- Find the values ofA(z)
Let A(ZJ) = area under probability curve from centerline (where X= JJL) to X=X 1 ; this area
= P(IJL <X<Xi). Obtain the values of A(Z) from the table of areas under the normal prob-
ability curve. Refer to Table 25, which is an excerpt from this table. Thus, if z = 1.60, A(Z)
= 0.44520; if z = 1.78, A(z) = 0.46246. Note that A(-zt) = A(Z) by symmetry. Record the
values of A(Z) in Table 24. - Compute the required probabilities
Refer to Fig. 21. Apply the areas in Table 24 to obtain these results: P(14 < X< 17) =
0.38493; P(12 <X< 16.2) = 0.28814 + 0.31057 = 0.59871; P(X< 10) = 0.5 - 0.44520 -
0.05480.
Related Calculations'. Many random variables that occur in nature have a nor-
mal probability distribution. For example, the height, weight, and intelligence of mem-
bers of a species have normal distributions. Although in theory this distribution applies
solely where the range of X values is infinite, in practice the distribution is applied as a
valid approximation where the range of lvalues is finite.
APPLICATION OF NORMAL DISTRIBUTIONThe time required to perform a manual operation is assumed to have a normal distribu-
tion. Studies of past performance disclose that the average time required is 5.80 h and the
standard deviation is 0.50 h. Find the probability (to three decimal places) that the opera-
tion will be performed within 5.25 h.
Calculation Procedure:- Compute the value of z corresponding to the boundary
value of X
Let X= time required to perform the operation, and refer to the preceding calculation pro-
cedure for the definition of z. For Jf- 5.25, z = (5.25 - 5.80)7050 = -1.10.
TABLE 25. Area under the Normal Curve
z .00 .01 .02 .03 .04
1.5 .4331 9 .4344 8 .4357 4 .4369 9 .4382 2
1.6 .4452 0 .4463 0 .4473 8 .4484 5 .4495 0
1.7 .4554 3 .4563 7 .4572 8 .4581 8 .4590 7z .0 5 .0 6 .0 7 .0 8 .0 9
1.5 .4394 3 .4406 2 .4417 9 .4429 5 .4440 8
1.6 .4505 3 .4515 4 .4525 4 .4535 2 .4544 9
1.7 .4599 4 .4608 0 .4616 4 .4624 6 .4632 7