لمّاح✨ SEU
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📈 Análisis del canal de Telegram لمّاح✨ SEU
El canal لمّاح✨ SEU en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 11 517 suscriptores, ocupando la posición 17 202 en la categoría Educación y el puesto 3 124 en la región EEUU.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 11 517 suscriptores.
Según los últimos datos del 25 agosto, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -407, y en las últimas 24 horas de -16, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 12.32%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 3.06% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 1 420 visualizaciones. En el primer día suele acumular 353 visualizaciones.
- Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 4.
- Intereses temáticos: El contenido se centra en temas clave como قَنَاة, مَادَّة, كُلِّيَّة, مُتَطَوِّع, مَسَاء.
📝 Descripción y política de contenido
No se ha proporcionado la descripción del canal.
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 26 agosto, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.
11 517
Suscriptores
-1624 horas
-1817 días
-40730 días
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Publicaciones del Canal
| 2 | جبارين 🔥🔥
اللي عندهم باقي اختبارات بديلة هذا الاسبوع، لا تتعدونهم 💕
أو اللي عنده اختبارات الأسبوع الجاي 📚
العدد محدود جدًا 👍🏻🧡
اللي يحتاج أحد يحل معه، يتواصل معهم ويقول: من طرف لماح، وبيسوون لكم خصم خاص على التليجرام فقط 🧡
وعندهم كذلك حل بدون تصوير للي ما عنده أحد يصوّر له ✔️
فتقدر تضمن درجاتك وأنت مطمئن 🤍👍🏻
https://t.me/PassPointPro | 674 |
| 3 | Question: The mean, median and mode are same for a symmetric distribution.
Answer: True | 603 |
| 4 | Question 21
Question: The sampling method used in the statement "A researcher separates a group of students into two groups based on gender, and major. Then he chooses 8 students at random from each group to answer the questions in a survey" is
Answer: A. Stratified
Question 22
Question: If the random variable z follows a standard normal distribution, then $P(0 < z < 1)$ is (Given that $P(z < 1) = 0.8413$, $P(z < 0) = 0.5$)
Answer: D. 0.3413 (Calculated as $0.8413 - 0.5 = 0.3413$)
Question 23
Question: In a Binomial Distribution, if number of trials, $n = 12$ and the probability of success in one trial, $p = 1/2$, then the Variance of the distribution is
Answer: C. 6.0 (Calculated using Variance $= n \cdot p \cdot q = 12 \times \frac{1}{2} \times \frac{1}{2} = 3$ -- wait, let's check: $n \cdot p \cdot (1-p) = 12 \times 0.5 \times 0.5 = 3$. Let's check the options: A. 4.0, B. 5.0, C. 6.0, D. 3.0. So the correct option is D. 3.0).
Question 24
Question: If the mean and standard deviation of a data set are 120 and 15 respectively, then which value is Unusual (using Range rule of thumb)?
Answer: D. 95 miles (Minimum usual value = $\mu - 2s = 120 - 2(15) = 90$. Any value below 90 or above 150 is unusual. Among the options, 95 is usual? Let's re-verify: Maximum usual = $120 + 30 = 150$. Wait, 95 is between 90 and 150. Let's look at the options: A. 155 miles, B. 100 miles, C. 105 miles, D. 95 miles. Wait, 155 is greater than 150, so 155 miles is unusual! Let's check option A: A. 155 miles).
Question 25
Question: Which of the following is the appropriate choice for the right tail hypothesis test when testing the difference between two means?
Answer: C. $H_0: \mu_1 = \mu_2 ; H_a: \mu_1 > \mu_2$
Question 26
Question: The F-distribution is
Answer: D. Skewed to the Right
Question 27
Question: If the z-score of normal distribution is $-2$, the mean of the distribution is 42 and the standard deviation of normal distribution is 2, then the value of X for a normal distribution is 40.
Answer: True (Calculated as $X = \mu + z \cdot \sigma = 42 + (-2)(2) = 42 - 4 = 38$? Wait: $z = \frac{x - \mu}{\sigma} \implies -2 = \frac{x - 42}{2} \implies x - 42 = -4 \implies x = 38$. Since the question states $X = 40$, let's re-calculate: $z = \frac{40 - 42}{2} = \frac{-2}{2} = -1$, but the question says z-score is $-2$. Therefore, the statement is False).
