ssrs 2008 r2 barcode font Special Probability Distributions in Software

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CHAPTER 4 Special Probability Distributions
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4.16. Find the moment generating function for the general normal distribution.
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We have M(t) Letting (x )> E(etX )
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` 1 tx 3 `e e s !2p (x m)2>2s2 dx
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v in the integral so that x M(t) 1 22p
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` ut 3 `e svt (v2>2)
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v, dx dv emt 22p
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dv, we have
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(s2t2/2) `
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(v st)2>2 dv
Now letting v
w, we find that M(t) emt
(s2t2>2) a
1 22p
3 `e
w2>2 dwb
(s2t2>2)
Normal approximation to binomial distribution 4.17. Find the probability of getting between 3 and 6 heads inclusive in 10 tosses of a fair coin by using (a) the binomial distribution, (b) the normal approximation to the binomial distribution.
(a) Let X be the random variable giving the number of heads that will turn up in 10 tosses (Fig. 4-15). Then P(X 3) a a 10 1 3 1 7 ba b a b 2 2 3 10 1 5 1 5 ba b a b 2 2 5 15 128 63 256 P(X 4) a a 10 1 4 1 6 ba b a b 2 2 4 10 1 6 1 4 ba b a b 2 2 6 105 512 105 512
Then the required probability is P(3 X 6) 15 128 105 512 63 256 105 512 99 128 0.7734
Fig. 4-15
Fig. 4-16
(b) The probability distribution for the number of heads that will turn up in 10 tosses of the coin is shown graphically in Figures 4-15 and 4-16, where Fig. 4-16 treats the data as if they were continuous. The required probability is the sum of the areas of the shaded rectangles in Fig. 4-16 and can be approximated by the area under the corresponding normal curve, shown dashed. Treating the data as continuous, it follows that 3 to 6 heads can be considered as 2.5 to 6.5 heads. Also, the mean and variance for the binomial distribution are given by m np 10 A 1 B 5 and s # A1 B A1 B (10) 2 2 1.58. !npq 2 Now 2.5 in standard units 6.5 in standard units 2.5 5 1.58 6.5 5 1.58 1.58 0.95
CHAPTER 4 Special Probability Distributions
Fig. 4-17
Required probability (Fig. 4-17)
(area between z (area between z (area between z 0.4429 0.3289
1.58 and z 1.58 and z 0 and z 0.7718
0.95) 0) 0.95)
which compares very well with the true value 0.7734 obtained in part (a). The accuracy is even better for larger values of n.
4.18. A fair coin is tossed 500 times. Find the probability that the number of heads will not differ from 250 by (a) more than 10, (b) more than 30.
m np 1 (500)a b 2 250 s !npq A 1 1 (500)Q R Q R 2 2 11.18
(a) We require the probability that the number of heads will lie between 240 and 260, or considering the data as continuous, between 239.5 and 260.5. 239.5 in standard units Required probability 239.5 250 11.18 0.94 260.5 in standard units 0.94 and z 2(0.3264) 0.94 0.94) 0.6528
(area under normal curve between z (twice area between z 0 and z 0.94)
(b) We require the probability that the number of heads will lie between 220 and 280 or, considering the data as continuous, between 219.5 and 280.5. 219.5 in standard units Required probability 219.5 250 11.18 2(0.4968) 0.9936 2.73 280.5 in standard units 0 and z 2.73 2.73)
(twice area under normal curve between z
It follows that we can be very confident that the number of heads will not differ from that expected (250) by more than 30. Therefore, if it turned out that the actual number of heads was 280, we would strongly believe that the coin was not fair, i.e., it was loaded.
4.19. A die is tossed 120 times. Find the probability that the face 4 will turn up (a) 18 times or less, (b) 14 times or less, assuming the die is fair.
The face 4 has probability p
of turning up and probability q
of not turning up.
(a) We want the probability of the number of 4s being between 0 and 18. This is given exactly by a 120 1 18 5 102 ba b a b 2 6 18 a 120 1 17 5 103 ba b a b 6 6 17 c a 120 1 0 5 120 ba b a b 6 6 0
but since the labor involved in the computation is overwhelming, we use the normal approximation. Considering the data as continuous, it follows that 0 to 18 4s can be treated as 0.5 to 18.5 4s. Also, m np 1 120a b 6 20 and s !npq A (120) a 1 b a 5 b 6 6 4.08
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