To find the binomial probability P(9) using a binomial probability table, we need to use the formula for binomial probability:
P(x) = nCx * p^x * q^(n-x)
Where:
- n is the number of trials (in this case, n = 13)
- p is the probability of success on a single trial (p = 0.7)
- q is the probability of failure on a single trial (q = 1 - p)
- x is the number of successful outcomes (in this case, x = 9)
Step 1: Calculate the binomial probability manually
Using the formula, we can calculate P(9) as follows:
P(9) = 13C9 * (0.7)^9 * (0.3)^(13-9)
Now, we need to calculate 13C9 (read as "13 choose 9"):
13C9 = (13!)/(9!(13-9)!)
The exclamation mark (!) denotes the factorial of a number.
Calculate the factorials:
13! = 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
(13-9)! = 4 * 3 * 2 * 1
Plug in the values:
13C9 = (13!)/(9!(13-9)!)
= (13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)/((9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)(4 * 3 * 2 * 1))
Simplify the expression:
13C9 = 13 * 12 * 11 * 10 / (4 * 3 * 2 * 1)
= 715
Now substitute the values back into the formula for binomial probability:
P(9) = 715 * (0.7)^9 * (0.3)^4
= 0.2572 (rounded to four decimal places)
Step 2: Determine if the normal approximation can be used
To determine if the normal approximation can be used, we need to check if np > 5 and nq > 5.
np = 13 * 0.7
= 9.1
nq = 13 * (1 - 0.7)
= 3.9
Since both np and nq are greater than 5 (9.1 > 5 and 3.9 > 5), the normal approximation can be used.
Step 3: Estimate the probability using the normal distribution approximation
To estimate the probability using the normal distribution approximation, we need to use the mean and standard deviation of the binomial distribution, which are:
Mean (μ) = np
Standard Deviation (σ) = √(npq)
μ = 13 * 0.7
= 9.1
σ = √(13 * 0.7 * 0.3)
= 1.7777 (rounded to four decimal places)
Now, we can use the normal distribution with the estimated mean and standard deviation:
P(9) ≈ P(8.5 < x < 9.5)
≈ P((8.5 - 9.1)/1.7777 < (x - 9.1)/1.7777 < (9.5 - 9.1)/1.7777)
≈ P(-0.3377 < z < 0.1126)
To find this probability using the standard normal distribution table (also known as the Z-table), we need to find the z-scores corresponding to -0.3377 and 0.1126 and find the probability between those values.
-0.3377 corresponds to a z-score of approximately -0.63
0.1126 corresponds to a z-score of approximately 0.85
From the Z-table, the probability between -0.63 and 0.85 is approximately 0.2158
Therefore, P(9) ≈ 0.2158 (rounded to four decimal places).
In summary:
- The binomial probability P(9) is approximately 0.2572 when calculated using the binomial probability formula.
- The binomial normal approximation P(9) is approximately 0.2158 when calculated using the normal distribution approximation.