Answer:
[tex]a. \ i. \ P(Y=3)=0.1665\\\ \ ii. \ P(Y\geq 3)=0.9005\\\ \ \ iii.\ \ P(Y\leq 3)=0.0995\\\\b.\ P(2\leq Y\leq 4)=0.4812\\\\c.\ \mu_Y=4.5\\\\d.\ \sigma_Y=1.5732[/tex]
Step-by-step explanation:a.-
a-Notice that this is a binomial probability distribution problem that is generally expressed as:
[tex]{n\choose x}p^x(1-p)^{n-x}[/tex]
Given that p=0.45, n=10, the probability of exactly 3 intoxicated drivers is:
[tex]P(X=3)={10\choose 3}0.45^3(1-0.45)^7\\\\=0.1665[/tex]
Hence, the probability of exactly three intoxicated drivers is 0.1665
ii. The probability of at least 3 intoxicated drivers:
We use the value of p=0.45 and n=10 and (1-p)=0.55:
[tex]P(X\geq 3)=1-P(X\leq 2)\\\\=1-[{10\choose 0}0.45^0(0.55)^{10}+{10\choose1}0.45^1(0.55)^9+{10\choose2}0.45^2(0.55)^8]\\\\=1-[0.0025+0.0207+0.0763]\\\\=0.9005[/tex]
Hence, the probability off at least three intoxicated drivers is 0.9005
iii. The probability of at most three intoxicated drivers:
-This is calculated the probabilities of between 0 to 3 as:
[tex]P(X\geq 3)=P(X=0)+P(X=1)+P(X=2)\\\\\\={10\choose0}0.45^0(0.55)^{10}+{10\choose1}0.45^1(0.55)^{9}+{10\choose2}0.45^2(0.55)^{8}\\\\\\=0.0025+0.0207+0.0763\\\\\\\\=0.0995[/tex]
Hence, the probability off at most three intoxicated drivers 0.0995
b. The probability of between tow and four, inclusive, is calculated by summing the exact probabilities of x=2,3 and x=4:
[tex]P(2\leq X\leq 4)=P(X=2)+P(X=3)+P(X=4)\\\\={10\choose2}0.45^2(0.55)^{8}+{10\choose3}0.45^3(0.55)^{7}+{10\choose4}0.45^4(0.55)^{6}\\\\=0.0763++0.1665+0.2384\\\\=0.4812[/tex]
Hence, the probability of between 2 and 4 is 0.4812
c. The mean of the binomial random variable is calculated as:
[tex]\mu=E(Y)=np, \ \ n=10, p=0.45\\\\=10\times 0.45\\\\4.5[/tex]
Hence, the mean of random variable Y is 4.5
d.Given that n=10 and p=0.45, the standard deviation of the binomial random variable Y is calculated as:
[tex]\sigma=\sqrt{np(1-p)}\\\\\\=\sqrt{0.45\times 0.55\times 10}\\\\\\=1.5732[/tex]
Hence, the standard deviation of the random variable Y is 1.5732
