Example Question - frequency distribution
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Identifying Modal Interval in Datasets
Para resolver esta pregunta, necesitamos identificar en cuál de los conjuntos de datos proporcionados el tercer intervalo es el intervalo modal. El intervalo modal es aquel que tiene la mayor frecuencia en un conjunto de datos. El intervalo modal es fácil de identificar mirando la frecuencia más alta en cada conjunto de intervalos: I) El tercer intervalo es 10 - 12 con una frecuencia de 2. Sin embargo, este no es el intervalo modal ya que el segundo intervalo (7 - 9) tiene una frecuencia más alta de 4. II) El tercer intervalo es [5, 7[ con una frecuencia de 12, la cual es la más alta de ese grupo de datos, haciendo de este intervalo el intervalo modal. III) El tercer intervalo es [20, 30[ con una frecuencia de 3, que no es la más alta ya que el primer intervalo ([10, 20[) tiene una frecuencia más alta de 5. La única serie de datos donde el tercer intervalo es el intervalo modal es la serie II. Por lo tanto, la respuesta correcta sería "Solo II", que corresponde a la opción B.
Analyzing Histogram Statements
Let's analyze the histogram in the image and address each statement provided: 1. There are 30 days represented on the histogram. To verify this, we need to sum up the number of days (frequency) for each temperature range shown on the histogram. 2. There were 8 days that had a daily high temperature of at least 85 degrees Fahrenheit. For this, we need to add up the number of days that fall within the temperature ranges that are 85 degrees Fahrenheit or higher. Looking at the histogram, this would include the last two bars on the right. 3. There were 27 days that had a daily high temperature lower than 90 degrees Fahrenheit. This would be the sum of the days represented in the temperature ranges that are below 90 degrees Fahrenheit. 4. If the median were to be calculated as 82.5 degrees, then a feasible mean would be 81.6. Without the raw data, we cannot calculate the mean or confirm this statement directly from the histogram. Additionally, histograms do not provide information about the mean or median values—only frequency distributions. We could estimate, but this would not be sufficient to confirm the statement's accuracy. 5. The data is skewed right. Skewness refers to asymmetry in the distribution of the data. A right-skewed distribution typically has a longer tail on the right side. We would determine this by looking at the shape of the histogram. The resolution and angle of the image makes it difficult for these statements to be irrefutably confirmed; however, examining a histogram's general shape typically allows for these kinds of observations. Please provide a clearer image or additional data so we could help you more accurately.
Calculating Frequency Distribution and Mean for English Test Scores
Para resolver esta pregunta, necesitamos elaborar una tabla de frecuencias que incluya la frecuencia absoluta, la frecuencia absoluta acumulada, la frecuencia relativa (expresada tanto en decimal como en porcentaje) y la frecuencia relativa acumulada (también en decimal y en porcentaje). Además, calcularemos la media de las notas. Aquí te muestro cómo podemos hacerlo: Primero, vamos a agrupar las notas de inglés que han sido proporcionadas para los 40 alumnos y contar cuántas veces aparece cada nota (frecuencia absoluta). Luego, vamos a sumar estas frecuencias para obtener la frecuencia absoluta acumulada. Notas: 1 2 3 4 5 6 7 8 Frecuencia absoluta (fa): 4 6 5 5 3 7 6 4 Frecuencia absoluta acumulada (Fa): 4 10 15 20 23 30 36 40 Ahora, debemos calcular la frecuencia relativa. Para ello, dividimos la frecuencia absoluta de cada nota entre el número total de alumnos (40). También convertiremos este número a porcentaje multiplicándolo por 100. Frecuencia relativa (fr) = fa / 40 Frecuencia relativa en % (fr%) = fr * 100 Hagamos el cálculo para cada nota: fr1 = 4/40 = 0.1 --> fr1% = 0.1 * 100 = 10% fr2 = 6/40 = 0.15 --> fr2% = 0.15 * 100 = 15% fr3 = 5/40 = 0.125 --> fr3% = 0.125 * 100 = 12.5% fr4 = 5/40 = 0.125 --> fr4% = 0.125 * 100 = 12.5% fr5 = 3/40 = 0.075 --> fr5% = 0.075 * 100 = 7.5% fr6 = 7/40 = 0.175 --> fr6% = 0.175 * 100 = 17.5% fr7 = 6/40 = 0.15 --> fr7% = 0.15 * 100 = 15% fr8 = 4/40 = 0.1 --> fr8% = 0.1 * 100 = 10% Para las frecuencias relativas acumuladas (FR), sumamos las frecuencias relativas hasta cada nota y lo mismo para obtener los porcentajes acumulados (FR%): FR1 = fr1 = 0.1 FR2 = fr1 + fr2 = 0.1 + 0.15 = 0.25 FR3 = FR2 + fr3 = 0.25 + 0.125 = 0.375 FR4 = FR3 + fr4 = 0.375 + 0.125 = 0.5 FR5 = FR4 + fr5 = 0.5 + 0.075 = 0.575 FR6 = FR5 + fr6 = 0.575 + 0.175 = 0.75 FR7 = FR6 + fr7 = 0.75 + 0.15 = 0.9 FR8 = FR7 + fr8 = 0.9 + 0.1 = 1 (o 100%) Finalmente, para calcular la media (promedio) de las notas, multiplicamos cada nota por su frecuencia absoluta y sumamos todos estos productos. Después, dividimos el resultado por el número total de notas (40): Media = (1*4 + 2*6 + 3*5 + 4*5 + 5*3 + 6*7 + 7*6 + 8*4) / 40 Haciendo los cálculos: Media = (4 + 12 + 15 + 20 + 15 + 42 + 42 + 32) / 40 Media = 182 / 40 Media = 4.55 La media de las notas de inglés de la clase es 4.55.
Probability of Drawing Tiles Analysis
To answer the questions provided in the image: 1. Will all the possible outcomes appear to be equally likely based on the observed frequencies? To determine if all outcomes are equally likely, we can compare the frequencies of each letter. The frequencies shown in the table are: - A: 6 - E: 12 - O: 8 - W: 14 The frequencies are not the same for each letter, so the possible outcomes do not appear to be equally likely based on the observed frequencies. 2. What is the experimental probability for drawing a W tile? To find the experimental probability of drawing a W tile, we use the frequency of the W outcome divided by the total number of outcomes. The total number of outcomes can be found by adding the frequencies of all the letters together: Total outcomes = 6 (A) + 12 (E) + 8 (O) + 14 (W) = 40 Now, we calculate the experimental probability for W: Experimental probability for W = Frequency of W / Total outcomes = 14 / 40 = 0.35 So, the experimental probability of drawing a W tile is 0.35, or 35%. 3. If there were a total of 100 tiles, what is the best estimate of the number of E tiles in the bag? We look at the frequency distribution to estimate the number of E tiles. From the total of 40 tiles that were drawn (as calculated above), E tiles were drawn 12 times. To find the best estimate for 100 tiles, we can set up a proportion based on the experimental results: 12 (frequency of E) / 40 (total outcomes) = x (estimated number of E tiles) / 100 (total tiles) Now, solve for x: (12 / 40) * 100 = x (0.3) * 100 = x x = 30 The best estimate for the number of E tiles in a bag of 100 tiles, based on the experimental probability, is 30.
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