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)
Answer:
11/13 x 4 = 3 5/13
Step-by-step explanation:
Answer:
x = 58
Step-by-step explanation:
The exterior angle is equal to the sum of the opposite interior angles
90 = 32+x
Subtract 32 from each side
90-32 = x
58 =x
Answer:
246.76$
Step-by-step explanation:
199 x .24 =
47.76
199 + 47.76 =
246.76
Answer:
About 53% percent of the students went to the cafeteria.
Step-by-step explanation:
there are 26 students and 14 left that can be put into the fraction 14/26.
To find the percentage you then divide 14 by 26 and turn it into a percent and you get something like 53.846... but you can simplify it to 53%
