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Neurological research has shown that in about 80% of people language abilities reside in the brings...

Neurological research has shown that in about 80% of people language abilities reside in the brings the left side another temper sent to display right bring language centers and the remaining 10% have two-sided language control.
1) assume that a freshman composition class contains 25 randomly selected people, what’s the probability that no more than 15 of them have the left to brain language control?
2) what are the mean and standard deviation of the number of these freshmen, who might be right brained in language abilities?
3) eat a randomly chosen group of five of these students,what’s the probability that no one has two sided language control?
4) in the entire freshman class of 12,000 students, how many would you expect to find of each type?
5) if and assumption of normality is justified use the 68–95–99.7 rule to describe how many students in the freshman class might have right brain language control.
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Answer #1

(1) n = 25, p = 0.8, q = 1 - p = 0.2

P(x) = C(n, x) p^x q^(n - x)

Binomial Distribution
n = 25
p = 0.8
Mean = 20.0000
Var = 4.0000
SD = 2.0000
0 0.0000
1 0.0000
2 0.0000
3 0.0000
4 0.0000
5 0.0000
6 0.0000
7 0.0000
8 0.0000
9 0.0000
10 0.0000
11 0.0001
12 0.0003
13 0.0012
14 0.0040
15 0.0118
16 0.0294
17 0.0623
18 0.1108
19 0.1633
20 0.1960
21 0.1867
22 0.1358
23 0.0708
24 0.0236
25 0.0038

P(x <= 15) = P(0) + P(1) + ... P(15) = 0 + 0 + ... + 0.0118 = 0.0173

(2) Mean = np = 20 and Standard deviation = sqrt(npq) = 2

(3) n = 5, p = 0.10, q = 1 - p = 0.90

P(x) = C(n, x) p^x q^(n - x)

Binomial Distribution
n = 5
p = 0.1
Mean = 0.5000
Var = 0.4500
SD = 0.6708
0 0.5905
1 0.3281
2 0.0729
3 0.0081
4 0.0005
5 0.0000

P(0) = 0.5905

(4) 12000 * 0.10 = 1200 students

(5) This question is not clear to me. I am trying to figure out what it means.

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