Question

Suppose that only 28% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible. If we randomly select 30 drivers that approach an intersection under these condition, what is the probability that at most 10 will come to a complete stop?

(current posted question is incorrect I believe)

Answer 1

Solution:

Here, we are given n = 30, p = 0.28

We have to find P(X≤10)

Here, we have to use normal approximation to binomial distribution.

First we have to check two conditions for normal approximation.

We have n = 30, p = 0.28, q = 1 – p = 1 – 0.28 = 0.72

np = 30*0.28 = 8.4

nq = 30*0.72 = 21.6

np and nq > 5

So, we can use normal approximation.

Now, we have to find mean and SD for normal approximation.

Mean = np = 30*0.28 = 8.4

SD = sqrt(npq) = sqrt(30*0.28*0.72) = 2.459268

We have to find P(X≤10) = P(X<10.5) [by using continuity correction]

Z = (X – mean) / SD

Z = (10.5 - 8.4) / 2.459268

Z = 0.853913

P(Z<0.853913) = P(X<10.5) = 0.803423

(by using z-table or excel)

Required probability = 0.803423