A. find the binomial probability p(x=5) where n=14 and p=0.30.
B. setup without solving the binomial probability p(x is at most 5) using probability notation C. how would you find normal approximation to the binomial probability p(x=5) in part A. please show how you would calculate mean and standard deviation in the form for normal approximation.
n=10, p=0.4, x=3
A. To find the binomial probability, p(x=5) where n=14 and p=0.30, we need to use the formula:
P(x=k) = C(n,k) * (p^k) * [(1-p)^(n-k)]
where:
- P(x=k) represents the probability of getting exactly k successes.
- C(n,k) is the number of ways to choose k successes from n trials, calculated using the combination formula.
In this case, we want to find P(x=5), so we substitute n=14, p=0.30, and k=5 into the formula:
P(x=5) = C(14, 5) * (0.30^5) * [(1-0.30)^(14-5)]
Calculating C(14, 5) may seem complicated, but fortunately many calculators and statistical software packages have built-in functions to calculate combinations. For example, using a calculator or software, C(14, 5) would be calculated as:
C(14, 5) = 2002
Substituting this value, along with the other values, we can now calculate P(x=5):
P(x=5) = 2002 * (0.30^5) * [(1-0.30)^(14-5)]
B. To set up the binomial probability p(x is at most 5) using probability notation, we use the notation:
P(x ≤ 5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5)
This represents the probability of getting 5 or fewer successes.
C. To find the normal approximation to the binomial probability P(x=5) (part A), we can use the normal distribution as an approximation when the sample size is large and the probability of success is not too close to 0 or 1.
For a binomial distribution, the formula for the mean (μ) is:
μ = n * p
And the formula for the standard deviation (σ) is:
σ = sqrt(n * p * (1-p))
In this case, n=14 and p=0.30. Substituting these values, we can calculate μ and σ:
μ = 14 * 0.30
σ = sqrt(14 * 0.30 * (1-0.30))
Therefore:
μ = 4.2
σ = sqrt(2.94) ≈ 1.71
So, for the normal approximation, we would approximate P(x=5) using the normal distribution with μ=4.2 and σ=1.71.