Answer:
c.
Step-by-step explanation:
Answer:
(number - mean)/standard deviation.
Plug in the 78 for the number, the 84.5 for the mean, the 3.1 for the standard deviation.
Step-by-step explanation:
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
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is the logical or/disjunction symbol.
![p\lor q](https://tex.z-dn.net/?f=p%5Clor%20q)
means "
![p](https://tex.z-dn.net/?f=p)
or
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". Disjunction is commutative, which means
![p\lor q](https://tex.z-dn.net/?f=p%5Clor%20q)
is equivalent to
![q\lor p](https://tex.z-dn.net/?f=q%5Clor%20p)
.
So if
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represents
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and
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represents
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, then
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means
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or
![x](https://tex.z-dn.net/?f=x%3C-3)
.