Question

According to a​ survey, 52​% of males between the ages of 18 and 24 lived at home in 2005​ (unmarried college students living in dorms are counted as living at​ home). A survey is administered at a community college to 19 randomly selected male students between the ages of 18 and 24​ years, and 16 of them respond that they live at home.
​(a) Based on the sample of 19 ​students, what proportion of community college males live at​ home?
​(b) Find the probability that 16 or more out of 19 community college male students live at​ home, assuming that the proportion who live at home is 52​%.
​(c) What might you conclude from this​ result?

Answer 1

Answer on Question #61151 – Math – Statistics and Probability

Question

According to a survey, 52% of males between the ages of 18 and 24 lived at home in 2005 (unmarried college students living in dorms are counted as living at home). A survey is administered at a community college to 19 randomly selected male students between the ages of 18 and 24 years, and 16 of them respond that they live at home.

(a) Based on the sample of 19 students, what proportion of community college males live at home?

(b) Find the probability that 16 or more out of 19 community college male students live at home, assuming that the proportion who live at home is 52%.

(c) What might you conclude from this result?

Solution

(a) Based on the sample of 19 students, what proportion of community college males live at home?

p = 16/19 = 0.8421 or 84.21%.

(b) Find the probability that 16 or more out of 19 community college male students live at home, assuming that the proportion who live at home is 52%.

Using binomial distribution with parameters n = 19, p = 0.52, q = 1 - p = 0.48, $P_n(k) = C_n^k p^k q^{n-k}$, where $C_n^k = \frac{n!}{k!(n-k)!}$, n! = 1 · 2 · 3 · ... · (n - 1) · n.

The probability that 16 or more out of 19 community college male students live at home is

P(16 or more) = P₁₉(16) + P₁₉(17) + P₁₉(18) + P₁₉(16) =

$$= C_{19}^{16} p^{16} q^3 + C_{19}^{17} p^{17} q^2 + C_{19}^{18} p^{18} q^1 + C_{19}^{19} p^{19} q^0 = 0.0037.$$

(c) What might you conclude from this result?

The sample of 19 might be biased or the original survey is no longer valid.

www.AssignmentExpert.com