hi pls help me answer the four check progress and the what can do (transfer) part. thank youuu

Convenience sampling is used in exploratory research for getting a gross esteratco' the results. withoutincurring the cost or lime .cauiron to selecta random samples Differentsampling is an extension of convenience Sampling For example, one may decide to draw the entire sample from one "representative" college. even though the population includes all colleges. Example: Identify the sampling technique illustrated below. 1.) A teacher assigned class numbers to each student and then spin a number roulette to identify the members of the sample. 2.) The teacher chose the top three students of the batch to represent the school in a regional quiz show. Answer: 1.) simple random sampling CHECK YOUR PROGRESS Do # 2, on the above example. WHAT IS IT Read "INTRODUCTION TO SAMPLING THEORY" pages 111 - 133 textbook (GEN. MATH - DIWA LEARNING SYSTEM INC. Copyright 2016) WHAT'S MORE SAMPLING DISTRIBUTION Is a distribution that shows the frequency with which values of statistics are observed when all the possible random samples are drawn from a given population. Example 1: A sample with size n = 3 is drawn from the set 5, 6, 8, 12, and 20. Construct the sampling distribution of the medians. Solution: Using combination, there are , Cran! / r! ( n- r ) 1 5 C3 = 5! / 31 ( 5 - 3 )! = 51 / 31 ( 21) = 20/2 = 10. There are 10 possible samples with size n = 3. The table below gives all these 10 samples and their corresponding medians. Sample Median 5, 6, 8 6 5, 6, 12 6 5, 6, 20 6 5, 8, 12 8 5, 8, 20 8 5, 12, 20 12 6, 8, 12 6, 8, 20 6, 12, 20 8, 12, 20Sample x Frequency 6 12 In sampling distribution, the standard deviation is usually called standard error. Example 2: Find the mean and standard error of the sampling distribution of the medians given in the example above. Solution: The mean of the sampling distribution is Hx = 6(3) + 8(4) + 12(3) = 86/10 = 8.6 10 Calculating the standard error of the distribution X X - Ux (X - (x ) 2 f ( x - [x ) ? 6 3 -2.6 6.76 20.28 8 4 -0.6 0.36 1.44 12 3 3.4 11.56 34.68 Total 10 56.4 o' = E f (x -[x)? / n= 56.4 / 10 = 5.64 0 = V5.64 = 2.37 The standard error of the sample medians is approximately 2.37. CHECK YOUR PROGRESS Random samples with size 4 are drawn from the population containing the values 14, 19, 26, 31, 48, and 53. a.) Construct a sampling distribution of the sample means b.) Find the mean of the sample means. c.) Compute the standard error of the sample means SAMPLING DISTRIBUTION OF THE SAMPLE MEANS The mean and the standard error of the distribution of sample means taken from a population with mean u and standard deviation o can be computed as Hs = H, Os = 0 / vn Example: Random samples with size 4 are drawn from the population containing the values 14, 19, 26, 31, 48, and 53. Find the mean and standard error of the sample means.Solution: The mean of the population is p = 14 +19 + 26 + 31 + 48 + 53 = 191/6 = 31.83 H = = 31.83 To get the standard error, compute first the population standard deviation: 0= (14 - 31 83) 4 (19 -31.83) + (26 -31 83) (31 -3181) + (48 -31 83) + (53 - 31.83) 0 - 14.3 Thus, the standard error of the mean is 0% = 0 / Vn = 14.3 / V4 - 7.15 CHECK YOUR PROGRESS A school has 900 junior high school students. The average height of these students is 68 in with a standard deviation of 6 in. Suppose you draw a random sample of 50 students. Find the mean, standard deviation, and variance of the distribution of all sample means that can be derived from the samples. DISTRIBUTION OF THE SAMPLE MEAN OF A NORMAL VARIABLE 2 = ( X - H ) o / vn THE CENTRAL LIMIT THEOREM Given a random variable x with mean u and variance o', then, regardless whether the population distribution of X Is normally distributed or not, as the sample size n gets larger, the shape of the distribution of the sample means taken from the population approaches a normal distribution, with mean u and standard deviation o / vn. Note: The emphasis of the central limit theorem is on the phrase " as the sample size gets larger". You can only assume the sampling distribution to be normal when the sample is large enough. Mostly, n 2 30 is sufficient to approximate the normality of the sampling distribution of the sample mean. Example: The average age of teachers in a certain town is 34 with a standard deviation of 4. If the principal of ABC College employs 100 teachers, find the probability that the average age of these teachers is less than 35.Solution: It is not given that the population is normally distributed but since n > 30, then you can assume that the sampling distribution of the mean ages of 100 teachers is normal according to Central Limit Theorem. Converting to standard normal score, Z = (35 - 34) / ( 4 / V100 ) = 10/4 = 2.5 Then, P ( X