# ssrs export to pdf barcode font Derive the coding formula (27), page 162, for the arithmetic mean. in Software Creating QR in Software Derive the coding formula (27), page 162, for the arithmetic mean.

5.34. Derive the coding formula (27), page 162, for the arithmetic mean.
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Let the jth class mark be xj. Then the deviation of xj, from some specified class mark a, which is xj a, will be equal to the class interval size c multiplied by some integer uj, i.e., xj a cuj or xj a cuj (also written briefly as x a cu). The mean is then given by x # a fj xj n a fj(a n a since n g fj. c cuj) a a fj n a cu c a fjuj n
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5.35. Use the coding formula of Problem 5.34 to find the mean height of the 100 male students at XYZ University (see Problem 5.33).
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The work may be arranged as in Table 5-9. The method is called the coding method and should be employed whenever possible. x # a a fu n c x 61 64 a S 67 70 73 67 a 15 b(3) 100 67.45 inches
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Table 5-9
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f 5 18 42 27 8 fu 10 18 0 27 16
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CHAPTER 5 Sampling Theory
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5.36. Find (a) the variance, (b) the standard deviation for the numbers in Problem 5.31.
(a) Method 1 As in Problem 5.31, we have x # s2 a (x n x)2 # (5
4.8. Then 4.8)2 (3 4.8)2 (6 4.8)2 20 (5 4.8)2 c (4 4.8)2
59.20 20 Method 2 s2 a f (x n x)2 # 6(5 59.20 20 (b) From (a), s2 2.96 and s
4.8)2 2.96
4.8)2
4.8)2 20
4.8)2
4.8)2
!2.96
5.37. Find the standard deviation of the weights of students in Problem 5.32.
s2 a f (x n x)2 # 15(162 4130 63 Then s algebra. !65.6 (lb)2 150)2 20(148 15 150)2 20 10(153 150)2 10 18 18(140 150)2
65.6 in units pounds square or (pounds)2 8.10 lb, where we have used the fact that units follow the usual laws of
!65.6 lb
5.38. Find the standard deviation of the heights of the 100 male students at XYZ University. See Problem 5.33.
# From Problem 5.33, X s 67.45 inches. The work can be arranged as in Table 5-10. a f (x n x)2 # 852.7500 100 28.5275 2.92 inches
Table 5-10
Height (inches) 60 62 63 65 66 68 69 71 72 74 Class Mark (x) 61 64 67 70 73 x x x # 67.45 6.45 3.45 0.45 2.55 5.55 (x x)2 # Frequency ( f ) 5 18 42 27 8 n gf 100 f (x x)2 #
41.6025 11.9025 0.2025 6.5025 30.8025
208.0125 214.2450 8.5050 175.5675 246.4200 g f (x x)2 # 852.7500
5.39. Derive the coding formula (28), page 162, for the variance.
As in Problem 5.34, we have xj a x # cuj and a c a fjuj n a cu #
CHAPTER 5 Sampling Theory
Then s2 1 n a fj(xj x)2 # 1 n a fj(cuj c2 n a fj(uj c2 2 n a fj(uj c2 2 n a fjuj c2 c2 a fju2 j n a fju2 j n a fu n
cu)2 # u)2 # 2uju # u2) # c2 #2 n a fju
2uc2 # n a fjuj 2u2c2 # c2 c2u2 #
c2 B
c2[u2 #
u2] #
a fu n R
a fjuj 2 n
5.40. Use the coding formula of Problem 5.39 to find the standard deviation of heights in Problem 5.33.
The work may be arranged as in Table 5-11. This enables us to find x as in Problem 5.35. From the last column # we then have s2 c2 B a fu n 97 100
(3)2 c and so s 2.92 inches.
15 2 b d 100
a fu n R
# c2( u2
u2) #
Table 5-11
x 61 64 a S 67 70 73 u 2 1 0 1 2 n f 5 18 42 27 8 gf 100 g fu fu 10 18 0 27 8 15 fu2 20 18 0 27 32 g fu 2 97
5.41. Find the first four moments about the mean for the height distribution of Problem 5.33.
Continuing the method of Problem 5.40, we obtain Table 5-12. Then, using the notation of page 162, we have Mr 1 Mr 2 a fu n a fu n
0.15 0.97
Mr 3 Mr 4
a fu n a fu n
CHAPTER 5 Sampling Theory
Table 5-12
x 61 64 67 70 73 u 2 1 0 1 2 n f 5 18 42 27 8 gf 100 fu 10 18 0 27 16 g fu 15 fu2 20 18 0 27 32 g fu 2 97 fu3 40 18 0 27 64 g fu 3 33 g fu 4 fu4 80 18 0 27 128 253