Answer:
a. P(X = 0)= 0.001
b. P(X = 1)= 0.001
c. P(X=2)= 0.044
d. P(X=3)= 0.117
e. P(X=4)= 0.205
f. P(X=5)= 0.246
g. P(X=6)= 0.205
h. P(X=7)= 0.117
i. P(X=8)= 0.044
j. P(X=9)= 0.001
k. P(X=10)= 0.001
Step-by-step explanation:
Hello!
You have the variable X with binomial distribution, the probability of success is 0.5 and the sample size is n= 10 (I suppose)
If the probability of success p=0.5 then the probability of failure is q= 1 - p= 1 - 0.5 ⇒ q= 0.5
You are asked to calculate the probabilities for each observed value of the variable. In this case is a discrete variable with definition between 0 and 10.
You have two ways of solving this excersice
1) Using the formula
[tex]P(X)= \frac{n!}{(n-X)!X!} * (p)^X * (q)^{n-X}[/tex]
2) Using a table of cummulative probabilities of the binomial distribution.
a. P(X = 0)
Formula:
[tex]P(X=0)= \frac{10!}{(10-0)!0!} * (0.5)^0 * (0.5)^{10-0}[/tex]
P(X = 0) = 0.00097 ≅ 0.001
Using the table:
P(X = 0) = P(X ≤ 0) = 0.0010
b. P(X = 1)
Formula
[tex]P(X=1)= \frac{10!}{(10-1)!1!} * (0.5)^1 * (0.5)^{10-1}[/tex]
P(X = 1) = 0.0097 ≅ 0.001
Using table:
P(X = 1) = P(X ≤ 1) - P(X ≤ 0) = 0.0107-0.0010= 0.0097 ≅ 0.001
c. P(X=2)
Formula
[tex]P(X=2)= \frac{10!}{(10-2)!2!} * (0.5)^2 * (0.5)^{10-2}[/tex]
P(X = 2) = 0.0439 ≅ 0.044
Using table:
P(X = 2) = P(X ≤ 2) - P(X ≤ 1) = 0.0547 - 0.0107= 0.044
d. P(X = 3)
Formula
[tex]P(X = 3)= \frac{10!}{(10-3)!3!} * (0.5)^3 * (0.5)^{10-3}[/tex]
P(X = 3)= 0.11718 ≅ 0.1172
Using table:
P(X = 3) = P(X ≤ 3) - P(X ≤ 2) = 0.1719 - 0.0547= 0.1172
e. P(X = 4)
Formula
[tex]P(X = 4)= \frac{10!}{(10-4)!4!} * (0.5)^4 * (0.5)^{10-4}[/tex]
P(X = 4)= 0.2051
Using table:
P(X = 4) = P(X ≤ 4) - P(X ≤ 3) = 0.3770 - 0.1719= 0.2051
f. P(X = 5)
Formula
[tex]P(X = 5)= \frac{10!}{(10-5)!5!} * (0.5)^5 * (0.5)^{10-5}[/tex]
P(X = 5)= 0.2461 ≅ 0.246
Using table:
P(X = 5) = P(X ≤ 5) - P(X ≤ 4) = 0.6230 - 0.3770= 0.246
g. P(X = 6)
Formula
[tex]P(X = 6)= \frac{10!}{(10-6)!6!} * (0.5)^6 * (0.5)^{10-6}[/tex]
P(X = 6)= 0.2051
Using table:
P(X = 6) = P(X ≤ 6) - P(X ≤ 5) = 0.8281 - 0.6230 = 0.2051
h. P(X = 7)
Formula
[tex]P(X = 7)= \frac{10!}{(10-7)!7!} * (0.5)^7 * (0.5)^{10-7}[/tex]
P(X = 7)= 0.11718 ≅ 0.1172
Using table:
P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.9453 - 0.8281= 0.1172
i. P(X = 8)
Formula
[tex]P(X = 8)= \frac{10!}{(10-8)!8!} * (0.5)^8 * (0.5)^{10-8}[/tex]
P(X = 8)= 0.0437 ≅ 0.044
Using table:
P(X = 8) = P(X ≤ 8) - P(X ≤ 7) = 0.9893 - 0.9453= 0.044
j. P(X = 9)
Formula
[tex]P(X = 9)= \frac{10!}{(10-9)!9!} * (0.5)^9 * (0.5)^{10-9}[/tex]
P(X = 9)=0.0097 ≅ 0.001
Using table:
P(X = 9) = P(X ≤ 9) - P(X ≤ 8) = 0.999 - 0.9893= 0.001
k. P(X = 10)
Formula
[tex]P(X = 10)= \frac{10!}{(10-10)!10!} * (0.5)^{10} * (0.5)^{10-10}[/tex]
P(X = 10)= 0.00097 ≅ 0.001
Using table:
P(X = 10) = P(X ≤ 10) - P(X ≤ 9) = 1 - 0.9990= 0.001
Note: since 10 is the max number this variable can take, the cummulated probability until it is 1.
I hope it helps!
