Unformatted text preview: Step 2: Calculate N(dl) and N(dl). We can best understand the values NM) and Nick) by exam
Figure 22. IO. The ﬁgure shows the normal distribution with an expected value of 0 and a s
deviation of i. This is frequently (rolled the standardized normal distribution. We mention
an earlier chapter that the probability that a drawing from this distribution will be between
+ I (within one standard deviation of its mean, in other words) is 68 .26 percent. Now? ii'et us ask a different question: What Is the probability that a drawing from the stando ='
normal distribution will be below a particular value? For example. the probability that a dr’a‘iiv
ll 'be below 0 is clearly 50 percent because the normal distribution is symmetric. Using statif __ Figure 22.10 Graph of Cumulative Probability Probability 0.37421 . ._. ,_. - .Iielue
Shaded area reproceniio cumulative probability. Because the
- . probability Is .6459 that a drawing from ﬂie‘otanderd n'orrnal
_ '. distribution will be below 3242'“ ”say that His-742) =' .0459.
. That is. the cumulative probability 01.3742 Iii .8450. terminology. we say that the cumulative probability of D is 50 percent. Statisticians also? say. that
N(O) = 50%. It turns. out that- - - .. .... . -' - " .. N(di)_=N(.3742)'- 6459'“
.797" 3331""? ft iii”! 245$ '.-N(d2)" .N(il527)-" “15607.. The ﬁrst value means that there Is a 64.59 peroent probability that a drawing i'rorn- the standardized 1. '
nomial distribution will be below .3742.T_he second value means that there is a 56.07 percent prob-
ability that a drawing from the standardized normal distribution will be below. [.527 More generally.
N(d) is the probability that a drawing from the standardized normal distribution; will be below 6. In
other words. N(d) is die cumulative probability of d. Note that d. and rig. in' our-- example are-slightly
above zero, so N(d.) and N(dl) are slightly greater ﬂ1en..-50 I
Perhaps the easiest way to determine Md.) and N(th) is from the EXCEL function NORMSDIST. In our amnple.INORMSDIST(.3742) and NORMSDlST(. I527) are .6459 and .5607. respective We can also deterﬁilﬁe the Eﬁfiiuisiw‘étrobibiiiiy frohiTable 22.3. For example. consider if .
_ This can be found In the table as .3 on the vertical and .07 on the horizontal. The value in the table fo d =37is. i443. Thisvalue is notthe cum—iiiativeproWi-of ..37 We mustﬁrstma'kea‘na'd -
to determine cumulative probabilityfihat is: - N(.37) = 50+. [443 = .6443
N(-. 37) = .so - 1.443 = .3557 Unfortunater our table handles only two significoht digits. whereas our value of .3742 has four-I:
signiﬁcant digits. Hence we must interpolate to ﬁnd N(. 3742). Because N(. 37)= .6443 and N(. 38) —.I
.6480 the difference between the two values is .0037 (= .6480 — 644.3) Since 3.742 is 42 percent of the way bemen .37 and .38. we Interpolate as: '
.34 N(.3742) = .6443 + .42 X .0037 = .6459 . ...
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- Spring '08