لمّاح✨ PMAU
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📈 Analytical overview of Telegram channel لمّاح✨ PMAU
Channel لمّاح✨ PMAU in the English language segment is an active participant. Currently, the community unites 11 174 subscribers, ranking 17 557 in the Education category and 3 132 in the USA region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 11 174 subscribers.
According to the latest data from 05 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -224 over the last 30 days and by -9 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 11.27%. Within the first 24 hours after publication, content typically collects 3.41% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 261 views. Within the first day, a publication typically gains 381 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
- Thematic interests: Content is focused on key topics such as قَنَاة, مَادَّة, كُلِّيَّة, مُتَطَوِّع, مَسَاء.
📝 Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 06 October, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.
11 174
Subscribers
-924 hours
-377 days
-22430 days
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| Date | Subscriber Growth | Mentions | Channels | |
| 06 October | 0 | |||
| 05 October | 0 | |||
| 04 October | 0 | |||
| 03 October | 0 | |||
| 02 October | 0 | |||
| 01 October | 0 |
Channel Posts
| 2 | ما هو مستواك الدراسي الحالي | 626 |
| 3 | ماهو مستواك الدراسي الحالي ⁉️ | 13 |
| 4 | تحضيري خطة A | 1 |
| 5 | متابعة قناة إعلامات كلية العلوم الإدارية و المالية على واتساب: https://whatsapp.com/channel/0029VbDjtVBCHDym7fFg092M | 1 022 |
| 6 |  No text... | 1 532 |
| 7 |  No text... | 1 |
| 8 | متابعة قناة ISLAM101 على واتساب: https://whatsapp.com/channel/0029Vb9MAitKrWQv4lw9Du1K
〰️〰️〰️〰️〰️〰️
متابعة قناة ISLAM102 على واتساب: https://whatsapp.com/channel/0029VbDbXdCAYlUSbJLDZ73T
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متابعة قناة ISLAM103 على واتساب: https://whatsapp.com/channel/0029Vb8jkPdIt5rp4W72Ja1c
〰️〰️〰️〰️〰️〰️〰️〰️
متابعة قناة ISLAM104 على واتساب: https://whatsapp.com/channel/0029VbDmyt2J3juw3v8f7Q2e | 2 000 |
| 9 | 3:00 To 5:00 💻
SCI101❌
MATH241❌
5:30 To 7:30💻
SCI201 ❌
ARB260 ✖️ | 2 090 |
| 10 | 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 | 1 562 |
| 11 | Question: The mean, median and mode are same for a symmetric distribution.
Answer: True | 1 033 |
| 12 | 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). | 945 |
| 13 |  No text... | 462 |
| 14 | 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 | 498 |
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