Answer:
a. <u>Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.</u>
b. <u>The probability is 0.2029 or 20.29%</u>
c. <u>The probability is 0.2029 or 20.29%</u>
d. <u>The probability is 0.1671 or 16.71%</u> 
e. <u>The probability is 0.9975 or 99.75%</u>
Step-by-step explanation:
<u>a. Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.</u>
b. Let's use the binomial distribution table, this way:
Binomial distribution (n=10, p=0.697)
  f(x)	F(x)	1 - F(x)
x	Pr[X = x]	Pr[X ≤ x]	
0	0.0000	0.0000	
1	0.0002	0.0002	
2	0.0016	0.0017	
3	0.0095	0.0112	
4	0.0384	0.0496	
5	0.1059	0.1555	
6	<u>0.2029</u> 0.3584	
7	0.2668	0.6252	
8	0.2301	0.8553	
9	0.1176	0.9729	
10	0.0271	1.0000	
<u>The probability is 0.2029 or 20.29%</u>
c. If 69.7% of 18-20 years old consumed alcoholic beverages in 2008, therefore, 30.3% did not and the binomial distribution table is:
Binomial distribution (n=10, p=0.303)
  f(x)	F(x)	1 - F(x)
x	Pr[X = x]	Pr[X ≤ x]
0	0.0271	0.0271	
1	0.1176	0.1447	
2	0.2301	0.3748	
3	0.2668	0.6416	
4	<u>0.2029</u> 0.8445	
5	0.1059	0.9504	
6	0.0384	0.9888	
7	0.0095	0.9983	
8	0.0016	0.9998	
9	0.0002	1.0000
10	0.0000	1.0000
<u>The probability is 0.2029 or 20.29%</u>
d. Let's use the binomial distribution table, this way:
Binomial distribution (n=5, p=0.697)
  f(x)	F(x)	1 - F(x)
x	Pr[X = x]	Pr[X ≤ x]
0	0.0026	0.0026	
1	0.0294	0.0319	
2	0.1351	<u>0.1671	</u>
3	0.3109	0.4779	
4	0.3576	0.8355	
5	0.1645	1.0000
P(0) + P(1) + P (2) = 0.0026 + 0.0294 + 0.1351	
<u>The probability is 0.1671 or 16.71%</u> 
e. Using the same binomial distribution table we used in d. we have:
P(1) + P (2) + P(3) + P(4) + P (5) = 0.0294 + 0.1351 + 0.3109 + 0.3576 + 0.1645
<u>The probability is 0.9975 or 99.75%</u>