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Chapter 4: Problem 13

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 \({ }^{\circ} \mathrm{F}\) ? [Hint: \({ }^{\circ} \mathrm{F}=1.8^{\circ} \mathrm{C}+32\).]

Short Answer

Expert verified
Mean: \(248^{\circ} \mathrm{F}\), Standard deviation: \(3.6^{\circ} \mathrm{F}\).

Step by step solution

01

Understand the Temperature Conversion Formula

The conversion formula from Celsius to Fahrenheit is \( ^\circ \, F = 1.8 \times ^\circ \, C + 32 \). This formula will help us convert both the mean and the standard deviation from Celsius to Fahrenheit.
02

Convert the Mean Temperature

Given the mean temperature in Celsius is \(120^{\circ} \mathrm{C}\), use the conversion formula: \[ \text{Mean in Fahrenheit} = 1.8 \times 120 + 32 \]. Calculate to find \(216 + 32 = 248^{\circ} \mathrm{F}\). Thus, the mean temperature in Fahrenheit is \(248^{\circ} \mathrm{F}\).
03

Convert the Standard Deviation

The standard deviation in Fahrenheit can be found by converting the Celsius standard deviation using the same multiplier from the conversion formula. Thus, it is \(1.8 \times 2^{\circ} \mathrm{C} = 3.6^{\circ} \mathrm{F}\). Note that the additive term (32) in the conversion formula does not affect the standard deviation.

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Key Concepts

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

Random Variables in Temperature
Understanding the concept of a random variable is crucial when dealing with temperature measurements of a substance. A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In the context of temperature, this can mean the unpredictable fluctuations in the temperature readings due to various environmental factors or measurement inaccuracies.

When we say that the melting temperature of a compound is a random variable, it implies that each time you measure it, the value might be slightly different due to minor uncontrollable variations. Yet, it's still centered around a certain average value, known as the mean.
  • Random variables are key for statistical analysis.
  • Helps model real-world situations where outcomes vary.
  • The melting point is not fixed but has a probability distribution.
Understanding Mean Temperature
The mean temperature is essentially the "central" or "average" temperature at which a compound melts, when measured multiple times as it shows consistency around a focal point despite variations. It is a statistical measure used to summarize a set of temperatures by a single representative number.

For instance, in our exercise, the mean temperature is given as 120掳C. This provides a reference point around which other temperature values are distributed. It helps gauge where most temperature readings will likely be, even as individual readings can differ slightly due to their random variable nature.
  • Mean provides an estimate for the central tendency of data.
  • Useful for comparing temperatures in different units, like Celsius and Fahrenheit.
  • Simplifies complex data sets into single figures.
Significance of Standard Deviation in Temperatures
Standard deviation is a statistic that measures the dispersion or spread of data points in a data set around the mean. A smaller standard deviation indicates that the temperatures are clustered closely to the mean, whereas a larger standard deviation implies more variation.

In our example, the standard deviation of the temperature in Celsius is given as 2掳C. This tells us that most of the temperature readings fall within 2掳C of the mean (120掳C) more often than not. After converting this to Fahrenheit, it becomes 3.6掳F, indicating the spread after unit conversion.
  • Standard deviation reflects variability from the mean.
  • Higher values suggest greater fluctuation and unpredictability.
  • Essential for understanding consistency and reliability of measurements.
Converting Celsius to Fahrenheit
Temperature conversion is vital for scientists and statisticians who need to work with data across different scales. The formula to convert Celsius to Fahrenheit is: \[ 掳F = 1.8 imes 掳C + 32 \] This formula helps convert both the average temperature and the fluctuations or standard deviation from Celsius to Fahrenheit scales. As demonstrated, when converting temperatures in this exercise, one only needs to apply the multiplier (1.8) for mean and standard deviation conversions, noting that the addition of 32 does not apply to standard deviation since it's related to central tendency shifts, not spread.
  • Conversion is crucial for maintaining data integrity across disciplines.
  • Allows for consistent comparison and analysis.
  • Important for recognizing changes in temperature scales while maintaining statistical properties.

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Most popular questions from this chapter

If \(X\) has an exponential distribution with parameter \(\lambda\), derive a general expression for the \((100 p)\) th percentile of the distribution. Then specialize to obtain the median.

An individual's credit score is a number calculated based on that person's credit history that helps a lender determine how much he/she should be loaned or what credit limit should be established for a credit card. An article in the Los Angeles Times gave data which suggested that a beta distribution with parameters \(A=150, B=850, \alpha=8\), \(\beta=2\) would provide a reasonable approximation to the distribution of American credit scores. [Note: credit scores are integer-valued]. a. Let \(X\) represent a randomly selected American credit score. What are the mean value and standard deviation of this random variable? What is the probability that \(X\) is within 1 standard deviation of its mean value? b. What is the approximate probability that a randomly selected score will exceed 750 (which lenders consider a very good score)?

Nonpoint source loads are chemical masses that travel to the main stem of a river and its tributaries in flows that are distributed over relatively long stream reaches, in contrast to those that enter at well-defined and regulated points. The article "Assessing Uncertainty in Mass Balance Calculation of River Nonpoint Source Loads" (J. of Envir. Engr., 2008: 247-258) suggested that for a certain time period and location, \(X\) = nonpoint source load of total dissolved solids could be modeled with a lognormal distribution having mean value \(10,281 \mathrm{~kg} / \mathrm{day} / \mathrm{km}\) and a coefficient of variation \(C V=.40\left(C V=\sigma_{X} / \mu_{X}\right)\). a. What are the mean value and standard deviation of \(\ln (X) ?\) b. What is the probability that \(X\) is at most 15,000 \(\mathrm{kg} / \mathrm{day} / \mathrm{km}\) ? c. What is the probability that \(X\) exceeds its mean value, and why is this probability not \(.5\) ? d. Is 17,000 the 95 th percentile of the distribution?

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 90th 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?

Suppose the reaction temperature \(X\) (in \({ }^{\circ} \mathrm{C}\) ) in a certain chemical process has a uniform distribution with \(A=-5\) and \(B=5\). a. Compute \(P(X<0)\). b. Compute \(P(-2.5
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