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junior college 1 | H1 Maths

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##### Audrey

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I got 199 for (i)

How can I do part (ii) and (iii)?

Part (ii), I tried as follows:
X - B(200, 0.97)
P(X =< 199) =

I was not able to get the answer.

Help is needed. Thanks.

Date Posted: 5 months ago
Views: 227
Boy Mow Chau
5 months ago
(ii)
205 tickets were sold, so no. of trials is 205, not 200.
... should be X~B(205,0.97).
since flight can hold 200 passengers, then total passengers that show up cannot exceed 200.

P(all show up can take flight)
= P(X<=200)
= 0.738681

(iii)
E(X) = np = 205x0.97 = 198.85
P(X>198.85 | X<=200)
= [P(X=199)+P(X=200)] / P(X<=200)
= (0.162715+0.157833) / 0.738681
= 0.433947

for (i), just to clarify ...
most probable number & expected number are not the same thing.
from GC ...
P(X=197) = 0.1095
P(X=198) = 0.1431
P(X=199) = 0.1627
P(X=200) = 0.1578
P(X=201)= 0.1269
since P(X=199) has highest probability, so most probable number of passengers is 199, whereas expected number is 198.85 .

### See 1 Answer

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##### Jia Earn Lim's answer
96 answers (Tutor Details)
1st
I haven’t done distributions in a while so do let me know if this is wrong :P But i believe you should be using n=205 here as the airline sold 205 tickets, hence there are 205 potential passengers. I hope my explanation in the working helps :)
Date Posted: 5 months ago