Thank you for visiting 11 The accepted temperature of a refrigerator is tex 38 0 circ F tex A temperature sensor is tested 10 times in the refrigerator yielding. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
We begin by analyzing the temperature measurements and then move on to the mass measurement.
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Step 1. Temperature Measurements
The refrigerator temperature is accepted as [tex]$38.0^\circ \text{F}$[/tex]. Ten tests produced the following temperatures (in degrees Fahrenheit):
[tex]$$37.8,\; 38.3,\; 38.1,\; 38.0,\; 37.6,\; 38.2,\; 38.0,\; 38.0,\; 37.4,\; 38.3.$$[/tex]
1.1. Calculate the Mean
The mean temperature [tex]$\bar{x}$[/tex] is given by
[tex]$$
\bar{x} = \frac{1}{10}\sum_{i=1}^{10} x_i.
$$[/tex]
Carrying out the calculation,
[tex]$$
\bar{x} \approx 37.97^\circ \text{F}.
$$[/tex]
Since the mean is very close to the accepted value of [tex]$38.0^\circ \text{F}$[/tex] (the difference is about [tex]$0.03^\circ \text{F}$[/tex]), we say that the measurements are accurate.
1.2. Calculate the Sample Standard Deviation
The sample standard deviation [tex]$s$[/tex] is calculated by
[tex]$$
s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2},
$$[/tex]
where [tex]$n=10$[/tex]. After computing the deviations and taking the square root, we obtain
[tex]$$
s \approx 0.295^\circ \text{F}.
$$[/tex]
A small standard deviation indicates that the values are tightly clustered around the mean. Therefore, the sensor is precise.
--------------------------------------------------
Step 2. Mass Measurement and Percent Error
A researcher measures the mass as [tex]$5.51\;\text{g}$[/tex], while the actual mass is [tex]$5.80\;\text{g}$[/tex].
2.1. Calculate the Percent Error
The percent error is determined by
[tex]$$
\text{Percent Error} = \left|\frac{\text{Measured Value} - \text{Actual Value}}{\text{Actual Value}}\right| \times 100\%.
$$[/tex]
Substitute the given values:
[tex]$$
\text{Percent Error} = \left|\frac{5.51 - 5.80}{5.80}\right| \times 100\% = \left|\frac{-0.29}{5.80}\right| \times 100\% \approx 5\%.
$$[/tex]
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Final Answer:
1. For the temperature measurements:
- The sensor is accurate: yes.
- The sensor is precise: yes.
2. For the mass measurement:
- The percent error is approximately [tex]$5\%$[/tex].
--------------------------------------------------
Step 1. Temperature Measurements
The refrigerator temperature is accepted as [tex]$38.0^\circ \text{F}$[/tex]. Ten tests produced the following temperatures (in degrees Fahrenheit):
[tex]$$37.8,\; 38.3,\; 38.1,\; 38.0,\; 37.6,\; 38.2,\; 38.0,\; 38.0,\; 37.4,\; 38.3.$$[/tex]
1.1. Calculate the Mean
The mean temperature [tex]$\bar{x}$[/tex] is given by
[tex]$$
\bar{x} = \frac{1}{10}\sum_{i=1}^{10} x_i.
$$[/tex]
Carrying out the calculation,
[tex]$$
\bar{x} \approx 37.97^\circ \text{F}.
$$[/tex]
Since the mean is very close to the accepted value of [tex]$38.0^\circ \text{F}$[/tex] (the difference is about [tex]$0.03^\circ \text{F}$[/tex]), we say that the measurements are accurate.
1.2. Calculate the Sample Standard Deviation
The sample standard deviation [tex]$s$[/tex] is calculated by
[tex]$$
s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2},
$$[/tex]
where [tex]$n=10$[/tex]. After computing the deviations and taking the square root, we obtain
[tex]$$
s \approx 0.295^\circ \text{F}.
$$[/tex]
A small standard deviation indicates that the values are tightly clustered around the mean. Therefore, the sensor is precise.
--------------------------------------------------
Step 2. Mass Measurement and Percent Error
A researcher measures the mass as [tex]$5.51\;\text{g}$[/tex], while the actual mass is [tex]$5.80\;\text{g}$[/tex].
2.1. Calculate the Percent Error
The percent error is determined by
[tex]$$
\text{Percent Error} = \left|\frac{\text{Measured Value} - \text{Actual Value}}{\text{Actual Value}}\right| \times 100\%.
$$[/tex]
Substitute the given values:
[tex]$$
\text{Percent Error} = \left|\frac{5.51 - 5.80}{5.80}\right| \times 100\% = \left|\frac{-0.29}{5.80}\right| \times 100\% \approx 5\%.
$$[/tex]
--------------------------------------------------
Final Answer:
1. For the temperature measurements:
- The sensor is accurate: yes.
- The sensor is precise: yes.
2. For the mass measurement:
- The percent error is approximately [tex]$5\%$[/tex].
Thank you for reading the article 11 The accepted temperature of a refrigerator is tex 38 0 circ F tex A temperature sensor is tested 10 times in the refrigerator yielding. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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Rewritten by : Jeany
