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].