Question

Use the binomial expression (p+q)^n to calculate a binomial distribution.n = 5 and p = 0.3I am so lost. Help please!

Answer 1

Answer:
`(p\text{ + q\rparen}^5\text{ = 1}`

Step-by-step explanation:

Given:

n = 5

p = 0.3

To find:

to use the binomial expression (p+q)^n to calculate a binomial distribution

The binomial expansion (p+q)^n can be written when we know the value of n

`\begin{gathered} To\text{ expand \lparen p+q\rparen}^n,\text{ we need to substitute n with the value given} \\ when\text{ n = 5} \\ (p+q)^n\text{ = \lparen p + q\rparen}^5 \\ To\text{ expand }(p\text{ + q\rparen}^5,\text{ the coefficient will be gotten using pascal's triangle} \\ For\text{ exponent of 5, the coeeficients are 1 5 10 10 5 1} \\ \\ The\text{ binomial expansion:} \\ (p\text{ + q\rparen}^5\text{ = p}^5\text{ + 5p}^(5-1)q^1+\text{ 10p}^(5-2)q^2\text{ + 10p}^(5-3)q^3+\text{ 5p}^(5-4)q^4\text{ + p}^(5-5)q^5 \\ NB:\text{ the sum of the exponents of each term will be equal to n = 5} \end{gathered}`
`(p+q)^5\text{ = p}^5\text{ + 5p}^4q\text{ + 10p}^3q^2\text{ + 10p}^2q^3\text{ + 5pq}^4\text{ + q}^5`
`\begin{gathered} Binomial\text{ distribution is written as:} \\ P(x\text{ =X\rparen= }^nC_xp^nq^(n-x) \end{gathered}`

p = 0.3

q = 1 - p = 1 - 0.3

q = 0.7

`\begin{gathered} (p+q)^5\text{ = p}^5\text{ + 5p}^4q\text{ + 10p}^3q^2\text{ + 10p}^2q^3\text{ + 5pq}^4\text{ + q}^5 \\ substitute\text{ the values of p and q:} \\ (p+q)^5\text{ = \lparen0.3\rparen}^5\text{ + 5\lparen0.3\rparen}^4(0.7)\text{ + 10\lparen0.3\rparen}^3(0.7)^2\text{ + 10\lparen0.3\rparen}^2(0.7)^3\text{ + 5\lparen0.3\rparen\lparen0.7\rparen}^4\text{ + \lparen0.7\rparen}^5 \\ \\ (p+q)^5\text{ = 1} \end{gathered}`