(a)
= 100
= 10
To find P(X<90):
Z = (90 - 100)/10
= - 1.00
By Technology, Cumulative Area Under Standard Normal Curve = 0.1587
So,
P(X<90):= 0.1587
So,
Answer is:
0.1587
(b)
= 100
= 10
To find P(X<70 OR X>125):
For X < 70:
Z = (70 - 100)/10
= - 3.00
By Technology, Cumulative Area Under Standard Normal Curve = 0.0013
So,
P(X < 70) = 0.0013
For X > 125::
Z = (125 - 100)/10
= 2.50
By Technology, Cumulative Area Under Standard Normal Curve = 0.9938
So
P(X > 125) = 1 - 0.9938 = 0.0062
So,
P(X<70 OR X>125) = 0.0013 +0.0062 = 0.0075
So,
Answer is:
0.0075
(c)
90% symmetrical corresponds to area = 0.90/2 = 0.45 on either side of mid value.
Table of Area Under Standard Normal Curve gives Z = 1.645
Low end:
Z = - 1.645 = (X - 100)/10
So,
X = 100 - (1.645 X 10)
= 100 - 16.45
= 83.55
High end:
Z = 1.645 = (X - 100)/10
So,
X = 100 + (1.645 X 10)
= 100 + 16.45
= 116.45
90% of values are greater than 83.55 and less than 116.45
6.2.5 2 of 10 (10 complete Given a normal distribution with ju= 100 and a =...
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