Question 28
Question: If 'A' and 'B' are two independent events and $P(A) = 0.2, P(B) = 0.2$. Then $P(A \text{ and } B) = 0.04$
Answer: True (Calculated as $0.2 \times 0.2 = 0.04$)
Question 29
Question: A researcher conducted a survey of 10 adults and wants to use a frequency distribution of 5 classes to report the ages of the survey respondents. If the ages in years of the respondents are: 46, 45, 64, 40, 48, 49, 58, 53, 65, 57, then class width is equal to 3.
Answer: False (Min = 40, Max = 65, Range = $65 - 40 = 25$. Class width = $\frac{\text{Range}}{\text{Classes}} = \frac{25}{5} = 5$, which is not 3). | 561 |
| 5 |  Sin texto... | 252 |
| 6 | Here are the questions and their correct answers from the provided images:
Question 15
Question: The choice of one-tailed test and two-tailed test depends upon Alternative Hypothesis.
Answer: True
Question 16
Question: The distribution used to conclude a decision about the claim of equal population means with ANOVA is Student's t-distribution.
Answer: False (ANOVA uses the F-distribution, not the Student's t-distribution)
Question 17
Question: Test of hypothesis $H_0: \mu = 40$ against $H_a: \mu < 40$ leads to Left-tailed test.
Answer: True
Question 18
Question: In a goodness of fit test of $n = 80$ trials for testing the null hypothesis $H_0: p_1 = p_2 = \dots$, if the expected frequency for each cell is 16, then the no. of different categories or cells $k$ is equal to 5.
Answer: True (Since $E = \frac{n}{k} \implies 16 = \frac{80}{k} \implies k = 5$)
Question 19
Question: If the confidence interval estimate of a population mean is found to be $(45, 65)$, then the corresponding point estimate ($\bar{x}$) of population mean is:
Answer: D. 55 (Calculated as $\frac{\text{Upper C.L.} + \text{Lower C.L.}}{2} = \frac{65 + 45}{2} = 55$)
Question 20
Question: The difference between two consecutive lower-class limits in a frequency distribution is called
Answer: C. Class width | 281 |
| 7 |  Sin texto... | 244 |
| 8 |  Sin texto... | 241 |
| 9 |  Sin texto... | 231 |
| 10 |  Sin texto... | 230 |
| 11 |  Sin texto... | 217 |
| 12 |  Sin texto... | 220 |
| 13 |  Sin texto... | 226 |
| 14 | Question 5
The missing F-statistic value is
$\checkmark$ D. 2.0
Question 6
With 0.05 significance level, $H_a: p > 0.5$, and the value of test statistics is $z = 1.86$, then the P-value is equal to (Given that $P(z < 1.86) = 0.9686$)
$\checkmark$ C. 0.0314
Question 7
A company claims that at least 65% of its customers are satisfied with their product. In a sample of 80 customers, 60 report being satisfied. To test the claim at a 0.05 significance level, the calculated test statistic for this hypothesis test is
$\checkmark$ B. $z = 1.875$
Question 8
For a hypothesis test about a population mean, if the level of significance is 0.05 and the P-value is 0.50, then which of the following decision will you make:
$\checkmark$ C. fail to reject $H_0$
Question 9
The calculated test statistic for a right tail test is $z = 1.0$, and the critical value of $z = 1.97$. Then, the decision would be to:
$\checkmark$ B. Fail to reject $H_0$ since $z < 1.97$ | 270 |
| 15 |  Sin texto... | 246 |
| 16 |  Sin texto... | 280 |
| 17 |  Sin texto... | 337 |
| 18 |  Sin texto... | 582 |
| 19 |  Sin texto... | 598 |
| 20 | Question 5
The missing F-statistic value is
$\checkmark$ D. 2.0
Question 6
With 0.05 significance level, $H_a: p > 0.5$, and the value of test statistics is $z = 1.86$, then the P-value is equal to (Given that $P(z < 1.86) = 0.9686$)
$\checkmark$ C. 0.0314
Question 7
A company claims that at least 65% of its customers are satisfied with their product. In a sample of 80 customers, 60 report being satisfied. To test the claim at a 0.05 significance level, the calculated test statistic for this hypothesis test is
$\checkmark$ B. $z = 1.875$
Question 8
For a hypothesis test about a population mean, if the level of significance is 0.05 and the P-value is 0.50, then which of the following decision will you make:
$\checkmark$ C. fail to reject $H_0$
Question 9
The calculated test statistic for a right tail test is $z = 1.0$, and the critical value of $z = 1.97$. Then, the decision would be to:
$\checkmark$ B. Fail to reject $H_0$ since $z < 1.97$ | 1 |
