- 84 Chap 4 Measures Of Dispersion And Skewness 4 Find The Standard Deviation And The Variance Using The 2 Methods With 1 (101.27 KiB) Viewed 32 times
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84 Chap 4 Measures of Dispersion and Skewness 4. Find the standard deviation and the variance (using the 2 methods) with the following grade distribution of 40 students. 5. Find the standard deviation and the variance (using the 3 methods) of the following weight distribution of 25 male students. Grades 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 5.00 6. A hospital administrator compared the age with the emergency admissions on given Monday with those Age Friday of the same week. 1-10 11-20 Calculate the standard deviation for each set of data and compare the age dispersion in the Monday and Friday admissions. 21-30 No. of Students 1 2 31-40 41-50 51-60 61-70 Total Monday (f) 7 9 10 9 5 3 3 Weight (kg) No. of Students 52-54 55-57 58-60 61-63 64-66 67-69 70-72 6 8 54 5 6 9 10 1 N3TE532 4 6 Friday (f) 7 8 10 14 12 6 3 60 4.3 tw va st a h
5.125 0.25 .875 0625 375 5875 3.5 On Using the second version of the formula for the standard deviation, we have: $2 nΣ fx² - (Σ fx)² n(n-1) Scores 5-7 8-10 11-13 14-16 17-19 20-22 23-25 Shortcut Formulas of Standard Deviation and Variance nΣ fd² - (fd)² n (n − 1) 2 4 14 9 6 3 2 - Σ , = 40 n = s² = c². S=C d -2 -1 29340 1560 = 18.80769 S = √18.80769 = 4.34 where c, the class size and d, the values of deviations, i.e. d = 0, ±1, ± 2, ... Consider the data of the preceding problem and compute the standard deviation using the shortcut formula. 0 1 2 3 4 40(8856) - (570)² 40(39) = nΣ fd² - (fa)² n (n − 1) - = fd -4 -4 0 9 12 9 8 Σ fd = 30 for variance for standard deviation 40(104) - (30)² 40(39) s'= √18.8076 4.34 d 4 1 0 1 4 9 16 Using the third version of the formula of the standard deviation, we have: nΣ fd² - (fd)² c² n(n-1) (3) ² = fd² 0 9 24 27 32 Σ fd² = 104 18.8076 8 4
M Age 11-20 21-30 31-40 41-50 51-60 61-70 71-80 ing the shortcut formula. f d 5 7 12 22 8 4 2 -2 -1 0 1 2 3 n=Σf=60 Using the formulas for s, we have: the standard deviation of the age distribution given in S² = Σfd=-19 fd S = -15 -14 -12 = 0 8 8 6 = 2.05 (10)² 60(127) - (-19)² 60(59) nΣ fd²-(fd)² n (n − 1) √2.05 (10) = 1.43 (10) 14.32 ď² 9 4 1 0 1 4 9 c² - 10² fa 45 28 12 0 8 16 18 Efdf = 127