Question 58759: Dear Sir,Madam,
Please can you help me to solve following?
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 between1,820 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 fewet 25 percent of passengers?
I would really appreciate your help as this is the only exercise I have been unable to answer.
Thanks very much,
Gerald Bremer
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.
a) what percent of the cruises will have between1,820 and 1,970 passengers?
P(1820<=X<=1970)=P((1820-1820))/120<=z<=(1970-1820))/120))=0.394
b) what percent of the cruises will have 1,970 passengers or more?
P(X>=1970)=P(z>=(1970-1820)/120)=0.1056
c) what percent of the cruises will have 1,600 or fewer passengers?
P(x<=1600)=P(z<=(1600-1820)/120))=0.0334
d) how many passengers are on the cruises with the fewet 25 percent of passengers?
Find the z-value that corresponds to the bottom 25% of population:
z=-0.6745
Then solve for X in:
-0.6745=(X-1820)/120
X-1820= -80.94
X<=1719 passengers are on the cruise.
Cheers,
Stan H.
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