082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

968 Chapter 29 Statistics and probability distributions
Table29.7
Cumulativenormal probabilities (continued).
z A(z) z A(z) z A(z) z A(z) z A(z) z A(z)
2.40 0.9918025 2.45 0.9928572 2.50 0.9937903 2.55 0.9946139 2.60 0.9953383 3.20 0.9993129
2.41 0.9920237 2.46 0.9930531 2.51 0.9939634 2.56 0.9947664 2.70 0.9965330 3.40 0.9996631
2.42 0.9922397 2.47 0.9932443 2.52 0.9941323 2.57 0.9949151 2.80 0.9974449 3.60 0.9998409
2.43 0.9924506 2.48 0.9934309 2.53 0.9942966 2.58 0.9950600 2.90 0.9981342 3.80 0.9999277
2.44 0.9926564 2.49 0.9936128 2.54 0.9944574 2.59 0.9952012 3.00 0.9986501 4.00 0.9999683
4.50 0.9999966
5.00 0.9999997
5.50 0.9999999
This table is condensed from Table 1 of theBiometrikaTablesforStatisticians, Vol. 1 (1st ed.), edited by E. S. Pearson
andH. O. Hartley. Reproduced with the kind permissionofE. S.Pearsonand the trusteesofBiometrika.
Source:StatisticsVol.1ProbabilityInferenceandDecisionbyHays,W.L.andWinkler,R.L.(HoltRinehart&Winston,
New York, 1970).
Example29.22 A random variable,h, has a normal distribution with mean 7 and standard deviation 2.
Calculate the probability that
(a)h>9 (b)h<6 (c)5<h<8
Solution (a) Applying the transformation gives
9 → 9 −7 = 1
2
Soh > 9 has the same probability asx > 1, wherexis a random variable with a
standard normal distribution:
P(h >9) =P(x >1) =1−P(x <1) =1−0.8413 =0.1587
(b) Applying the transformation gives
6 → 6 −7 = −0.5
2
Soh < 6 has the same probability asx < −0.5:
P(x < −0.5) =P(x > 0.5) = 1 −P(x < 0.5) = 0.3085
(c) Applying the transformation to5and 8 gives
5 → 5 −7 =−1 8→ 8 −7 = 0.5
2 2
and so werequireP(−1 <x<0.5). Therefore
and then
P(x < 0.5) = 0.6915 P(x < −1) = 0.1587
P(−1 <x<0.5) = 0.6915 −0.1587 = 0.5328