ssrs 2008 r2 barcode font Setting this equal to 1 > 2, we find that 2a4 from which a2 36 2(36)2 2(2) 4(2)(81) 36 36a2 81 in Software

Encoding QR Code JIS X 0510 in Software Setting this equal to 1 > 2, we find that 2a4 from which a2 36 2(36)2 2(2) 4(2)(81) 36 36a2 81

Setting this equal to 1 > 2, we find that 2a4 from which a2 36 2(36)2 2(2) 4(2)(81) 36 36a2 81
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Therefore, the required median, which must lie between 0 and 3, is given by a2 from which a (c) 1.62 approx. E(X ) 4 3 2 x (9 81 30 x2) dx 9 9 22 2 4 3x3 81 x5 2 3 5 0
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which is practically equal to the median. The mode, median, and mean are shown in Fig. 3-6.
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Fig. 3-6
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3.34. A discrete random variable has probability function f(x) (b) the median, and (c) compare them with the mean.
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1 > 2x where x
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1, 2, . . . . Find (a) the mode,
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1, for which the
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(a) The mode is the value x having largest associated probability. In this case it is x probability is 1 > 2.
(b) If x is any value between 1 and 2, P(X x) 1 and P(X x) 1. Therefore, any number between 1 and 2 2 2 could represent the median. For convenience, we choose the midpoint of the interval, i.e., 3 > 2. (c) As found in Problem 3.3, m Problem 3.33. 2. Therefore, the ordering of the three measures is just the reverse of that in
CHAPTER 3 Mathematical Expectation
Percentiles 3.35. Determine the (a) 10th, (b) 25th, (c) 75th percentile values for the distribution of Problem 3.33.
From Problem 3.33(b) we have P(X a) 4 9a2 81 2 a4 4
18a2 a4 81 a4) > 81 0.10.
(a) The 10th percentile is the value of a for which P(X a) 0.10, i.e., the solution of (18a2 Using the method of Problem 3.33, we find a 0.68 approx. (b) The 25th percentile is the value of a such that (18a2 (c) The 75th percentile is the value of a such that (18a2 a4) > 81 a4) > 81 0.25, and we find a 0.75, and we find a
1.098 approx. 2.121 approx.
Other measures of dispersion 3.36. Determine, (a) the semi-interquartile range, (b) the mean deviation for the distribution of Problem 3.33.
(a) By Problem 3.35 the 25th and 75th percentile values are 1.098 and 2.121, respectively. Therefore, Semi-interquartile range (b) From Problem 3.33 the mean is m Mean deviation 2 30 x
3 8>5
2.121 2 8>5. Then
0.51 approx.
M.D.5E(uX
0.555 approx.
8 2 4x B (9 5 81 x B
4x (9 81
x2) R dx
mu f (x) dx
x2) R dx
38>5 x
8 4x B (9 5 81
x2) R dx
Skewness and kurtosis 3.37. Find the coefficient of (a) skewness, (b) kurtosis for the distribution of Problem 3.19.
From Problem 3.19(b) we have s2 (a) Coefficient of skewness (b) Coefficient of kurtosis a3 a4 m3 s3 m4 s4 11 25 m3 0.1253 2.172 32 875 m4 3693 8750
It follows that there is a moderate skewness to the left, as is indicated in Fig. 3-6. Also the distribution is somewhat less peaked than the normal distribution, which has a kurtosis of 3.
Miscellaneous problems 3.38. If M(t) is the moment generating function for a random variable X, prove that the mean is m the variance is s2 M s (0) [Mr(0)]2.
From (32), page 79, we have on letting r 1 and r mr 1 Then from (27) m Mr(0) m2 s2 Ms(0) [Mr(0)]2 2, mr 2 Ms(0)
Mr(0) and
Mr(0)
CHAPTER 3 Mathematical Expectation
3.39. Let X be a random variable that takes on the values xk k with probabilities pk where k (a) Find the characteristic function f(v) of X, (b) obtain pk in terms of f(v).
(a) The characteristic function is
ikv a pke
1, . . . ,
f(v)
E(eivX)
ivx a e k pk
(b) Multiply both sides of the expression in (a) by e
and integrate with respect to v from 0 to 2p. Then
ijvf(v) dv
3v 0e
since
i(k 3v 0 e
j)v dv
Therefore,
1 2p e 2p 3v 0
ei(k j)v 2p 2 i(k j) 0
a pk 3
ei(k
j)v dv
2ppj
k2j k j
ijvf(v) dv
or, replacing j by k, pk 1 2p e 2p 3v 0
ikvf(v) dv
We often call g n n pkeikv (where n can theoretically be infinite) the Fourier series of f(v) and pk the k Fourier coefficients. For a continuous random variable, the Fourier series is replaced by the Fourier integral (see page 81).
3.40. Use Problem 3.39 to obtain the probability distribution of a random variable X whose characteristic function is f(v) cos v.
From Problem 3.39 pk 1 2p e 2p 3v 0 1 2p e 2p 3v 0
ikv cos v dv iv ikv B e
1 2p i(1 e 4p 3v 0 If k 1, we find p1 1; if k 2 random variable is given by
k)v dv
1 2p e 4p 3v 0
1 2.
R dv
i(1 k)v dv
1, we find p e
For all other values of k, we have pk
0. Therefore, the
X As a check, see Problem 3.20.
probability 1>2 probability 1>2
3.41. Find the coefficient of (a) skewness, (b) kurtosis of the distribution defined by the normal curve, having density f (x) 1 e 22p
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