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Use the given data values (a sample of female arm circumferences in centimeters) to identify the corresponding z scores that are used for a normal quantile plot, then identify the coordinates of each point in the normal quantile plot. Construct the normal quantile plot, then determine whether the data appear to be from a population with a normal distribution. \( 32.3,44.9,40.5,34.2,38.1 \) List the z scores for the normal quantile plot. \( \square . \square, \square . \square \) (Round to two decimal places as needed. Use ascending order.)

Ask by Edwards Mcfarlane. in the United States
Mar 10,2025

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The z scores for the normal quantile plot are: -1.28, -0.52, 0.00, 0.52, 1.28. The coordinates of each point in the normal quantile plot are: (-1.28, 32.3), (-0.52, 34.2), (0.00, 38.1), (0.52, 40.5), (1.28, 44.9). When constructing the normal quantile plot, these points should lie approximately along a straight line, indicating that the data likely come from a normally distributed population.

Solution

1. Sort the data in ascending order: \[ 32.3,\;34.2,\;38.1,\;40.5,\;44.9. \] 2. For a sample of \( n=5 \) observations, the plotting positions are given by \[ p_i = \frac{i-0.5}{n}, \quad i=1,2,\ldots,5. \] Thus, \[ \begin{array}{rl} p_1 & = \frac{1-0.5}{5} = \frac{0.5}{5} = 0.10,\\[1mm] p_2 & = \frac{2-0.5}{5} = \frac{1.5}{5} = 0.30,\\[1mm] p_3 & = \frac{3-0.5}{5} = \frac{2.5}{5} = 0.50,\\[1mm] p_4 & = \frac{4-0.5}{5} = \frac{3.5}{5} = 0.70,\\[1mm] p_5 & = \frac{5-0.5}{5} = \frac{4.5}{5} = 0.90. \end{array} \] 3. To obtain the corresponding \( z \) scores, we use the inverse standard normal distribution function: \[ z_i = \Phi^{-1}(p_i). \] This gives: \[ \begin{array}{rl} z_1 & = \Phi^{-1}(0.10) \approx -1.28,\\[1mm] z_2 & = \Phi^{-1}(0.30) \approx -0.52,\\[1mm] z_3 & = \Phi^{-1}(0.50) = 0,\\[1mm] z_4 & = \Phi^{-1}(0.70) \approx 0.52,\\[1mm] z_5 & = \Phi^{-1}(0.90) \approx 1.28. \end{array} \] 4. The coordinates of the points in the normal quantile plot are the pairs: \[ (-1.28,\, 32.3),\quad (-0.52,\, 34.2),\quad (0.00,\, 38.1),\quad (0.52,\, 40.5),\quad (1.28,\, 44.9). \] 5. To construct the normal quantile plot, plot these points with the \( z \)-scores on the horizontal axis and the corresponding data values on the vertical axis. If the data come from a normal distribution, the points should approximately lie along a straight line. 6. Inspection of the plot (or examining the linearity of the above coordinates) would suggest that these data points do indeed appear to come from a population with a normal distribution. The z scores (in ascending order) for the normal quantile plot are: \[ -1.28,\ -0.52,\ 0.00,\ 0.52,\ 1.28. \]

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To calculate the z-scores for the given data values, we first need to compute the mean and standard deviation of the sample. 1. **Calculate the mean (μ)**: \[ \text{Mean} = \frac{32.3 + 44.9 + 40.5 + 34.2 + 38.1}{5} = \frac{189.5}{5} = 37.9 \] 2. **Calculate the standard deviation (σ)**: \[ \text{Variance} = \frac{(32.3 - 37.9)^2 + (44.9 - 37.9)^2 + (40.5 - 37.9)^2 + (34.2 - 37.9)^2 + (38.1 - 37.9)^2}{5 - 1} \] \[ = \frac{(5.6)^2 + (7)^2 + (2.6)^2 + (3.7)^2 + (0.2)^2}{4} \] \[ = \frac{31.36 + 49 + 6.76 + 13.69 + 0.04}{4} = \frac{100.85}{4} = 25.2125 \] \[ \text{Standard deviation} = \sqrt{25.2125} ≈ 5.02 \] 3. **Calculate z-scores for each data value**: \[ z = \frac{x - \mu}{\sigma} \] For 32.3: \[ z_1 = \frac{32.3 - 37.9}{5.02} ≈ \frac{-5.6}{5.02} ≈ -1.11 \] For 44.9: \[ z_2 = \frac{44.9 - 37.9}{5.02} ≈ \frac{7}{5.02} ≈ 1.39 \] For 40.5: \[ z_3 = \frac{40.5 - 37.9}{5.02} ≈ \frac{2.6}{5.02} ≈ 0.52 \] For 34.2: \[ z_4 = \frac{34.2 - 37.9}{5.02} ≈ \frac{-3.7}{5.02} ≈ -0.74 \] For 38.1: \[ z_5 = \frac{38.1 - 37.9}{5.02} ≈ \frac{0.2}{5.02} ≈ 0.04 \] 4. **List of z-scores in ascending order**: \[ -1.11, -0.74, 0.04, 0.52, 1.39 \] Thus, the z-scores for the normal quantile plot are: \[ -1.11, -0.74, 0.04, 0.52, 1.39 \] Coordinates of each point in the normal quantile plot would correspond to the z-scores paired with the sorted data values: 1. (z_1, 32.3) 2. (z_4, 34.2) 3. (z_5, 38.1) 4. (z_3, 40.5) 5. (z_2, 44.9) The final answer for the z-scores is: \( \square -1.11, \square -0.74, \square 0.04, \square 0.52, \square 1.39 \)

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