the probability of at least 90

Question

According to a newly published report approximately 43% of adults say filing their tax paperwork is worse than a trip to the dentist. If a random sample of 200 is chosen, the probability that at least 90 of them share this opinion is what number?

Answer 1

This is a probability problem and not a Math problem

 

p=0.43, q=1-p=0.57

n=200

 

Mean = np = 200*0.43=86

Standard Variation = √npq = 200*0.43*0.57 = 7

 

P(X>90)= P[(X-Mean/Standard Variation)>((90-86)/7]=P(Z>0.57)=1-P(Z<0.57)

 

From Normal Distribution table we can find  P(Z<0.57)=0.7088

 

P(X>90)=1-0.7088=0.2912 (29.12%)

Answer 2

Start with the proportion: 43/100 = x/200

 

100x=8600

x=86

(So, out of 200, approximately 86 people will share the opinion.)

 

Therefore, the probability that at least 90 of the 200 will share the opinion is 0.