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

966 Chapter 29 Statistics and probability distributions
Example29.21 Thecontinuousrandomvariable vhasastandardnormaldistribution.Calculatetheprobability
that
(a)0<v<1 (b) −1<v<1 (c) −0.5v2
Solution (a) Figure 29.17 shows the area (probability) required.
P(v < 1) = 0.8413 using Table 29.7
P(v < 0) = 0.5
using symmetry
P(0 < v < 1) = 0.8413 −0.5 = 0.3413
(b) Figure 29.18 shows the area (probability) required.
P(−1 < v <1) = 2 ×P(0 < v <1)
= 2 ×0.3413 = 0.6826
usingsymmetry
This tells us that 68.3% of the values of v are within one standard deviation of the
mean.
(c) Figure 29.19 shows the area (probability) required.
P(v 2) = 0.9772
P(v −0.5) =P(v >0.5) =1−P(v <0.5)
= 1 −0.6915 = 0.3085
P(−0.5 v 2) = 0.9772 −0.3085 = 0.6687
Note that whether or not inequalities defining v are strict is of no consequence in
calculating the probabilities.
N(v)
N(y)
N(y)
Figure29.17
P(0 < v < 1) = 0.3413.
0 1 v
–1 0 1 y
Figure29.18
P(−1 < v < 1) = 0.6826.
–0.5 0 2 y
Figure29.19
P(−0.5 v 2) = 0.6687.
29.14.2 Non-standardnormal
Table 29.7 allows us to calculate probabilities for a random variable with a standard
normal distribution. This section show us how to use the same table when the variable
has a non-standard distribution. A non-standard normal has a mean value other than 0
and/or a standard deviation other than 1. The non-standard normal is changed into a
standard normal by application of a simple rule. Suppose the non-standard distribution
hasamean µandastandarddeviation σ.Thenallnon-standardvaluesaretransformed
tostandard values using
non-standard → standard
X → X − µ
σ