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4. (08.06)The following data show the height, in inches, of 11 different plants in a garden:94109522210335After removing the outlier, what does the mean absolute deviation of this data set represent? (4 points)On average, the height of a plant varies 2.8 inches from the mean of 6 inches.On average, the height of a plant varies 3.2 inches from the mean of 5 inches.On average, the height of a plant varies 2.8 inches from the mean of 5 inches.On average, the height of a plant varies 3.2 inches from the mean of 6 inches.

Question

4. (08.06)The following data show the height, in inches, of 11 different plants in a garden:94109522210335After removing the outlier, what does the mean absolute deviation of this data set represent? (4 points)On average, the height of a plant varies 2.8 inches from the mean of 6 inches.On average, the height of a plant varies 3.2 inches from the mean of 5 inches.On average, the height of a plant varies 2.8 inches from the mean of 5 inches.On average, the height of a plant varies 3.2 inches from the mean of 6 inches.

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Solution

Primero, identifiquemos el valor atípico en el conjunto de datos. Los datos son: 9, 4, 10, 9, 5, 2, 2, 10, 3, 3, 5. El valor atípico es 22, ya que es significativamente mayor que los otros valores.

Eliminamos el valor atípico y nos quedamos con los siguientes datos: 9, 4, 10, 9, 5, 2, 2, 10, 3, 3, 5.

Ahora, calculamos la media de estos datos: Media=9+4+10+9+5+2+2+10+3+3+511=62115.64 \text{Media} = \frac{9 + 4 + 10 + 9 + 5 + 2 + 2 + 10 + 3 + 3 + 5}{11} = \frac{62}{11} \approx 5.64

Luego, calculamos la desviación absoluta de cada valor respecto a la media: 95.643.36 |9 - 5.64| \approx 3.36 45.641.64 |4 - 5.64| \approx 1.64 105.644.36 |10 - 5.64| \approx 4.36 95.643.36 |9 - 5.64| \approx 3.36 55.640.64 |5 - 5.64| \approx 0.64 25.643.64 |2 - 5.64| \approx 3.64 25.643.64 |2 - 5.64| \approx 3.64 105.644.36 |10 - 5.64| \approx 4.36 35.642.64 |3 - 5.64| \approx 2.64 35.642.64 |3 - 5.64| \approx 2.64 55.640.64 |5 - 5.64| \approx 0.64

Sumamos todas las desviaciones absolutas: 3.36+1.64+4.36+3.36+0.64+3.64+3.64+4.36+2.64+2.64+0.64=30.52 3.36 + 1.64 + 4.36 + 3.36 + 0.64 + 3.64 + 3.64 + 4.36 + 2.64 + 2.64 + 0.64 = 30.52

Finalmente, calculamos la desviación media absoluta dividiendo la suma de las desviaciones absolutas entre el número de datos: Desviacioˊn Media Absoluta=30.52112.77 \text{Desviación Media Absoluta} = \frac{30.52}{11} \approx 2.77

Por lo tanto, la opción correcta es: En promedio, la altura de una planta varıˊa 2.8 pulgadas de la media de 5 pulgadas. \text{En promedio, la altura de una planta varía 2.8 pulgadas de la media de 5 pulgadas.}

This problem has been solved

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