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Find the value of k in given a standard normal distribution such that
The problem of finding the value of k in a given standard normal distribution has been solved in very easy steps. You only may need to have the table for z score or table of area under normal curve.
Some examples are given that how to find the value of k in a given standard normal distribution.
Given a standard normal distribution find the value of k such that
(a) P( Z < k ) = 0.0427
(b) P ( Z > k ) = 0.2946
(c) P(-0.93 < Z < k ) = 0.7235
Solution. (a)
P( Z < k ) = 0.0427
table(k) = 0.0427
from area table
k = -1.72
Solution (b)
P( Z > k ) = 0.2946
P( Z > k ) = 1-P( Z <= k ) = 0.2946
P( Z <= k ) = 1-0.2946
table(k)=0.7054
from area table
k = 0.54
Solution(c)
P(-0.93 < Z < k ) = 0.7235
table(k) - table(-0.93) = 0.7235
table(k) - 0.1762 = 0.7235
table(k) = 0.7235 + 0.1762
table(k) = 0.8997
from area table
k = 1.28
Some examples are given that how to find the value of k in a given standard normal distribution.
Given a standard normal distribution find the value of k such that
(a) P( Z < k ) = 0.0427
(b) P ( Z > k ) = 0.2946
(c) P(-0.93 < Z < k ) = 0.7235
Solution. (a)
P( Z < k ) = 0.0427
table(k) = 0.0427
from area table
k = -1.72
Solution (b)
P( Z > k ) = 0.2946
P( Z > k ) = 1-P( Z <= k ) = 0.2946
P( Z <= k ) = 1-0.2946
table(k)=0.7054
from area table
k = 0.54
Solution(c)
P(-0.93 < Z < k ) = 0.7235
table(k) - table(-0.93) = 0.7235
table(k) - 0.1762 = 0.7235
table(k) = 0.7235 + 0.1762
table(k) = 0.8997
from area table
k = 1.28
Table for area under normal curve |
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