Answer :
Answer:
a) Let X the random variable of interest "Number of students who have read the first Harry Potter book", on this case we now that:
[tex]X \sim Binom(n=12, p=0.64)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
b) [tex] E(X) = \mu_X = np = 12*0.64= 7.68[/tex]
c) [tex] Var(X) = \sigma_X = np(1-p) = 12*0.64(1-0.64)= 2.76[/tex]
d) [tex]P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988[/tex]
e) [tex] P(X\geq 2) =1- [0.00000474+0.000101]=0.9999[/tex]
f)[tex] P(X>5) = 1- P(X \leq 5)=0.9030 [/tex]
And since we want "NO more than 5" we can do this 1-0.9030= 0.097
g) [tex] P(2\leq X\leq 7) =0.4459[/tex]
h) [tex] P(X>7) = 1-P(X\leq 7)=1-0.4458=0.5541 [/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Part a
Let X the random variable of interest "Number of students who have read the first Harry Potter book", on this case we now that:
[tex]X \sim Binom(n=12, p=0.64)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part b
The expected value is given by the following expression:
[tex] E(X) = \mu_X = np = 12*0.64= 7.68[/tex]
Part c
The variance is given by the following expression:
[tex] Var(X) = \sigma_X = np(1-p) = 12*0.64(1-0.64)= 2.76[/tex]
Part d
For this case we want this probability
[tex]P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988[/tex]
Part e
For this case we want this probability:
[tex] P(X\geq 2) = 1-P(X<2) = 1-P(X\leq 1)=1- [P(X=0)+P(X=1)][/tex]
[tex]P(X=0)=(12C0)(0.64)^0 (1-0.64)^{12-0}=0.00000474[/tex]
[tex]P(X=1)=(12C1)(0.64)^1 (1-0.64)^{12-1}=0.000101[/tex]
[tex] P(X\geq 2) =1- [0.00000474+0.000101]=0.9999[/tex]
Part f
For this case we can find this probability first:
[tex] P(X>5) = 1- P(X \leq 5) [/tex]
[tex]P(X=0)=(12C0)(0.64)^0 (1-0.64)^{12-0}=0.00000474[/tex]
[tex]P(X=1)=(12C1)(0.64)^1 (1-0.64)^{12-1}=0.000101[/tex]
[tex]P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988[/tex]
[tex]P(X=3)=(12C3)(0.64)^3 (1-0.64)^{12-3}=0.0059[/tex]
[tex]P(X=4)=(12C4)(0.64)^4 (1-0.64)^{12-4}=0.0234[/tex]
[tex]P(X=5)=(12C5)(0.64)^5 (1-0.64)^{12-5}=0.0666[/tex]
[tex] P(X>5) = 1- P(X \leq 5)=0.9030 [/tex]
And since we want "NO more than 5" we can do this 1-0.9030= 0.097
Part g
For this case we can find this probability first:
[tex] P(2\leq X\leq 7) [/tex]
[tex]P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988[/tex]
[tex]P(X=3)=(12C3)(0.64)^3 (1-0.64)^{12-3}=0.0059[/tex]
[tex]P(X=4)=(12C4)(0.64)^4 (1-0.64)^{12-4}=0.0234[/tex]
[tex]P(X=5)=(12C5)(0.64)^5 (1-0.64)^{12-5}=0.0666[/tex]
[tex]P(X=6)=(12C6)(0.64)^6 (1-0.64)^{12-6}=0.1382[/tex]
[tex]P(X=7)=(12C7)(0.64)^7 (1-0.64)^{12-7}=0.2106[/tex]
Adding all the values we got:
[tex] P(2\leq X\leq 7) =0.4457[/tex]
Part h
For this case we can find this probability first:
[tex] P(X>7) = 1-P(X\leq 7) [/tex]
[tex]P(X=0)=(12C0)(0.64)^0 (1-0.64)^{12-0}=0.00000474[/tex]
[tex]P(X=1)=(12C1)(0.64)^1 (1-0.64)^{12-1}=0.000101[/tex]
[tex]P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988[/tex]
[tex]P(X=3)=(12C3)(0.64)^3 (1-0.64)^{12-3}=0.0059[/tex]
[tex]P(X=4)=(12C4)(0.64)^4 (1-0.64)^{12-4}=0.0234[/tex]
[tex]P(X=5)=(12C5)(0.64)^5 (1-0.64)^{12-5}=0.0666[/tex]
[tex]P(X=6)=(12C6)(0.64)^6 (1-0.64)^{12-6}=0.1382[/tex]
[tex]P(X=7)=(12C7)(0.64)^7 (1-0.64)^{12-7}=0.2106[/tex]
[tex] P(X>7) = 1-P(X\leq 7)=1-0.4458=0.5541 [/tex]