e Let Us Try! 0' Hello! I hope you are having a good day. Before proceeding to Lesson 1, try to read and answer the activity below rst. Direction: Choose the letter of the correct answer and write the correct letter on the separate sheet of paper. 1. What do you call to the set of all people, objects, events, or ideas researchers want to investigate? A. data B. population C. sample D. statistics 2. What do you call a number that describes a population characteristic? A. data B. parameter C. samples D. statistic 3. What do you call a number that describes a sample characteristic? A. data B. parameter C. samples D. statistic 4. What distribution pertains to the frequency distribution of the sample mean from all the possible random samples of a particular sample size n taken from the given population? A. frequency C. normal C. population D. sampling 5. Which of the following is NOT a stepin creating sampling distribution of the sample mean? A. Determine the number of sets of all possible random samples. B. Compute for the standard deviation and variance of the samples. C. Construct a frequency distribution table of the sample mean and probability. D. List all the possible random samples and solve for the sample mean of each set of samples. Let Us Study Before we start with our lesson, let us rst read the selection below. RANDOM SAMPLING In research, collecting data can either be done in the entire population or the subset of this population called sample. Population-includes all its elements from a set of data. Sampleconsists of one or more data drawn from the population. It is a subset, or an incomplete set taken from a population of objects or Example: As of May 6, 2020, the Department of Health recorded 131,786 unique individuals who got tested for the COVID19 out of approximately 109,369,019 Filipinos. 1. Identify the population. A11 Filipinos 2. How many elements are there in the population? 109 369 019 3. Who are the samples? Filipinos tested for Covid 19 4. How many samples are selected? 131,7 86 5. Did the situation illustrate random sampling? Why or why not? Noz because not all Filipinos have egual chance of being selected. Random Sampling is a sampling method of choosing representatives from the population wherein every sample has an equal chance of being selected. Types of Random Sampling Types of random sampling techniques Denition, sample, illustration Simple random sampling this technique is the most basic random sampling wherein each element in the population has an equal probability of being selected. They are usually represented by a unique identication number that is written on equalsized and shaped papers and then selection of samples is possible through the lottery method. Systematic sampling a random sampling that uses a list of all the elements in the population and then elements are being selected based on the kth consistent intervals. To get the kth interval, divide the population size by the sample size. POPULATION SAMPLE iiiiii ... iiiiii I??? iitiisi; *i iiiiii iiiiiiii 15 Stratified sampling a random sampling wherein the population is divided into different strata or divisions. The number of samples will be proportionately picked in each stratum that is why all strata are represented in the samples. gAZEPULATION SAMPLE iiiii +43; $36; 1' E E 15' a 8 wswrwvwi iiiisi Cluster sampling a random sampling wherein population is divided into clusters or groups and then the clusters are randomly selected. All elements of the clusters randomly selected are considered the samples of the study. POPULATION SAMPLE W. 5 7 B iiii'if'i w . 5 , B m? .. .. . .. .. uni-in 'I'l'i'i'l'i'l'i' w SESEEZsegee; l'g'la'l'e''l' T''''' mr'lmmrmr 4% iii\" Probability sampling is a sampling technique that involve random selection. Likewise, simple random, systematic, and stratied and cluster sampling are all probability sampling techniques. Non-probability sampling is a sampling technique that do not involve random selection of data. - Convenience sampling is being used by persons giving questionnaires on the streets to ask the passers-by. - Purposive sampling the respondents are being selected based on the goal of the studies of the researcher. PARAMETER AND STATISTIC A parameter is a measure that is used to describe the population while statistic is a measure that is used to describe the sample. Example: A study is conducted to a readytoharvest bananas in an experimental eld and getting the individual weights of bananas in the eld. If the weight of all readytoharvest bananas in the eld are included in the data, then it is a population. The average weight of all the readytoharvest bananas in the eld is a PARAMETER. A panel of tasters rates the bananas according to categories: Poor, Acceptable and Good. If only some of the bananas are included in the taste test, then it is a sample. The number of bananas in the sample under the classication of \"good\" is a STATISTIC. SAMPLING DISTRIBUTION OF STATISTICS Sampling distribution of the sample means is a frequency distribution using the computed sample mean from all the possible random samples of a particular sample size taken from the given population. Steps to follow in making a sampling distribution of the sample mean: 1. Determine the number of sets of all possible random samples that can be drawn from the given population by using the formula, NCn, where N is the population size and n is the sample size. Example: A population of Senior High School consists of numbers 1, 2, 3, 4, and 5. In this activity, we are given a population of 1, 2, 3, 4, and 5 and sample size of 3, therefore we have, N! 5! 5! 54-3-24 10 NC\": 5C3 = [n!(Nn)!J : [3!(53)\" : [3!2u _ (344x21) _ *or use your calculator function nCr. 2. List all the possible random samples and solve for the sample mean of each set of samples. Sample Mean 1, 2, 3 2 1, 2, 4 2.33 1, 2, 5 2.67 1, 3, 4 2.67 1, 3, 5 3 1, 4, 5 3.33 2, 3, 4 3 2, 3, 5 3.33 2, 4, 5 3.67 3, 4, 5 4 3. Construct a frequency and probability distribution table of the sample means indicating its number of occurrence or the frequency and probability. Sample Means Frequency Probability P(X) 1 2 1 E20.10 2.33 1 i=o.10 10 \fLet Us Practice . You did great on your rst day! Now, let us try what you have learned. Are you ready? ACTIVITY 1: Identify the random sampling technique used in each item. Situation Answer 1. You decide to survey every 10th student Systematic sampling on the list and ask them the organization that they belong. 2. You divide the population into two groups, male and female, and randomly pick respondents from each of the group. 3. You assign numbers to the members of the population and then use draw lots to obtain your samples to answer your survey on the most popular festivals in the country. 4. You randomly pick ve out of fteen barangays to conduct your survey in your municipality or city about their best environmentfriendly practices. 5. You write the names of each student in pieces of paper, shufes, and then draw eight names to answer a survey on their ethical media practices. ACTIVITY 2: A population consists of the values (1, 4, 3, 2). Consider samples of size 2 that can be drawn from this population. a. Determine the number of sets of all possible random samples N! 4-! 4-! 4-3-24 Ncn = 402 = E [mmmu : [21(44):] 2 [212:] _ (HXZ-l) _ b. List down all the possible samples and corresponding sample mean. Sample Mean | 1.5 | 1, 3 | 2.5 | 2.5 | 2, 4 | 3, 4 0. Construct the sampling distribution of the sample means. Sample Means Frequency Probability P(X) 1.5 1 2 1=0.17 6 2.5 220.33 a 3 3. 5 1 1:017 6 Let Us Assess Analyze the given situation below and then identify the type of random sampling methods being illustrated in each item. Write the letter of your answer in a separate answer sheet. 1. From the given populations, which of the following sample is most likely to be representatives of the population from which it is drawn? A. population: SAP beneficiaries sample: car owners B. population: lawmakers sample: students C. population: online shoppers sample: cellphone users D. population: graduating students sample: grade 8 students 2. A National High School has 2,000 first year high school students. Mrs. Mogol, the school principal, wants to obtain information from these students about last year's lesson that has not been tackled. What is the target population in her study? A. all students in her school B. parents of all students in her school C. first year high school students in her school D. parents of first year high school in her school 3. In random sampling, the sample should be the population.A. as large as C. not taken from B. different from D. representative of . What do you call a number that describes a sample characteristic? A. data C. sample B. parameter D. statistic . What distribution pertains to the frequency distribution of the sample mean from all the possible random samples of a particular sample size n taken from the given population? A. frequency B. normal C. population D. sampling Let Us Try! 0' Hello! I hope you are having a good day. Before proceeding to Lesson 2, try to read and answer the activity below rst. Direction: Choose the letter of the correct answer and write the correct letter on the separate sheet of paper. 1. Which of the following is the notation used for the mean of the sampling distribution of sample means? A. M B. m? C. 0 D. 0x 2. A population consists of ve (5) measurements 2, 3, 6, 5, and 7. What is the mean of the population? A. 3.44 B. 4.60 C. 4.92 D. 5.20 3. Which of the following measures the dispersion of the sampling distribution of the sample mean and is calculated using the relation 2(Xu)2- Pm or 2ng - P(x)] we? A. mean of the sampling distribution of sample mean B. variance of the sampling distribution of the sample mean C. standard deviation of the sampling distribution of the sample mean D. average deviation of the sampling distribution of the sample mean 4. This distribution is used to determine the probability of an event by transforming the mean of the sample to an approximately standard normal variable if the population variance is known and sample size is greater than 30. A. tDistribution B. ZDistribution C. ChiSquare Distribution D. Pearson Correlation Coefcient 0 Let Us Study Before we start with our lesson, let us first read the selection below. A. FINDING THE MEAN AND VARIANCE OF THE SAMPLING DISTRIBUTION OF MEANS Statisticians do not just dene the discrepancy of the individual data values about the mean of the population. They are also concerned to know how the means of the samples of the same size taken from the same group vary about the population mean. Illustrative example: Mark is conducting a survey on grade 12 students of Nasyonalismo High School. He found out that there are only few students who knew about the makers of the Philippine ag consisting of 1, 2, 3, 4, and 5 SHS students from 5 sections. Suppose that the sample size of 2 sections were drawn from this population (without replacement), describe the sampling distribution of the sample means. 1. What is the mean and variance of the sampling distribution of the sample means? 2. Compare these values to the mean and variance of the population. Steps Solution Compute the mean of the population using the formula 1+2+3+4+5 = 3= 11 \"gm. (1) u N 5 3.00 5. This distribution is used to estimate population parameters when the population variance is unknown, and the sample size is less than 30. A. ZDistribution B. tDistribution C. ChiSquare Distribution D. Pearson Correlation Coefcient 2. Compute the variance of x x-u (Ix-11F the population using the 1 2 4 2 _ 206-102 2 1 1 formula 0 1v . 3 0 O 4 1 1 5 2 4 2(X u)2= 10 (2) O2 = 2(x \")2 N = L0 5 3 So, the variance of the population is 2. 3. Determine the number of Use the formula \"Cu. Here N = 5 and n = 2. possible samples of size 2 (Without replacement). Use 5C2 = M1311\")! = 2!(:!_2)! = 10 the combination formula C2: n'( It\")! where st the So, there are 10 possible samples of size 2 . . . that can be drawn. population Slze and n IS the sample size. 4. List all possible samples Samples Mean and compute the 1% -g corresponding means. 1,4 2.5 1,5 3.0 2,3 2.5 2,4 3.0 2,5 3.5 3,4 3.5 3,5 4.0 4,5 4.5 5. Construct the sampling Sampling Distribution of Sample distribution of the sample Means . Sample Frequen Probabllity PW) means. _ Mean x cy 1.5 l l 10 2.0 1 1 / 10 2.5 2 1 / 5 3.0 2 1 / 5 3.5 2 1 /5 4.0 1 1 / 10 4.5 | 1 1 1/10 Total 1 1o 1 1.00 6. Compute the mean of the [6) Sample Probability (a) E - Pa) sampling distribution of the Mean Y Pa) sample means. 1.5 1/10 0.15 2.0 1 / 10 0.20 2.5 1 / 5 0.50 3.0 1 / 5 0.60 3.5 1 /5 0.70 4.0 1/10 0.40 4.5 1/10 0.45 Total 1.00 3.00 7. Compute the variance of Salm ifba S $11) (3% P b If? ((S . . . . pe 11ty . ea . roaiity- . the sampling distribution of Mea PG) n _ ean_ Mean_ P.Mean)2) the sample means using the n r P.Mean P.Me PG) - (E \")2 formula 0% : Z[P()_() - (X i _\" (8;)2 I02] u)2 1.5 1/1 1.5 2.25 0.225 0 2.0 1/1 1.0 1.00 0.100 0 2.5 1/5 0.5 0.25 0.050 3.0 1/5 0.0 0.00 0.000 3.5 1/5 0.25 0.25 0.050 4.0 1/1 1.00 1.00 0.100 0 4.5 1/1 2.25 2.25 0.225 0 Tot 1.00 0.750 a1 Note: S. Mean = sample mean, P. Mean= population mean So, the variance of the sampling distribution is 0.75 The mean of the population is equal to the sample means. This happens because all samples obtained of n size came from the same population. The variance of the population is greater than the variance of the sample means, since a sample of n is less than the population N then the data vary by a small amount as compared to population. The standard error of the mean is the standard deviation of the sampling distribution of the mean. Given a population with a nite mean u and a nite nonzero variance 02, the sampling distribution of the mean . . . - . a" approaches a normal distribution w1th a mean of u and a variance of 7 as N, the sample size, increases. B. DISTRIBUTION OF THE SAMPLE MEAN FOR NORMAL POPULATION 1. Population variance 02 is known The population has a mean u and variance of 02, the distribution of the sample mean is (at least approximately) normal and standard error of the mean a where o is the population standard deviation and n is the _ _ L X _ x/Tl' sample size. To determine the probability of a certain event, we can use the 2dllrz'nbm'z'oiz by transforming the mean of the sample data to an 2'. This 45 distribution is best applied for large sample sizes, say n 2 30. x approximately normal variable 2, using the relation 2 = 2. Population variance 09 is unknown The standard error of the mean becomes 5- = where s is the point S 3 estimate of 0' (population standard deviation) or the sample standard deviation and n is the sample size. To estimate the population parameters, we can use tdistributions by using the formula t = f?\" . Remember that as m n the sample size is very large, the standard deviation 5 is almost indistinguishable from the population standard deviation 0 and therefore, t and z distributions are essentially identical. Remember that, we use the t distributions for small sample size, say n 123). 4. Find P(X > 123). Solution: 1. By the Central Limit theorem, it; = [.1 = 122 and a; = i z 6.45497 x/E 2. According to CLT, )7 is approximately normally distributed so we use 2 = X7,\" to solve P (E > 120) and P (E 120), solve for P (E > 124), Z = X- H 120-122 Z = 50 Z = - 124- 122 50 V60 60 -2 2 Z= 6.45497 6.45497 Z = -0.31 Z = 0.31 Using the z table, let us look for the areas of normal curve of z = -0.31 0.00 0.01 0.02 0.03 0.04 0.05 0.0 0.07 0.08 0.09 0.0000 0.00005 0.00004 0.00004 0.00004 0.00004 0.00004 0.00003 0.00003 To use the table, locate -3.8 0.09607 0.00007 0.00007 0.00006 0.00006 0.00006 0.00005 0.00005 0.00005 3.7 100011 0.00010 0.00010 0.00010 0.00009 0.00009 0.00008 0.00008 0.00008 0.00008 The first two digit -3.6 .0001 00015 0.00015 0.00014 0 00014 0.00013 0.00013 0.00012 0.00012 0.00011 -3.5 0.00023 0.00022 0.00022 0.00021 0.00020 0.00019 0.00019 0.00018 0.00017 0.00017 vertically then the 0.00034 0.00032 0.00031 0.00030 0.00029 0.00028 0.00027 0.00026 0.00025 0.00024 0.00048 0.00047 0.00045 0.00043 0.00042 0.00040 0.00039 .00038 0.00036 0.00035 Last digit horizontally. -3.2 0.00069 0.00066 0.00064 0.00062 0.00060 0.00058 0.00056 0.00054 0.00052 0.00050 0.00097 0.00094 0.00090 0.00087 0.00084 0.00082 0.00079 0.00076 0.00074 0.00071 Then locate the 3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 2.9 0-00187 0.00181 0.00175 0 00169 0.00164 0.00159 0.00154 0.00149 0.00144 0.00139 2.8 .0025 0.00248 0.00240 0.00233 0.00226 0.00219 0.00212 0.00205 0.00199 0.00193 Intersection. 2.7 0.00347 0.00336 0.00326 0.00317 0.00307 0.00298 0.00289 0.00280 0.00272 0.00264 -2.6 0.00466 0. 00453 0.00440 0.00427 0.00415 0.00402 0.00391 0.00379 0.00368 0.0035 -2.5 0.00621 0.00587 0.00570 0.00554 0.00539 0.00523 0.00508 0.00494 0.0048 -2.4 0.00820 0.00798 0.00776 0.00755 0.00734 0.00714 0.00695 0.00676 0.00657 0.00639 Example: - 0 . 3 1 -2.3 0.0107 0.01044 0.01017 0.0099 0.00964 0.00939 0.00914 0.00889 0.00866 0.00842 2.2 0.01390 0.01355 0.01321 0.01287 0.01255 .01222 0.01191 0.01160 0.01130 0.01101 2.1 0.01786 0.01743 0.01700 0.01659 0.01618 0.01578 0.01539 0.01500 0.01463 0.01426 2.0 0.02275 0.02222 0.02169 0.02118 0.02068 0.02018 0.01970 0.01923 0.01876 0.01831 -1.9 0.02872 0.02807 0.02743 0.02680 0.02619 0.02559 0.02500 0.02442 0.02385 0.02330 -1.8 0.03593 .03515 0.03438 0.03362 0.03288 0.03216 0.03144 0.03074 0.03005 0.02938 -1.7 0.04457 0.04363 0.04272 0.04182 0.04093 0.04006 0.03920 0.03836 0.03754 0.03673 Note: -1.6 0.0548 0.05370 0.05262 0.05155 0.05050 0.04947 0.04846 0.04746 0.04648 0.04551 0.06681 0.06552 0.06426 0.06301 0.06178 0.06057 0.05938 0.05821 0.05705 0.05592 the table is 1.4 0.08076 0.07927 0.07780 0.07636 0.07493 0.07353 0.07215 0.07078 0.06944 0.06811 1.3 0.09680 0.09510 0.09342 0.09176 0.09012 0.08851 0.08691 0.08534 0.08379 0.08226 negative since -1.2 0.11507 0.11314 0.11123 0.10935 0.10749 0.10565 0.10383 0.10204 0.10027 0.09853 -1.1 0.13567 .13350 0.13136 0.12924 0.12714 0.1250 0.12302 0.12100 0.11900 0.11702 -1.0 0.15866 0.15625 0.15386 0.15151 0.14917 0.14686 0.14457 0.14231 0.14007 0.13786 the value is negative. -0.9 0.18406 0.18141 0.17879 0.17619 0.17361 0.17106 0.16853 0.16602 0.16354 0.16109 -0.8 0.2118 0.20897 0.20611 0.20327 0.20045 .1976 0.19489 0.19215 0.18943 0.18673 0.7 0.24196 0.23885 0.23576 0.23270 0.22965 0.22663 0.22363 0.22065 0.21770 0.21476 0.6 0.2742 0.27093 0.26763 0.26435 0.26109 0.25785 0.25463 0.25143 0.24825 0.24510 0.5 0.30854 0.30503 0.30153 0.29806 0.29460 0.29116 0.28774 0.28434 0.28096 0.27760 04 0 34458 0 34090 0.33724 0.33360 0.32636 0.32276 0.31918 0.31561 0.3120 -0.3 0.38209 0.37828 0.37448 0.37070 0.36693 0.36317 0.35942 0.35569 0.35197 0.34827 -U.2 0.42074 0.41683 0.41294 0.40905 0.40517 0.40129 0.39743 0.39358 0.38974 0.38591 0.1 0.46017 0.45620 0.45224 0.44828 0.44433 0.44038 0.43644 0.43251 0.42858 0.42465 0.0 0.50000 0.49601 0.49202 0.48803 0.48405 0.48006 0.47608 0.47210 0.46812 0.46414 P (x > 120) = P(z > - 0.31) = 0.37828 or 37.83%Using the z table, let us look for the areas of normal curve of z = 0.31 Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.50000 0.50798 0.51197 0.51595 0.51994 0.52392 0.52790 0.53188 0.53586 To use the table, 0.1 0.53983 54380 0.54776 0.55172 0.55567 0.55962 0.56356 0.56749 0.57142 0.57535 0 57026 0 58317 0.58706 0.59095 0.59483 0.59871 0.60257 0.60642 0.61026 0.61409 locate the first 0.3 0.61791 0.62172 0.62552 0.62930 0.63307 0.63683 0.64058 0.64431 0.64803 0.65173 0.4 0 65342 0.03970 0.66276 0.66640 0.67003 0.67364 0.67724 0.68082 0.68439 0.68793 two digit vertically 0.5 0.69146 0.69497 0.69847 0.70194 0.70540 0.70884 0.71226 0.71566 0.71904 0.72240 0.6 0.72575 0.72907 0.73237 0.73565 0.73891 0.74215 0.74537 0.74857 0.75175 0.75490 then the ).75804 0.76115 0.76424 0.76730 0.77035 0.77337 0.77637 0.77935 0.78230 0.78524 0.78814 0.79103 0.79389 0.79673 0.79955 0.80234 0.80511 0.80785 0.81057 0.81327 Last digit horizontally. 0.9 0.81594 0.81859 0.82121 0.82381 0.82639 0.82894 0.83147 0.83398 0.83646 0.83891 1.0 0.84134 0.84375 0.84614 0.84849 0.85083 0.85314 0.85543 0.85769 0.85993 0.86214 Then locate the 1.1 0.86433 0.86650 0.86864 0.87076 0.87286 0.87493 0.87698 0.87900 0.88100 0.88298 1.2 0.88493 0.88686 0.88877 0.89065 0.89251 0.89435 0.89617 0.89796 0.89973 0.90147 Intersection. 