Suppose Doris & Sakar know the specification limits for a dog food bag filling m
ID: 3225969 • Letter: S
Question
Suppose Doris & Sakar know the specification limits for a dog food bag filling machine are 40.0 +/- 1.25 pounds. In other words, the filling process is considered to be functioning at an appropriate level (functioning ``in control") if the amount of fill in the dog food bags is between 38.75 pounds and 41.25 pounds. Suppose the filling process is normally distributed with a mean of 40 pounds and a standard deviation of 0.50 pounds. Doris & Sakar select a bag of dog food at random from the assembly line. Let X = amount of fill (amount of dog food) in the bag.
Find the probability that the bag is under-filled; i.e., the fill is less than 38.75 pounds.
a. 0.0035
b. 0.0062
c. 0.9938
d. 0.9965
e. None of the above.
Find the probability that the bag is not under-filled; i.e., the fill is at least 38.75 pounds.
0.0035
0.0062
0.9938
0.9965
None of the above.
Find the probability that the bag has a fill amount that meets specifications, i.e., has a fill amount that is between, and, including, 38.75 and 41.25 pounds.
0.0064
0.0124
0.9876
0.9938
None of the above.
Find the probability that the bag has a fill amount that meets specifications, i.e., has a fill amount that is between, and, including, 38.75 and 41.25 pounds.
0.0064
0.0124
0.9876
0.9938
None of the above.
If 10,000 bags are filled, approximately how many bags would you expect to fail to meet a minimum fill amount of 38.75 pounds?
1.0
6.2
62
None of the above
Find the 95th percentile (i.e., find x such that P(X <= x) = 0.95).
1.645
39.1775
40.4112
40.8225
None of the above.
Please answer ALL questions for a thumbs-up!
a.0.0035
b.0.0062
c.0.9938
d.0.9965
e.None of the above.
Explanation / Answer
mean = 40 , s = 0.50
a)
P(X <38.75)
z = ( x -mean) / s
= ( 38.75 - 40) / 0.50
= -2.5
we need to find P(Z < -2.5)
P(X < 38.75) = P(Z < -2.5) = 0.0062
b)
P(X >38.75)
z = ( x -mean) / s
= ( 38.75 - 40) / 0.50
= -2.5
we need to find P(Z > -2.5)
P(X > 38.75) = P(Z > -2.5) = 0.9938
c)
P(38.75 < x < 41.25)
P(X < 38.75)
z = ( x -mean) / s
= ( 38.75 - 40) / 0.50
= -2.5
P(X < 41.25)
z = ( x -mean) / s
= ( 41.25 - 40) / 0.50
= 2.5
P( 38.75 < X < 41.25) = P(-2.5 < Z < 2.5) = 0.9876
e)
From the table we see that 95% of the area under a standard normal curve is to the left of 1.65.
1.65 = ( x -40 ) / 0.50
x = 40.8225
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