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Pumpkin pie A can of pumpkin pie mix contains a mean of 30 ounces and a standard deviation of 2 ...
tlc_71111
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Pumpkin pie A can of pumpkin pie mix contains a mean of 30 ounces and a standard deviation of 2 ...
Pumpkin pie
A can of pumpkin pie mix contains a mean of 30 ounces and a standard deviation of 2 ounces. The contents of the cans are normally distributed. What is the probability that four randomly selected cans of pumpkin pie mix contain a total of more than 126 ounces?
Textbook
Stats: Modeling the World
Edition:
4
^{th}
Authors:
Bock, Velleman, De Veaux
Read 42 times
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Antoinette12
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Two methods are shown below to solve this problem:
Method 1:
Let P = one can of pumpkin pie mix and T = four cans of pumpkin pie mix.
We are told that the contents of the cans are normally distributed, and can assume that the content amounts are independent from can to can.
E(T
) =
E(
P
_{1}
+ P
_{2}
+ P
_{3}
+ P
_{4}
) =
E
(P
_{1}
) +
E
(P
_{2}
) +
E
(P
_{3}
) +
E
(P
_{4}
) = 120 ounces
Since the content amounts are independent,
Var(T
) =
Var(
P
_{1}
+ P
_{2}
+ P
_{3}
+ P
_{4}
) =
Var
(P
_{1}
) +
Var
(P
_{2}
) +
Var
(P
_{3}
) +
Var
(P
_{4}
) = 16
SD
(
T
) =
=
= 4 ounces
We model T with
N
(120, 4)
z =
= 1.5
P = P(T
> 126) =
P
(
z
> 1.5) = 0.067
There is a 6.7% chance that four randomly selected cans of pumpkin pie mix contain more than 126 ounces.
Method 2:
Using the Central Limit Theorem approach, let
= average content of cans in sample
Since the contents are Normally distributed,
is modeled by
N
.
P
=
P
(
> 31.5) =
P
=
P
(
z
> 1.5) = 0.067
There is about a 6.7% chance that 4 randomly selected cans will contain a total of over 126 ounces.
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