Pregunta
upstudy study bank question image url

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

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

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.

Solución

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: \[ -1.28,\ -0.52,\ 0.00,\ 0.52,\ 1.28. \]

Respondido por UpStudy AI y revisado por un tutor profesional

error msg
Explicar
Simplifique esta solución

Mind Expander

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: \( \square -1.11, \square -0.74, \square 0.04, \square 0.52, \square 1.39 \)

preguntas relacionadas

Question 12(Mulliple Choice Warth 5 points) \[ (04.06 \mathrm{HC}) \] A researcher wants to test the claim that the proportion of juniors who watch television regularly is greater than the proportion of seniors who watch television regularly She finds that 56 of 70 randomly selected juniors and 47 of 85 randomly selected seniors report watching television regularly. Construct \( 95 \% \) confidence intervals for each population proportion. Which of the statemente gives the correct outcome of the research or's tert of the dalim? The \( 95 \% \) confidence interval for juniors is (706, 894), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is (721, 879), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( 706,894 ), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( \( 721, .879 \) ), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.

Latest Statistics Questions

¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad