Question

1. Statistical Parameters: it all the time. The data set presented in Table 1 represents a sample of stream flows in a perennial stream for a 45-day period, These stream flows were measured using a flow meter that has a precision of t0.1 Cfs. Find the following statistical parameters for the sample data given in Table 1. A perennial stream is a stream that has water flowing in a. Central Tendency i. Mean the average value of the random variable. ii. Median the value of the random variable that is in the middle of the sample set. Mode frequently iii. the value of the random variable that occurs most b. Dispersion i. Variance squared an average measure of the deviation about the mean ii. Standard deviation an average measure of the deviation about the mean. c. Asymmetry i. Skew a measure of the symmetry or asymmetry of the sample data about the mean. Graphical Representation of Data: Given the data presented in Table 1, develop the following 2. Histogram- use an interval of 2 Cfs Box Plot- are there any outliers? a. b. Discrete vs, Continuous Random Variable: You will notice that there is most likely no mode for the sample data set presented in Table 1. The reason there are not multiple instances of exactly the same value of the random variable is due to the continuous nature of the random variable. Continuous random variables are random variables whose value can take on infinitely many values- the value of which is typically governed by the precision of the device used to measure the random variable. A discrete random variable, on the other hand, is a random variable that can only take on a finite value. The number of people in a car is an example of a discrete random variable. 3. Find the mean, median and mode of the stream flows if we use a flow meter whose precision is 5 Cfs. For example, for the stream flow for March 1, 2018 would be 855 Cfs and the stream flow for March 22, 2018 would be 1375 Cfs. a.
Table 1 Date 31 31-Mar-18 3-Mar-18 4-Mar-18 5-Mar-18 6-Mar-18 1-Apr-18 2-Apr-18 18 4-Apr-18 18 18 20 20-Mar-18 18 6 30 30-Mar-18
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Answer #1

1. a, b and c. And 3. a.

1 Flow(Cfs)(0.1 precision) 856.5 1169.8 826 922 547.2 1318 519.5 Flow(Cfs)(5 precision) 855 1170 825 920 545 1320 520 720 1120 790 1490 1030 965 1110 815 1250 920 780 4 6 10 12 13 14 15 16 17 18 19 20 21 1119.7 791.1 1492.4 1028.3 965.7 1112.3 814.7 1249.6 922.1 1332.1 501.1 1330 500 Sheet1 Sheet2 Sheet3

34 35 36 37 38 39 40 41 42 43 1098.2 1347 1473.6 516 1096 501.6 797.4 743.3 719.3 650.8 1154.8 695.8 1277.2 1100 1345 1475 515 1095 500 795 745 720 650 1155 695 1275 45 46 47 48 49 50 43870 Surm 974.3388889 Mean 4385 974 5555556 920 1345 93930.4798 306.4807984 0,047317808 922.1 Median #N/A Mode 93781.31419 Variance 06.2373494 Std.deviation 0.045609136 Skew

2.

a. Histogram:

GH Bin Frequency Histogranm 400 500 600 700 800 900 1000 0 5 500-600 6 600-700 7 700-800 8 800-900 9 900-1000 10 1000-1100 1100 11 1100-1200 12006 12 1200-13001300 13 1300-1400 1400 14 1400-1500 1500 15 16 Frequency 8 Bin More

b. Box plot:

Lowest =501.1

Q1 =731.3

Median =Q2 =922.1

Q3 =1263.4

Highest =1497.9

49.3 1203.+ 331.3

Outliers are the values that are greater than Q3+1.5(Q3 - Q1) =2061.55 (there are no values greater than this as the highest is 1497.9) and the values that are less than Q1 - 1.5(Q3 - Q1) =532.1 (there are outliers as the minimum is 501.1).

Hence, all the values that are less than 532.1 and upto 501.1(including) are the outliers.

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