Splet09. jun. 2024 · The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to : A 10 B 36 C 43 D 60 Check Answer 2 JEE Main 2024 (Online) 26th June … SpletThe mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The correct (new) mean is : A) 35.2 B) 36.1 C) 36.5 D) 39.1 Correct Answer: C) 36.5 Description for Correct answer: According to the question, Correct observation = 50 × 36 + 48 − 23 50 = 1800 + 25 50 = 1825 50 = 36.5
The mean of 50 observations was 36. It was found later …
SpletTo calculate "within 1 standard deviation," you need to subtract 1 standard deviation from the mean, then add 1 standard deviation to the mean. That will give you the range for 68% of the data values. 285− 37 = 248 285 − 37 = 248 285+ 37 = 322 285 + 37 = 322 The range of numbers is 248 to 322. The second part of the empirical rule states ... Splet06. apr. 2024 · In the following table, find the missing frequencies f1 and f2 if mean of 50 observations given below is 36.4 . C.I.0−1010−2024−3030−4040−5050−6060−70f35fl10f285 Heights of the peopl. Solution For 9. In the following table, find the missing frequencies f1 and f2 if mean of 50 … denver coffee shops per capita
The mean of 100 observations is 50. If one of the observations …
Spletk-means clustering is a method of vector quantization, originally from signal processing, that aims to partition n observations into k clusters in which each observation belongs to the cluster with the nearest mean (cluster centers or cluster centroid ), serving as a prototype of the cluster. This results in a partitioning of the data space ... SpletThe mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is : Answer: B) 36.5 Explanation: Correct Sum = (36 … SpletFind the probability that the mean of a sample of size 36 will be within 10 units of the population mean, that is, between 118 and 138. ... Find the probability that in a sample of 50 returns requesting a refund, the mean such time will be more than 50 days. Scores on a common final exam in a large enrollment, multiple-section freshman course ... fgo servants with skill seal