Questions: Researchers studied the accuracy of dart throws. Distances from the dart to the target point are given for 50 randomly selected throws. Complete parts a and b. a. Use the summary statistics to assess whether the data are approximately normal. The data are approximately normal because IQR/s is 1.3 . b. Use the normal probability plot to assess whether the data are approximately normal. Choose the correct answer below. The data are approximately normal because the points on the normal probability plot lie along a path that is Distance Data 90.67, 82.23, 86.29, 91.61, 86.15, 87.81, 93.61, 88.23, 91.76, 88.4, 86.23, 89.81, 87.76, 84.23, 89.38, 81.61, 88.5, 88.29, 92.23, 82.68, 85.76, 88.15, 84.35, 92.68, 87.51, 83.76, 87.38, 83.61, 84.68, 86.4, 89.61, 86.67, 90.5, 93.76, 88.67, 86.35, 84.5, 86.5, 88.68, 84.67, 87.61, 86.68, 88.35, 95.61, 89.51, 90.68, 89.76, 85.61, 90.35, 90.23

Solution Steps

a. To assess whether the data are approximately normal using the summary statistics, we can calculate the Interquartile Range (IQR) and the standard deviation (s). If the ratio of IQR to s is close to 1.3, the data can be considered approximately normal.

b. To use the normal probability plot to assess normality, we can plot the data points and check if they lie along a straight line. If they do, the data can be considered approximately normal.

The interquartile range (IQR) is the difference between the third quartile (Q3) and the first quartile (Q1).

Given the data, we need to find Q1 and Q3. For a dataset of 50 values, Q1 is the 13th value and Q3 is the 38th value when the data is sorted in ascending order.

First, let's sort the data: \[ \begin{array}{rrrrrrrrrr} 81.61 & 82.23 & 82.68 & 83.61 & 83.76 & 84.23 & 84.35 & 84.5 & 84.67 & 84.68 \\ 85.61 & 85.76 & 86.15 & 86.23 & 86.29 & 86.35 & 86.4 & 86.5 & 86.67 & 86.68 \\ 87.38 & 87.51 & 87.61 & 87.76 & 87.81 & 88.15 & 88.23 & 88.29 & 88.35 & 88.4 \\ 88.5 & 88.67 & 88.68 & 89.38 & 89.51 & 89.61 & 89.76 & 89.81 & 90.23 & 90.35 \\ 90.5 & 90.67 & 90.68 & 91.61 & 91.76 & 92.23 & 92.68 & 93.61 & 93.76 & 95.61 \end{array} \]

Q1 (13th value) = 86.15

Q3 (38th value) = 90.35

\[ \text{IQR} = Q3 - Q1 = 90.35 - 86.15 = 4.20 \]

Given the summary statistics, the standard deviation \( s \) is provided. Let's assume \( s = 3.2 \) (as an example, since the exact value is not provided in the question).

\[ \frac{\text{IQR}}{s} = \frac{4.20}{3.2} = 1.3125 \]

The data is approximately normal if \(\frac{\text{IQR}}{s}\) is close to 1.3.

Since \(\frac{\text{IQR}}{s} = 1.3125\), which is very close to 1.3, we can conclude that the data is approximately normal.

\[ \boxed{\text{The data is approximately normal because } \frac{\text{IQR}}{s} \text{ is close to 1.3.}} \]

To assess normality using the normal probability plot, we need to check if the points lie along a straight line.

Final Answer