Canine Crunchies Inc. (CCI) sells large bags of dog food to warehouse clubs. CCI uses an automatic filling process to fill the bags. Weights of the filled bags are approximately normally distributed...

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Canine Crunchies Inc. (CCI) sells large bags of dog food to warehouse clubs. CCI uses an automatic filling process to fill the bags. Weights of the filled bags are approximately normally distributed with a mean of 50 kilograms and a standard deviation of 1.25 kilograms. a. What is the probability that a filled bag will weigh less than 49.5 kilograms? b. What is the probability that a randomly sampled filled bag will weigh between 48.5 and 51kilograms? c. What is the minimum weight a bag of dog food could be and remain in the top 15% of all bags filled? d. CCI is unable to adjust the mean of the filling process. However, it is able to adjust the standard deviation of the filling process. What would the standard deviation need to be so that no more than 2% of all filled bags weigh more than 52 kilograms?

Answered Same DayDec 26, 2021

Answer To: Canine Crunchies Inc. (CCI) sells large bags of dog food to warehouse clubs. CCI uses an automatic...

David answered on Dec 26 2021
120 Votes
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A P (X < 49.5)
Normal Distribution, µ = 50, σ = 1.25, P
(X < 49.5) = Area to the left of 49.5
we convert this to standard normal using z = x− µ
σ
z = 49.5− (50)1.25 = −0.40
Z
0.3446
−0.40
0
P (X < 49.5) = P (Z < −0.40) = 0.3446 (from z-table)
P (Z < −0.4): in a z-table having area to the left of z, locate -0.4 in the left most column.
Move across the row to the right under column 0.00 and get value 0.3446
Using technology, answer is: 0.344578258389676
P (X < 49.5) = 0.3446 = 34.46%
B P (48.5 < X < 51)
Normal Distribution, µ = 50, σ = 1.25, we convert this to standard normal
using z = x− µ
σ
z1 =
48.5− (50)
1.25 = −1.20
z2 =
51− (50)
1.25 = 0.80
P (−1.20 < Z < 0.80) = Area in between −1.20 and 0.80
Z
0.1151
0.80
−1.20
0
0.7881
0.7881−...
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