What does "a visitors comes by bus a restaurant meal" mean?
it is known that half the visitors to a certain game resort come by bus and 80% buy a meal ia a restaurant. Three resort's tourism officer's, speculating about the probability that a visitors comes by bus a restaurant meal, suggest values of 0.2,0.4, 0.6.
1.Show that two of these values are impossible.
2.Assuming the third to be correct, what is the probability that a visitor who does not come by bus will buy a meal in a restaurant.
I think it means 'The probability that a visitor who comes by bus buys a restaurant meal'. Let's call this p.
1)
Suppose that all visitors who do not come by bus buy a restaurant meal. If we look at the percentage of all guests buying a restaurant meal:
p = 0.2. Then: 1 x 0.5 + 0.2 x 0.5 = 0.6 < 0.8
p = 0.4. Then: 1 x 0.5 + 0.4 x 0.5 = 0.7 < 0.8
So even if all other guests buy a meal, the percentage of the guests that buy a meal will never be 80% if only 20% or 40% of the 'bus guests' buy a meal.
2)
If we take the third probablity:
p = 0.6. Then: q x 0.5 + 0.6 x 0.5 = 0.8.
Here, q denotes the probability that a guest who does not arrive by bus buys a meal. If we invert this equality:
q = (0.8 - 0.6 x 0.5) x 2 = 1.
This means that, if p = 0.6, all visitiors who do not come by bus buy a restaurant meal, hence the probability is 1.
Dear All,
I would like to request to solve the problem. The problem is:
1)Consider the pdf
f(x)= B xB-1 0<x<1
0 else
Use Monte Carlo method (MCM)to generate and print 10 observations from this distribution, with Beta(B)=.5. You will write a simple R code for any value of B. Generate but donot print 10,000 observations, compute their sample mena and compare it to the true mean(of the distribution above).
2)Same question as above for the so-called Weibull distribution, with h=1.5find first F -1.)You will write down a simple R code for any value of h.
f(x) = (3/h3 )x2 e(-x 3 /h2 ) 0<x<infinity
0 else