Question

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal places.)

a. x=5
b. x <= 5
c. x>=6

Answer 1

`\mathbf{P(X=5) =0.0888}`

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Explanation:

The probability of a binomial mass distribution can be expressed with the formula:

`\mathtt{P(X=x) =(^(n)_(x) ) \ \pi^x \ (1-\pi)^(n-x)}`

`\mathtt{P(X=x) =((n!)/(x!(n-x)!) ) \ \pi^x \ (1-\pi)^(n-x)}`

where;

n = 8 and π = 0.36

For x = 5

The probability
`\mathtt{P(X=5) =((8!)/(5!(8-5)!) ) \ 0.36^5 \ (1-0.36)^(8-5)}`

`\mathtt{P(X=5) =((8!)/(5!(3)!) ) \ 0.36^5 \ (0.64)^(3)}`

`\mathtt{P(X=5) =((8 * 7 * 6 * 5!)/(5!(3)!) ) * \ 0.0060466 \ * 0.262144}`

`\mathtt{P(X=5) =((8 * 7 * 6 )/(3 * 2 * 1) ) * \ 0.0060466 \ * 0.262144}`

`\mathtt{P(X=5) =({8 * 7 } ) * \ 0.0060466 \ * 0.262144}`

`\mathtt{P(X=5) =0.0887645}`

`\mathbf{P(X=5) =0.0888}` to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)
`\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})`

`{P(x \leq 5) = ( (8!)/(0!(8!)) * (0.36)^0 * (1-0.36)^8 \ ) + (8!)/(1!(7!)) * (0.36)^1 * (1-0.36)^7 \ +`
`(8!)/(2!(6!)) * (0.36)^2 * (1-0.36)^6 \ + (8!)/(3!(5!)) * (0.36)^3 * (1-0.36)^5 + (8!)/(4!(4!)) * (0.36)^4 * (1-0.36)^4 \ + (8!)/(5!(3!)) * (0.36)^5 * (1-0.36)^3 \ )`

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )

P ( x ≥ 6) = 1 - 0.9707

P ( x ≥ 6) = 0.0293