The fill amount in 2-liter soft drink bottles is normally distributed with a mea

Question

The fill amount in 2-liter soft drink bottles is normally distributed with a mean of 2.0 liters and a standard deviation of 0.05 liter. If bottles contain less than 94% of the listed net content (1.88 liters, in this case), the manufacturer may be subject to penalty by the state office of consumer affairs. Bottles that have a net content above 2.09 liters may cause excess spillage upon opening. Complete parts (a) through (e) below. a. What proportion of bottles contain between 1.88 and liters? b. What proportion of bottles contain between 1.88 and 2.09 liters? c. WI proportion of bottles contain below 1.88 liters or above 2.09 liters? d. At least how much soft drink is contained in 99% of the bottles? liter (s) e. Ninety-nine percent of the bottles contain an amount that is between which two values (symmetrically distributed)around the mean? Between liter(s) and liter (s)

Explanation / Answer

given Mean ( u ) =2
Standard Deviation ( sd )=0.05
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1.88) = (1.88-2)/0.05
= -0.12/0.05 = -2.4
= P ( Z <-2.4) From Standard Normal Table
= 0.0082
P(X < 2) = (2-2)/0.05
= 0/0.05 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(1.88 < X < 2) = 0.5-0.0082 = 0.4918
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1.88) = (1.88-2)/0.05
= -0.12/0.05 = -2.4
= P ( Z <-2.4) From Standard Normal Table
= 0.0082
P(X < 2.09) = (2.09-2)/0.05
= 0.09/0.05 = 1.8
= P ( Z <2.09) From Standard Normal Table
= 0.9641
P(1.9 < X < 2.1) = 0.9641-0.0082 = 0.9559   
c.

To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1.88) = (1.88-2)/0.05
= -0.12/0.05 = -2.4
= P ( Z <-2.4) From Standard Normal Table
= 0.0082

P(X < 2.09) = (2.09-2)/0.05
= 0.09/0.05 = 1.8
= P ( Z <2.09) From Standard Normal Table
= 0.9641
P(1.9 < X < 2.1) = 0.9641-0.0082 = 0.9559      
  
d.
P ( Z < x ) = 0.99
Value of z to the cumulative probability of 0.99 from normal table is 2.326
P( x-u/s.d < x - 2/0.05 ) = 0.99
That is, ( x - 2/0.05 ) = 2.33
--> x = 2.33 * 0.05 + 2 = 2.1163

e)

the z-values framing 99% around the mean.
+-invNorm(0.005) = +-2.5758

Lower limit:: x-bar = -2.5758*(0.05/5)+2 = 1.9742
Upper limit:: x-bar = +2.5758*(0.05/5)+2 = 2.0258

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