Answer:
a. 0.9931
b. 0.3423
c. 0.3907
d. 0.2670
e. 3.15
f. 1.0796
Step-by-step explanation:
The probability of the variable that said the number of left-handed people follow a Binomial distribution, so the probability is:
![P(x)=nCx*p^{x}*(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28x%29%3DnCx%2Ap%5E%7Bx%7D%2A%281-p%29%5E%7Bn-x%7D)
nCx is calculated as:
![nCx=\frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=nCx%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
Where x is the number of left-handed people, n is the number of people selected at random and p is the probability that a person is left--handed. So P(x) is:
![P(x)=5Cx*0.63^{x}*(1-0.63)^{5-x}](https://tex.z-dn.net/?f=P%28x%29%3D5Cx%2A0.63%5E%7Bx%7D%2A%281-0.63%29%5E%7B5-x%7D)
Then the probabilities P(0), P(1), P(2), P(3), P(4) and P(5) are:
![P(0)=5C0*0.63^{0}*(1-0.63)^{5-0}=0.0069](https://tex.z-dn.net/?f=P%280%29%3D5C0%2A0.63%5E%7B0%7D%2A%281-0.63%29%5E%7B5-0%7D%3D0.0069)
![P(1)=5C1*0.63^{1}*(1-0.63)^{5-1}=0.0590](https://tex.z-dn.net/?f=P%281%29%3D5C1%2A0.63%5E%7B1%7D%2A%281-0.63%29%5E%7B5-1%7D%3D0.0590)
![P(2)=5C2*0.63^{2}*(1-0.63)^{5-2}=0.2011](https://tex.z-dn.net/?f=P%282%29%3D5C2%2A0.63%5E%7B2%7D%2A%281-0.63%29%5E%7B5-2%7D%3D0.2011)
![P(3)=5C3*0.63^{3}*(1-0.63)^{5-3}=0.3423](https://tex.z-dn.net/?f=P%283%29%3D5C3%2A0.63%5E%7B3%7D%2A%281-0.63%29%5E%7B5-3%7D%3D0.3423)
![P(x)=5C4*0.63^{4}*(1-0.63)^{5-4}=0.2914](https://tex.z-dn.net/?f=P%28x%29%3D5C4%2A0.63%5E%7B4%7D%2A%281-0.63%29%5E%7B5-4%7D%3D0.2914)
![P(x)=5C5*0.63^{5}*(1-0.63)^{5-5}=0.0993](https://tex.z-dn.net/?f=P%28x%29%3D5C5%2A0.63%5E%7B5%7D%2A%281-0.63%29%5E%7B5-5%7D%3D0.0993)
Then, the probability P(x≥1) that there are some lefties among the 5 people is:
P(x≥1) = P(1) + P(2) + P(3) + P(4) + P(5)
P(x≥1) = 0.0590 + 0.2011 + 0.3423 + 0.2914 + 0.0993 = 0.9931
The probability P(3) that there are exactly 3 lefties in the group is:
P(3) = 0.3423
The probability P(x≥4) that there are at least 4 lefties in the group is:
P(x≥4) = P(4) + P(5) = 0.2914 + 0.0993 = 0.3907
The probability P(x≤2) that there are no more than 2 lefties in the group is:
P(x≤2) = P(0) + P(1) + P(2) = 0.0069 + 0.0590 + 0.2011 = 0.2670
On the other hand, the expected value E(x) and standard deviation S(x) of the variable that follows a binomial distribution is:
![E(x)=np=5(0.63)=3.15\\S(x)=\sqrt{np(1-p)}=\sqrt{5(0.63)(1-0.63)}=1.0796](https://tex.z-dn.net/?f=E%28x%29%3Dnp%3D5%280.63%29%3D3.15%5C%5CS%28x%29%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B5%280.63%29%281-0.63%29%7D%3D1.0796)