ssrs 2008 r2 barcode font Special Probability Distributions in Software

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CHAPTER 4 Special Probability Distributions
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Required proportion
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area between z (area between z (area between z 0.3413 0.0675
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1 and z 1 and z
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0.17 and z 0.2738 27%
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Fig. 4-20
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Fig. 4-21
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(b) Let x1 be the required maximum grade and z1 its equivalent in standard units. From Fig. 4-21 the area to the left of z1 is 10% 0.10; hence, Area between z1 and 0 and z1 1.28 (very closely). Then z1 (x1 6.7)> 1.2 1.28 and x1 0.40
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5.2 or 5 to the nearest integer.
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(c) Let x2 be the required minimum grade and z2 the same grade in standard units. From (b), by symmetry, z2 1.28. Then (x2 6.7)> 1.2 1.28, and x2 8.2 or 8 to the nearest integer.
4.59. A Geiger counter is used to count the arrivals of radioactive particles. Find the probability that in time t no particles will be counted.
Let Fig. 4-22 represent the time axis with O as the origin. The probability that a particle is counted in a small time t is proportional to t and so can be written as t. Therefore, the probability of no count in time t is 1 t. More precisely, there will be additional terms involving ( t)2 and higher orders, but these are negligible if t is small.
Fig. 4-22
Let P0(t) be the probability of no count in time t. Then P0(t t) is the probability of no count in time t t. If the arrivals of the particles are assumed to be independent events, the probability of no count in time t t is the product of the probability of no count in time t and the probability of no count in time t. Therefore, neglecting terms involving ( t)2 and higher, we have (1) From (1) we obtain (2) i.e., dP0 dt Solving (3) by integration we obtain (3) ln P0 lP0 or dP0 P0 or P0(t) l dt lim P0(t
t S0
P0(t
t) t) t
P0(t)[l P0(t)
lP0(t)
To determine c, note that if t 0, P0(0) c is the probability of no counts in time zero, which is of course 1. Thus c 1 and the required probability is (4) P0(t) e
CHAPTER 4 Special Probability Distributions
4.60. Referring to Problem 4.59, find the probability of exactly one count in time t.
Let P1(t) be the probability of 1 count in time t, so that P1(t t) is the probability of 1 count in time t Now we will have 1 count in time t t in the following two mutually exclusive cases: (i) 1 count in time t and 0 counts in time t (ii) 0 counts in time t and 1 count in time t The probability of (i) is P1(t)(1 t). The probability of (ii) is P0(t) t. Thus, apart from terms involving ( t)2 and higher, (1) This can be written (2) Taking the limit as (3) or (4) Multiplying by e t, this can be written (5) which yields on integrating (6) P1(t) te
P1(t P1(t
P1(t)(1 P1(t)
P0(t)
t) t dP1 dt dP1 dt
lP0(t)
lP1(t)
t S 0 and using the expression for P0(t) obtained in Problem 4.59, this becomes le
d lt (e P1) dt
If t 0, P1 (0) is the probability of 1 count in time 0, which is zero. Using this in (6), we find c2 Therefore, (7) P1(t) te
By continuing in this manner, we can show that the probability of exactly n counts in time t is given by (lt)n e lt (8) Pn(t) n! which is the Poisson distribution.
SUPPLEMENTARY PROBLEMS
The binomial distribution
4.61. Find the probability that in tossing a fair coin 6 times, there will appear (a) 0, (b) 1, (c) 2, (d) 3, (e) 4, (f) 5, (g) 6 heads. 4.62. Find the probability of (a) 2 or more heads, (b) fewer than 4 heads, in a single toss of 6 fair coins. 4.63. If X denotes the number of heads in a single toss of 4 fair coins, find (a) P(X (c) P(X 2), (d) P(1 X 3). 3), (b) P(X 2),
4.64. Out of 800 families with 5 children each, how many would you expect to have (a) 3 boys, (b) 5 girls, (c) either 2 or 3 boys Assume equal probabilities for boys and girls. 4.65. Find the probability of getting a total of 11 (a) once, (b) twice, in two tosses of a pair of fair dice.
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