Respuesta :
Let [tex]X_1[/tex] and [tex]X_2[/tex] represent the weights of the two oranges.
Let [tex]d=X_1-X_2[/tex] represent the difference in weight.
We're given that
[tex]X_1[/tex] ~ N(12, 1.2^2)
[tex]X_2[/tex] ~ N(12, 1.2^2)
Then, [tex]d~ N(12-12, 1.2^2 + 1.2^2) or N(0, 2*1.2^2)
So the standard deviation of d is [tex] \sqrt{2 \cdot 1.2^2} = 1.6971 [/tex]
Let's find
[tex]P(d\ \textgreater \ 3) = 1 - \phi ( \frac{3-0}{1.6971}) = 1- 0.9615 = 0.0385 [/tex]
So the answer is 0.0385
Let [tex]d=X_1-X_2[/tex] represent the difference in weight.
We're given that
[tex]X_1[/tex] ~ N(12, 1.2^2)
[tex]X_2[/tex] ~ N(12, 1.2^2)
Then, [tex]d~ N(12-12, 1.2^2 + 1.2^2) or N(0, 2*1.2^2)
So the standard deviation of d is [tex] \sqrt{2 \cdot 1.2^2} = 1.6971 [/tex]
Let's find
[tex]P(d\ \textgreater \ 3) = 1 - \phi ( \frac{3-0}{1.6971}) = 1- 0.9615 = 0.0385 [/tex]
So the answer is 0.0385
Answer:
0.0392
Step-by-step explanation:
The weight is normally distributed with :
Mean (\mu)= 12 ounces
Standard deviation (\sigma ) = 1.2 ounces
Let X_1 be the weight of first orange and X_2 be the weight of second orange.
Let X be the difference of the weight of two oranges (X_1-X_2)
X_1(12,1.2^{2}) and X_2(12,1.2^{2})
X(12-12,√1.2^{2}+1.2^{2}) = X(0,1.6971)
We need to calculate P(X>3) = 1-P(X≤3)
For calculating P(X≤3) in normal distribution, we use
Z = [tex]\frac{X-\mu}{\sigma }[/tex]
= [tex]\frac{3-0}{1.6971}[/tex]
= 1.7677
P(Z≤ 1.7677) = 0.9608 (From normal distribution table)
P(X > 3) = 1-0.9608
= 0.0392
