Question

Generation Y has been defined as those individuals who were born between 1981 and 1991. A 2010 survey by a credit counseling foundation found that 55​% of the young adults in Generation Y pay their monthly bills on​ time) Suppose we take a random sample of 180 people from Generation Y. Complete part e below

e. What is the probability that between 94 and 106 of them will pay their monthly bills on​ time? ​__________​(Round to four decimal places as​ needed.)

Answer 1

here,total number of people in sample (n) = 180,

proportion found to pay bills on time (p)= 0.55

As n is large and p > 0.55, we can approximate this binomial distribution to normal.

mean = n*p = 180*0.55 = 99

and standard deviation = sqrt(n*p*(1-p)) = sqrt(180*0.55*0.45) = 6.67

so, the probability that between 94 and 106 of them will pay their monthly bills on​ time:-

[Z=(X-99) / 6.67]

[ using STANDARD NORMAL TABLE]

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