Suppose it is known that 83% of motorists wear a seatbelt while driving. The police stop a random sample of 200 drivers. What is the probability that more than 80% of them are wearing a seat belt?

~ 84.86%

assuming the sample consist in independant controls.

p = Sum (k=161, k=200, (200 k)*0.83^k*0.17^(200-k))

where (200 k) is the combinatory coefficient = number of choice of subsets of k elements in a set of 200.

This post was edited by feanur on Apr 11 2013 02:42pm

where did you get that k= 161

80% of 200 = 160

More than 80% is 161 to 200.

but the answer is 0.8708.
so confused

I got ~ 88.78% for "80% or more wear a seat belt".
I don't believe in 87.08%. Where does it come from ?

Eh... for this question you should use the normal approximation to binomial probabilities....
have you gone over binomial probailities and the normal function in class?

I got the answer from answer key.

yes i have gonne over bionomial probabilities.
but i dont't know how to calculate it

so let W be the number of people wearing seatbelts
W~Bin(n=200, p=0.83)
then you use Z transformation

Z = W-np/sqroot(npq) ~ N(0,1)

then you want Z>0.8


So if you plug in the number...
W>0.8*200+0.5 <- (the 0.5 is the correction facter as this is not a continuous function)
=Z>(0.8*200+0.5-0.83*200)/sqroot(200*0.83*0.17)
=Z>-1.0353
find that in the table
=0.5+0.3508
=0.8508.


eh... well, that's how I do these questions...