Question

According to a​ study, 73​% of all males between the ages of 18 and 24 live at home. ​ (Unmarried college students living in a dorm are counted as living at​ home.) Suppose that a survey is administered and 184 of 223 respondents indicated that they live at home.​

(a) Use the normal approximation to the binomial to approximate the probability that at least 184 respondents live at home.​

(b) Do the results from part​ (a) contradict the​ study?

Answer 1

a)

X ~ Bin ( n , p)

= n p = 223 * 0.73 = 162.79

= sqrt [ n p( 1 - p) ] = sqrt [ 223 * 0.73 ( 1 - 0.73) ] = 6.6297

Using normal approximation ,

P ( X < x) = P ( (Z < X - µ ) / σ )

With continuity correction,

P(X >= 184 ) = P(X > 183.5)

P ( X > 183.5 ) = P(Z > (183.5 - 162.79 ) / 6.6297 )

= P ( Z > 3.12 )

= 1 - P ( Z < 3.12 )

= 1 - 0.9991 (From Z table)

= 0.0009

b)

Since this probabiltiy is less than 0.05, the results from part​ (a) contradict the​ study.