1 Q1) Both undergraduates and postgraduates can use the university cafeteria. Each diner can choose between buying a meal or bringing a packed lunch. (Everyone has exactly one meal each, no more and...

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1 Q1) Both undergraduates and postgraduates can use the university cafeteria. Each diner can choose between buying a meal or bringing a packed lunch. (Everyone has exactly one meal each, no more and no less). The cafeteria offers a daily choice between a hot meal or a cold meal. A survey of undergraduate diners finds that 40% of them bring their own food. Overall, only 25% of the diners bring their own food. Postgraduates make up one fifth of the diners in the cafeteria.


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1 Both undergraduates and postgraduates can use the university Q1) cafeteria. Each diner can choose between buying a meal or bringing a packed lunch. (Everyone has exactly one meal each, no more and no less). The cafeteria offers a daily choice between a hot meal or a cold meal. A survey of undergraduate diners finds that 40% of them bring their own food. Overall, only 25% of the diners bring their own food. Postgraduates make up one fifth of the diners in the cafeteria. a) What is the probability that a diner is an undergraduate and has bought a meal? [8] b) What is the probability that someone who has bought a meal is a postgraduate? [8] c) Give two different examples from the question of mutually exclusive events. [4] Q2) Two types of vehicle use the university central bus station: Uni-link buses and National Express coaches. a) On average, a National Express coach arrives every two hours. What is the probability that only one bus will arrive in a four hour period? [5] b) On the U1 Uni-link bus route, buses either go to the airport or to the docks. Other Uni-link routes do not go to the airport. At the bus station, half of the U1 buses which arrive are going to the airport. At midday, 10% of the Uni-link buses are full. Of these particular buses, 30% are airport buses on the U1 route. U1 buses make up 40% of the Uni-link arrivals at the bus station. What is the probability that a bus is full at midday, given that it is an Airport U1 bus? [6] c) On average, a certain number, k, of National Express coaches go to Bournemouth every day. It is found that the probability of n coaches going - k . to Bournemouth in a day is e What is n?Q3) If P(A) = 0.2 and P(AnB) = 0.4 a) find (5) P(BA) b) Can P(B) = 0? Explain your answer. (5) c) What is the smallest value that P(AB)can take? Explain your answer. (5) d) What would P(AB)have to be, for the complement...



Answered Same DayDec 23, 2021

Answer To: 1 Q1) Both undergraduates and postgraduates can use the university cafeteria. Each diner can choose...

David answered on Dec 23 2021
129 Votes
Solution 1:
a) P (P) = 1/5 or 0.2, P (U) = 4/5 or 0.8
P (U/M) = P (U and M)/ P (M)
P (U and B) = 0.40
P (U and M) = (1 - 0.60)* (1 -0.75) = 0.45
P (B) = 0.25
b) P (B) = 1 – P (M) = 0.75
P (M and P) =
0.2*0.75
P (M/P) = P (M and P)/ P (P)
= 0.15/0.2
= 0.75
c) P (U and B) and P (U and M) both are not equal to zero.
Solution 2:
a) We have λ =2/4 = 0.5
Using Poisson distribution,
P [X = 1] = e-λ λ x/x!
= e-0.5 (0.5) 1/1!
= 0.9098
b) P(full at midday | airport U1 bus)
= P(Airport U1 bus | full at midday) . P(full at midday) / P(airport U1 bus)
P(airport U1 bus) = P(AirportNU1 bus)
P(Airport N U1 bus) = P(Airport | U1 bus) * P(U1 bus)
P(Airport n U1 bus) = 0.5 x 0.4
=0.2
therefore P(airport U1 bus | full at midday = 0.3, P(full at midday) = 0.1
0.3 x 0.1 / 0.2 = 0.15
c) We have λ =2/4 = 0.5
Using Poisson distribution
e-λ λ n/n!= e-k
e-0.5 (0.5)n/n! = e-k
e-0.5 (0.5)n/n! = e-k
n = 1
Solution 3:
We have P ( ̅ ( 1 – 0.2 = 0.8
a) ( ̅
( ̅
( ̅

= 0.4/0.8
= 0.5
b) If P (B) = 0, then P ( ̅ which is not considered to be certainly true. P ( ̅ B) is
0.4 which indicates that event B occurs and hence cannot be zero.
c)
( ̅ ( ̅ (
0.4 *P (B)
P (B) = 0.5
( ̅
( ̅
(
= 0.4/0.8
= 0.5
d) ( ̅ ( ̅ = P (A) = 0.8 (because they are independent)
Solution 4:
a) The expected value of sample mean would be equal to the population mean i.e. 25
b) Yes, because the sample size is large.
c) Z = (X-bar - µ)/ (√σ²/n)
The respective Z-score with p > 0.1587 is 1
1 = (X-bar – 25)/ √22/n)
1 = (27 – 25)/ √22/n)
√22/n = 2
Squaring both sides, we get
22/n = 4
n = 22/4 = 5.5
d) According to chi-squared distribution, the Mean, µ = n = 25 and
Variance, σ² = 2n
2n = 22
n = 11
Solution 5:
a) We have n = 9, X-bar = 5 and s = 9
80% confidence interval is given by:-
X-bar ± t (α/2, n – 1)*s/√n
5...
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