a)
HERE on a typicall day 75% chances are their for north and 25% for south
so out of 1030 total cars 729 from north and 301 from south.
but theoretically they should be 75% and 25% of 1030 i.e 772 and 258
so clearly these number are very uncertain theoritically.
b)
let X be the number of cars from north out of 1030
then X~binomial(n,p) n 1030 p as 0.75
P(|X-mean|>= |729-mean|)
i.e putting mean as n*p i.e 772.5 also 772.5-729 is43.5
P(X>=816 or x<=729)
(pbinom(729,1030,.75)) + 1-(pbinom(815,1030,.75))
0.00196941
required probavility is 0.00196941
At 8:14 am on a typical day, 75.0% of the cars entering York University fron Keele...