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TUGAS
Statistika dasar“CHAPTER 6.2 Standard NORMAL PROBABILITY DISTRIBUTION“
Di susun Oleh :Kelompok 3
Dessy Sagita 150610090020Agung Ramandha 150610090029Aisa Permatadewi150610090039
Amelia Novi Heliana 150610090040
PROGRAM STUDI AGRIBISNIS
FAKULTAS PERTANIAN
UNIVERSITAS PADJADJARANStandard Normal Probability Distribution
The letter z is commonly used to designate this particular normal random
variable. It has same general appearance as other normal distributions, but with the
special properties of μ = 0 and σ = 1
To find the probability that a normal random variable is within any specific
interval, we must compute the area under the normal curve over the interval.
Example, z assumes a value of -0,50 or larger, that is , P (z ≥ -0,50). We note that the
probability we are seeking can be written as the sum of two probabilities : P (z ≥ -
0,50) = P (-0,50 ≤ z ≤ 0,00) + P (x ≥ 0,00). From the table we d=find that P (0,00 ≤ z
≤ 0,50) = 0,1915. Therefore P(z ≥ -0,50) = 0,1915 + 0,5000 = 0,6915.
In another example, probability of obtaining a z value between 1,00 and 1,58,
that is P (1,00 ≤z ≤ 1,58). We know that probability of a z value between z = 0,00 and
1,00 is 0,3413 and that probability of a value between z = 0,00 and z = 1,58 is 0,4429,
so there must be a 0,44209 – 0,3413 = 0,1016. Thus, P (1,00 ≤ z ≤ 1,58) = 0,1016
Recall that the body of table 6.1 gives the area under the curve between the
mean and a particular z value. We have been given the information that the area in the
upper tail of the curve ia .10. Hence, we must determine how much of the area is
between the mean and the z value of interest. Because we know that .5000 of the area
is above the mean , .5000-.1000-.4000must be the area under the curve between the
mean and the desired z value. Scanning the body of the table, we find .3997 as the
probability value closest to .4000. the section of the table providing this result follows
Z .06 .07 .08 .09
-
-
-
1.0 .3554 .3577 .3599 .3621
1.1 .3770 .3790 .3810 .3830
1.2 .3962 .3980 .3994 .4015
1.3 4131 .4147 .4162 .4177
1.4 4279 .4292 .4306 .4319
-
-
.3994 area value in body of table closest to .9000
Reading the z value from the left-most column and the top row of the table,
we find that the corresponding z value is 1.28.Thus, area of
approximately .4000( actually .3997) will be between the mean and z =1.28.*in terms
of the question originally asked , the probability is approximately .10 that the z value
will be larger than 1.28.
The examples illustrate that the table of areas for the stansard normal
probability distribution can be used to find probabilities associated with value of the
standatd normal random variable z. to types of question can be aksed. The first type
of question specifies a value.of values, for z and asks us to use the table to determine
the corresponding areas, or probabilities. The second type of question provides an
area, or probability,and asks us to use the table of determine the corresponding z
value. Thus, we need to be flexible in using the standard normal probability table to
answer the desired probability question.
Exercise :
14. Given that z is a standard normal random variable, find z for each situation.
a. The area between 0 and z is .4750 = 1,96
b. The area between 0 and z is .2291 = .61
c. The area to the right of z is .1341 = .1293 = .33
d. The area to the left of z is .6700 = .335 = .96
15. Given that z is a standard normal random variable, find z for each situation.
a. The area to the left of z is .2119 = .55
b. The area between -z and z is .9030 = .9030
2 = .4515 between -1,66 and 1,66
c. The area between -z and z is .2052 = .2052
2 = .1025 between -.25 and .25
d. The area to the left of z is .9948 = .9948
2 = .4974 = 2,79
e. The area to the right of z is .6915 = 6915
2 = .3457 = 1,01
16. Given that z is a standard normal random variable, find z for each situation .
a. The area to the right of z is .01 = .0040
b. The area to the right of z is .025 = .0987
c. The area to the right of z is .05 = .0199
d. The area to the right of z is .10 = .0398
