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.
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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<x<1970)=P(0<z<1.25) = 0.3943
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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<x<1600) = P (-15.1667<z<-1.8333 ) =0.0334
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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.