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>
