Question

According to a​ study,

52%

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

153

of

249

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

153153

respondents live at home.​ (b) Do the results from part​ (a) contradict the​ study?

​(a)

​P(X≥153)=

Answer 1

X ~ Bin ( n , p)

Where n = 249 , p = 0.52

Using Normal Approximation to Binomial
Mean = n * P = ( 249 * 0.52 ) = 129.48
Variance = n * P * Q = ( 249 * 0.52 * 0.48 ) = 62.1504
Standard deviation = √(variance) = √(62.1504) = 7.8836

P(X < x) = P(Z < ( x - mean) / SD )

a)

with continuity correction ,

P(X >= 153) = P(X > 152.5)

P ( X > 152.5 ) = P(Z > (152.5 - 129.48 ) / 7.8836 )
= P ( Z > 2.92 )
= 1 - P ( Z < 2.92 )
= 1 - 0.9982
= 0.0018

b)

Since this probability is less than 0.05, the event isunusual.

The results from part​ (a) contradict the​ study.