The resistance for resistors of a certain type random variab

The resistance for resistors of a certain type random variable X having the normal distribution with mean 9 ohms standard deviation 0.4 ohms. A resistor is between 8.6 and 9.8 ohms.

a)what is prob. that a randomly chosen resistor is acceptable?

b) unacceptable?

c)what is the 75th percentile?

d)what is prob. that out of 4 ranom & independent selected resistors, 2 are acceptable?

Solution

Normal Distribution
Mean ( u ) =9
Standard Deviation ( sd )=0.4
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)              
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 8.6) = (8.6-9)/0.4
= -0.4/0.4 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(X < 9.8) = (9.8-9)/0.4
= 0.8/0.4 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(8.6 < X < 9.8) = 0.97725-0.15866 = 0.8186                  

b)
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 8.6) = (8.6-9)/0.4
= -0.4/0.4= -1
= P ( Z <-1) From Standard Normal Table
= 0.1587
P(X > 9.8) = (9.8-9)/0.4
= 0.8/0.4 = 2
= P ( Z >2) From Standard Normal Table
= 0.0228
P( X < 8.6 OR X > 9.8) = 0.1587+0.0228 = 0.1814
c)
P ( Z < x ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.674
P( x-u/s.d < x - 9/0.4 ) = 0.75
That is, ( x - 9/0.4 ) = 0.67
--> x = 0.67 * 0.4 + 9 = 9.2696                  

d)
P( X = 2 ) = ( 4 2 ) * ( 0.8186^2) * ( 1 - 0.8186 )^2
= 0.1323