10 Q2 Suppose the current measurements in a strip of wire fo

(10] Q.2. Suppose the current measurements in a strip of wire follow a normal distribution with a mean of 10 milli amperes and a variance of 4 (milliamperes)2. (a) What is the probability that a measurement will exceed 13 milliamperes? (b) Determine the value for which the probability that a current measurement is below this value is 0.98.

Solution

Normal Distribution
Mean ( u ) =10
Standard Deviation ( sd )=4
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 13) = (13-10)/4
= 3/4 = 0.75
= P ( Z >0.75) From Standard Normal Table
= 0.2266                  
b)
P ( Z < x ) = 0.98
Value of z to the cumulative probability of 0.98 from normal table is 2.054
P( x-u/s.d < x - 10/4 ) = 0.98
That is, ( x - 10/4 ) = 2.05
--> x = 2.05 * 4 + 10 = 18.216