Question 135564: 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! 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.
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You need to convert these numbers to z-scores then use your chart of calculator
to find the percentages.
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a. What percent of the cruises will have between 1,820 and 1,970 passengers?
P(1820
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b. What percent of the cruises will have 1,970 passengers or more?
P(x>=1970) = 0.5-0.3943 = 0.1056
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c. What percent of the cruises will have 1,600 or fewer passengers?
P(0
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d. How many passengers are on the cruises with the fewest 25 percent of passengers?
The z-value corresponding to lowest 25% is -0.6745
Use z=(x-u)/s to find the corresponding x-score.
-0.6745 = (x-1820)/120
x-1820 = 120*-0.6745
x-1820= -80.9388
x = 1539 (rounded down)
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Cheers,
Stan H.
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