Hefty Hamburger advertises their famous Quarter Packin Pound

Hefty Hamburger advertises their famous “Quarter Packin’ Pounder” hamburger. Suppose, in fact, an inspector finds that the weight of each hamburger is not exactly as advertised, but is normally distributed with mean 0.25 pounds and standard deviation 0.102 pounds.

1. Find the probability a hamburger weighs exactly as advertised (0.25 pounds).   (2 points)

2. What is the probability that a hamburger weighs between 0.12 and 0.3 pounds? (3 points)

3. Any of the hamburgers that weigh less than 0.18 pounds must be thrown away. What is the probability that a hamburger will NOT be thrown away?    (3 points)

4. If the z-score of a hamburger is -0.25, what is the weight of the hamburger? Would it be thrown away? (3 points)

5. Solve to fill in the blank:

90% of all hamburgers weigh less than _________ pounds. SHOW WORK!!!   (4 points)

Solution

Mean ( u ) =0.25
Standard Deviation ( sd )=0.102
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
1)
TO find P(X=25)= P(0.245 < X < 0.255)
P(a < = Z < = b) = F(b) - F(a)
P(X < 0.245) = (0.245-0.25)/0.102
= -0.005/0.102 = -0.049
= P ( Z <-0.049) From Standard Normal Table
= 0.48045
P(X < 0.255) = (0.255-0.25)/0.102
= 0.005/0.102 = 0.049
= P ( Z <0.049) From Standard Normal Table
= 0.51955
P(0.245 < X < 0.255) = 0.51955-0.48045 = 0.0391                  

2)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.12) = (0.12-0.25)/0.102
= -0.13/0.102 = -1.2745
= P ( Z <-1.2745) From Standard Normal Table
= 0.10124
P(X < 0.3) = (0.3-0.25)/0.102
= 0.05/0.102 = 0.4902
= P ( Z <0.4902) From Standard Normal Table
= 0.688
P(0.12 < X < 0.3) = 0.688-0.10124 = 0.5868                  
3)
P(X > 0.18) = (0.18-0.25)/0.102
= -0.07/0.102 = -0.6863
= P ( Z >-0.686) From Standard Normal Table
= 0.7537                  
5)
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 0.25/0.102 ) = 0.9
That is, ( x - 0.25/0.102 ) = 1.28
--> x = 1.28 * 0.102 + 0.25 = 0.3808