1.3 0.90320 0.90490 0.90658 0.90824 0.90988 0.91149 0.91309 0.91466 0.91621 0.91774 0.91924 0.92073 0.92220 0.92364 0.92507 0.92647 0.92785 0.92922 0.93056 0.93189 1.5 0.93319 0.93448 0.93574 0.93699 0.93822 0.93943 0.94062 0.94179 0.94295 0.94408 1.6 0.94520 0.94630 0.94738 0.94845 0.94950 0.95053 0.95154 0.95254 0.95352 0.95449 Example: 0 . 3 1 1.7 0.95543 0.95637 0.95728 0.95818 0.95907 0.95994 0.96080 0.96164 0.96246 0.96327 1.8 0.96407 0.96485 0.96562 0.96638 0.96712 0.96784 0.96856 0.96926 0.96995 0.97062 1.9 0.97128 0.97193 0.97257 0.97320 0.97381 0.97441 0.97500 0.97558 0.97615 0.97070 2.0 0.9772 0.97778 0.97831 0.97882 0.97932 0.97982 0.98030 0.98077 0.98124 0.98169 0.98214 0.98257 0.98300 0.98341 0.98382 0.98422 0.98461 0.98500 0.98537 0.98574 2.2 0.98610 0.98645 0.98679 0.98713 0.98745 0.98778 0.98809 0.98 840 0.98870 0.98899 P (x 123) To find the probability solve using, Z = Z =_ 123-122 50 60 -1 Z= 6.45497 Z = -0.15P (E > 123): P(z > 0.15) Using the 2 table above 2 0.44038 or 44.04% Therefore, the probability that X assumes a value greater than 123 is 44.04%. 4. Find P (X > 123) The distribution of X is unknown, so it cannot be solved. Let Us Practice . Let us check your understanding about sampling distribution of the sample means and central limit theorem. Complete the following table and solve for the unknowns. The birth weight of baby boys who were delivered at 9 months is normally distributed with a mean of 7348.20 g with a standard deviation of 817.99 g. Suppose there were nine boy babies born on a given day and the mean birth weight is calculated. Compute the following: a. the mean and the standard deviation of the sample mean b. Find the probability that the mean weight of the nine baby boys born was less than 7160 g. c. Find the probability that the mean weight of the nine babies born was greater than 71 13.1 g. Complete the table by supplying the missing values inside the box. a. Mean: #5 = = Cl Standard deviation 0X 2 81.3599 z|:| b. Use the formula z = Xi" to solve P (E 7113.1) 5 z : 2:," T; _ 7113.1 7348.20 :| _ 235.1 272.66 Using the 2 table (refer to the table on the example), look for the areas of normal curve of z = 0.86 P (Y > 7113.1) 2 P(z > 0.86) = |:| Therefore, the probability that X assumes a value greater than |:| is |:| G Let Us Assess Let us check if you learned from the discussion. Read the questions carefully and write your answer on the clean sheet of paper. 1. A population consist of the following values {3, 5, 7}. What would be the possible samples size of 2 which can be drawn with replacement? A. {(33), (5, 7)} 13- {(3,5),(3,7),(5,7)} C. {(3,3),(3,5),(3,7) ,(5,3),(5,5),(5,7)} D. {(3,3),(3,5),(3,7),(5,3),(5,5),(5,7),(7,2),(7,5),(7,7)} For questions 23, refer to the situation below. The average yearly Medicare Hospital Insurance benefit per person is Php 5000. If the benets are normally distributed with a standard deviation of php 540, 2. Find the probability that the mean benet for a random sample of 17 patients is Less than php 4800. a. 6.3% b. 60.30% c. 30.60% d. 3.6% . Find the probability that the mean benet for a random sample of 17 patients is More than php 5100. a. 77.64% b. 76.47% c. 67.47% d. 64.77% 4. Which of the following statements best describe the population mean if the mean of the sampling distribution of the mean is 9.12? a. The population mean is greater than 9.12. b. The population mean is less than 9.12. C. The population mean is equal to 9.12. d. The population mean and mean of the sampling distribution of the means cannot be compared. 5. A study found out that the average of cell phones lifetime is 24.3 months with a standard deviation is 2.6 months. If a company provides its 43 employees with a cell phone, assume cell phone life is normally distributed, find the probability that the mean lifetime of these phones will be less than 23.8 months. a. 15.07% b. 10.57% c. 57.10% (:1. 7.15%