Consider the following data set. $$ 52,60,57,54,56,59,55,53,60,59,59,57,57,54,57 $$ Use the data entry method in your scientific calculator to enter the measurements. Recall the proper memories to find the mean. Recall the proper memories to find the standard deviation. (Round your answer to four decimal places.) Calculate the range. The range is approximately how many standard deviations? (Round your answer to one decimal place.) The range is approximately Tandard deviations.

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The given data set is  
$$
52,\,60,\,57,\,54,\,56,\,59,\,55,\,53,\,60,\,59,\,59,\,57,\,57,\,54,\,57
$$

**Step 1. Calculate the Mean**

To find the mean, sum all the measurements and divide by the total number of measurements.

There are $$ n = 15 $$ measurements. Their sum is

$$
52 + 60 + 57 + 54 + 56 + 59 + 55 + 53 + 60 + 59 + 59 + 57 + 57 + 54 + 57 = 849.
$$

Thus, the mean is

$$
\bar{x} = \frac{849}{15} = 56.6.
$$

---

**Step 2. Calculate the Standard Deviation**

We first find the squared deviation from the mean for each measurement, then sum those values. The deviations are computed as follows:

\[
\begin{array}{rclcl}
52 - 56.6 &=& -4.6 &\quad \Rightarrow \quad (-4.6)^2 &=& 21.16,$$1mm]
60 - 56.6 &=& 3.4  &\quad \Rightarrow \quad (3.4)^2  &=& 11.56,\[1mm]
57 - 56.6 &=& 0.4  &\quad \Rightarrow \quad (0.4)^2  &=& 0.16,\[1mm]
54 - 56.6 &=& -2.6 &\quad \Rightarrow \quad (-2.6)^2 &=& 6.76,\[1mm]
56 - 56.6 &=& -0.6 &\quad \Rightarrow \quad (-0.6)^2 &=& 0.36,\[1mm]
59 - 56.6 &=& 2.4  &\quad \Rightarrow \quad (2.4)^2  &=& 5.76,\[1mm]
55 - 56.6 &=& -1.6 &\quad \Rightarrow \quad (-1.6)^2 &=& 2.56,\[1mm]
53 - 56.6 &=& -3.6 &\quad \Rightarrow \quad (-3.6)^2 &=& 12.96,\[1mm]
60 - 56.6 &=& 3.4  &\quad \Rightarrow \quad (3.4)^2  &=& 11.56,\[1mm]
59 - 56.6 &=& 2.4  &\quad \Rightarrow \quad (2.4)^2  &=& 5.76,\[1mm]
59 - 56.6 &=& 2.4  &\quad \Rightarrow \quad (2.4)^2  &=& 5.76,\[1mm]
57 - 56.6 &=& 0.4  &\quad \Rightarrow \quad (0.4)^2  &=& 0.16,\[1mm]
57 - 56.6 &=& 0.4  &\quad \Rightarrow \quad (0.4)^2  &=& 0.16,\[1mm]
54 - 56.6 &=& -2.6 &\quad \Rightarrow \quad (-2.6)^2 &=& 6.76,\[1mm]
57 - 56.6 &=& 0.4  &\quad \Rightarrow \quad (0.4)^2  &=& 0.16.
\end{array}
$$

Summing these squared deviations gives

$$
21.16 + 11.56 + 0.16 + 6.76 + 0.36 + 5.76 + 2.56 + 12.96 + 11.56 + 5.76 + 5.76 + 0.16 + 0.16 + 6.76 + 0.16 = 91.60.
$$

Since we consider these measurements as a sample, the sample variance is

$$
s^2 = \frac{91.60}{n-1} = \frac{91.60}{14} \approx 6.5429.
$$

Thus, the sample standard deviation is

$$
s = \sqrt{6.5429} \approx 2.5560 \quad (\text{rounded to four decimal places}).
$$

---

**Step 3. Calculate the Range**

The range is the difference between the largest and smallest values. Here, the smallest value is $$ 52 $$ and the largest is $$ 60 $$.

$$
\text{Range} = 60 - 52 = 8.
$$

---

**Step 4. Calculate How Many Standard Deviations the Range Represents**

To determine how many standard deviations the range spans, divide the range by the standard deviation:

$$
\text{Number of Standard Deviations} = \frac{\text{Range}}{s} = \frac{8}{2.5560} \approx 3.127.
$$

Rounded to one decimal place, the range is approximately

$$
3.1 \text{ standard deviations}.
$$

---

Final Answers:

- Mean: $$56.6$$
- Standard Deviation: $$2.5560$$
- Range: $$8$$
- The range is approximately $$3.1$$ standard deviations.
Mean: 56.6 Standard Deviation: 2.5560 Range: 8 The range is approximately 3.1 standard deviations.

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