Question 134149: 38. The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution.
a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect?
b. What percent of the garages take between 29 hours and 34 hours to erect?
c. What percent of the garages take 28.7 hours or less to erect?
d. Of the garages, 5 percent take how many hours or more to erect?
44. The number of passengers on the Carnival Sensation during one-week cruises in the Caribbean follows the normal distribution. The mean number of passengers per cruise is 1,820 and the standard deviation is 120.
a. What percent of the cruises will have between 1,820 and 1,970 passengers?
b. What percent of the cruises will have 1,970 passengers or more?
c. What percent of the cruises will have 1,600 or fewer passengers?
d. How many passengers are on the cruises with the fewest 25 percent of passengers?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 38. The accounting department at Weston Materials, Inc., a national manufacturer of unattached garagess, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution.
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a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect?
z(29) = (29-32)/2 = -3/2
z(34) = (34-32)/2 = 1
z(32) = 0
P(32 < x < 34) = P(0< z < 1) = 0.34
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b. What percent of the garages take between 29 hours and 34 hours to erect?
P(29 < x < 34) = P(-1.5 < z < 1) = 0.7745
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c. What percent of the garages take 28.7 hours or less to erect?
z(28.7) = (28.7-32)/2 = -3.3
P(0 < x < 28.7) = P (-10 < z < -3.3) = 0.00048348...
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d. Of the garages, 5 percent take how many hours or more to erect?
find the z-value corresponding to an area of 95% to the left and only
5% to the right under the curve.
Use your z-chart or InvNor(0.95) = 1.645 on your calculator.
Now find the x-value corresponding to that z-value.
1.645 = (x-32)/2
x-32 = 2*1.645
x= 35.29 hours
5% of the houses require 35.29 or more hours to erect.
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44. The number of passengers on the Carnival Sensation during one-week cruises in the Caribbean follows the normal distribution. The mean number of passengers per cruise is 1,820 and the standard deviation is 120.
a. What percent of the cruises will have between 1,820 and 1,970 passengers?
Find the z-values; Then find the percentage between those z-values
to get P(1820 < x < 1970) = 0.3944
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b. What percent of the cruises will have 1,970 passengers or more?
Same procedure:
P(x > 1970) = 0.1056
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c. What percent of the cruises will have 1,600 or fewer passengers?
Same procedure:
P(x < 1600) = 0.3334
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d. How many passengers are on the cruises with the fewest 25 percent of passengers?
Find InvNor(0.25)=-0.6745
Find the x-value associated with z=-0.6745
-0.6745 = (x-1820)/120
x= 1820-0.6745*120
x <= 1739 passengers on the cruises with the fewest 25% of passengers(rounded down)
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Cheers,
Stan H.
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