Find the probability that fewer than 16 are satisfied

Question

A poll reported that 70% of adults were satisified with the job the major airlines were doing. Suppose 20 adults were selected at random and the number who are satisfied are recoreded. What is the probability that fewer than 16 were satisfied with the airlines

Answer 1

Cannot use normal distribution because the sample size is not large enough , as  n(1-p) <= 20*0.2 = 4 < 10

 

Will have to use t-distribution.

 

test stat:   (p - P)/ D where D = sqrt( P(1-P)/n) and P is the population proportion and p is sample proportion

 

             p - P = 0.8 - 0.7 = 0.1

 

 D  = sqrt( 0.7 * 0.3 / 20) = sqrt(  0.21/20) = 0.102

 

 the test stat is 0.1/0.102 = 0.9804

 

 

0.8366 = 83.6% <--- normal distribution table

 

0.8304 = 83.04% <---- t-table