Question

The accompanying table contains the overall download and upload speeds in mbps for nine carriers in a particular country. Click the icon to view the data. For the download speed and the upload speed separately: a. Compute the mean and median. b. Compute the variance, standard deviation, range, and coefficient of variation. c. Based on the mean and median, are the data skewed? If so, how? d. Based on the results of (a) through (C), what conclusions can you reach concerning the download and upload speed of various carriers? a. Compute the mean and median. The mean download speed is (Type an integer or decimal rounded to two decimal places as needed.) The mean upload speed is (Type an integer or decimal rounded to two decimal places as needed.) The median download speed is. (Type an integer or decimal rounded to two decimal places as needed.) The median upload speed is (Type an integer or decimal rounded to two decimal places as needed.) b. Compute the variance, standard deviation, range, and coefficient of variation. The variance for the download speeds is

Answer 1

For Download speed:

Solution: Х x2 22.7 515.29 24.8 615.04 20.5 420.25 16.8 282.24 10.9 118.81 11.2 125.44 10.4 108.16 6.8 46.24 4.3 18.49 { x = 128.4 {x2 = 2249.96 2x Meană = n 22.7 + 24.8 + 20.5 + 16.8 + 10.9 + 11.2 + 10.4 + 6.8 + 4.3 9 128.4 9 = 14.2667 Median : Observations in the ascending order are : 4.3, 6.8, 10.4, 10.9, 11.2, 16.8, 20.5, 22.7, 24.8 Here, n = 9 is odd. M = value of (472)" n + 1 )th observation 2

= value of (**) 9 +1 th observation 2 = value of 5th observation = 11.2 Mode : In the given data, no observation occurs more than once. Hence the mode of the observations does not exist, means mode=0. {x2 (Ex)2 n Population Variance o2 II n (128.4) 2 2249.96 - 9 9 2249.96 - 1831.84 9 418.12 9 46.4578

(2x)2 Ex2 n Population Standard deviation o = n (128.4) 2 2249.96 - 9 9 12249.96 - 1831.84 9 1418.12 = 9 = 746.4578 = 6.816 o Co-efficient of Variation (Population) = · 100 % 6.816 = 100 % 14.2667 = 47.78 %

For Upload Speed:

Solution: Х x2 13.4 179.56 14.4 207.36 8.7 75.69 12 144 6.4 40.96 5.6 31.36 6.4 40.96 3.2 10.24 4.2 17.64 { x = 74.3 x2 = 747.77 ΣΧ Mean a = n 13.4 + 14.4 + 8.7 + 12 + 6.4 + 5.6 + 6.4 + 3.2 + 4.2 Il 9 74.3 = 9 = 8.2556 Median : Observations in the ascending order are : 3.2, 4.2, 5.6, 6.4, 6.4, 8.7, 12, 13.4, 14.4 Here, n = 9 is odd. M value of (6) n + 1 )th observation 2 9 +1 th = value of E)" observation

= value of () 9 +1th observation 2 = value of 5th observation = = 6.4 Mode : In the given data, the observation 6.4 occurs maximum numbe of times (2) :: Z = 6.4 (Ex)2 Σχ2. n Population Variance o2 = n (74.3)2 747.77 - 9 = 9 747.77 - 613.3878 = 9 134.3822 = 9 = 14.9314

(2x)2 { x2 n Population Standard deviation o = n (74.3) 2 1747.77 - 9 II 9 1747.77 - 613.3878 = 9 1134.3822 = 9 II 14.9314 = 3.8641 Co-efficient of Variation (Population) = 100 % X 3.8641 = 100 % 8.2556 = 46.81 %

The download speed is left skewed

The upload speed is symmetrical since median= mode

I have given my best to solve your problem. Please like the answer if you are satisfied with it. ?