Question

A preliminary survey shows that 32% of college students smoke. In a class of 36 students, what is the probability that more than 47% of the students smoke? Use Appendix B.1 for the z-values. (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Probability

Answer 1

solution:-

given that p = 0.32 , n = 36

standard deviation = sqrt(p*(1-p)/n) = sqrt(0.32*(1-0.32)/36)

   = 0.0777

P(x > 0.47) = P(z > (0.47-0.32)/0.0777)

= P(z > 1.93)

= 1-P(z < 1.93)

= 1 - 0.9732

= 0.0268