrenheit context, converting the mean keeps the interpretation accurate.After converting the individual temperatures, you find the mean Fahrenheit by transforming the mean Celsius directly, as calculation correctness hinges on proportionate scaling mirrored in the conversion:\[ F = \left(\frac{9}{5}\right)C + 32 \]This ensures temperatures remain contextually relevant when society or documentation predominantly uses a different scale, such as those adhering to Fahrenheit standards.

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

The article "The Load-Life Relationship for M50 Bearings with Silicon Nitride Ceramic Balls" (Lubrication Engr., 1984: 153-159) reports the accompanying data on bearing load life (million revs.) for bearings tested at a \(6.45 \mathrm{kN}\) load. $$ \begin{array}{lrrrrrr} 47.1 & 68.1 & 68.1 & 90.8 & 103.6 & 106.0 & 115.0 \\ 126.0 & 146.6 & 229.0 & 240.0 & 240.0 & 278.0 & 278.0 \\ 289.0 & 289.0 & 367.0 & 385.9 & 392.0 & 505.0 & \end{array} $$ a. Construct a normal probability plot. Is normality plausible? b. Construct a Weibull probability plot. Is the Weibull distribution family plausible?

Let \(X\) be the temperature in \({ }^{\circ} \mathrm{C}\) at which a certain chemical reaction takes place, and let \(Y\) be the temperature in \({ }^{\circ} \mathrm{F}\) (so \(Y=1.8 X+32\) ). a. If the median of the \(X\) distribution is \(\tilde{\mu}\), show that \(1.8 \tilde{\mu}+32\) is the median of the \(Y\) distribution. b. How is the 90 th percentile of the \(Y\) distribution related to the 90 th percentile of the \(X\) distribution? Verify your conjecture. c. More generally, if \(Y=a X+b\), how is any particular percentile of the \(Y\) distribution related to the corresponding percentile of the \(X\) distribution?

Mopeds (small motorcycles with an engine capacity below \(50 \mathrm{~cm}^{3}\) ) are very popular in Europe because of their mobility, ease of operation, and low cost. The article "Procedure to Verify the Maximum Speed of Automatic Transmission Mopeds in Periodic Motor Vehicle Inspections" (J. of Automobile Engr., 2008: 1615-1623) described a rolling bench test for determining maximum vehicle speed. A normal distribution with mean value \(46.8 \mathrm{~km} / \mathrm{h}\) and standard deviation \(1.75 \mathrm{~km} / \mathrm{h}\) is postulated. Consider randomly selecting a single such moped. a. What is the probability that maximum speed is at most \(50 \mathrm{~km} / \mathrm{h}\) ? b. What is the probability that maximum speed is at least \(48 \mathrm{~km} / \mathrm{h}\) ? c. What is the probability that maximum speed differs from the mean value by at most \(1.5\) standard deviations?

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cl} 2\left(1-\frac{1}{x^{2}}\right) & 1 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Compute the cdf of \(X\). b. Obtain an expression for the \((100 p)\) th percentile. What is the value of \(\tilde{\mu}\) ? c. Compute \(E(X)\) and \(V(X)\). d. If \(1.5\) thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the \(1.5\) thousand gallons is expected to be left at the end of the week? [Hint: Let \(h(x)=\) amount left when demand \(=x\).]

A family of pdf's that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, \(k\) and \(\theta\), both \(>0\), and the pdf is $$ f(x ; k, \theta)=\left\\{\begin{array}{cl} \frac{k \cdot \theta^{k}}{x^{k+1}} & x \geq \theta \\ 0 & x<\theta \end{array}\right. $$ a. Sketch the graph of \(f(x ; k, \theta)\). b. Verify that the total area under the graph equals 1 . c. If the rv \(X\) has pdf \(f(x ; k, \theta)\), for any fixed \(b>\theta\), obtain an expression for \(P(X \leq b)\). d. For \(\theta

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.