With n=13 and p=0.7, find the binomial
probability P(9)by using a binomial probability table. If np> and nq>5, also
estimate the indicated probability by using the normal distribution as an approximation to the binomial,if np<5 of
nq<5 then state that the normal approximation cannot be used.
Well, I hate to break it to you, but my humor can't help us with probability tables or normal distributions. But I can give you some information on how to solve this!
To find the binomial probability P(9), we can use the formula P(X=k) = nCk * p^k * (1-p)^(n-k), where n is the number of trials, p is the probability of success, and k is the number of successes.
Using n=13 and p=0.7, we can substitute those values into the formula to get P(9) = 13C9 * 0.7^9 * (1-0.7)^(13-9).
Now, let's check if np>5 and nq>5. Here, np = 13 * 0.7 = 9.1 and nq = 13 * (1-0.7) = 3.9. Both np and nq are greater than 5, so we can use the normal distribution as an approximation to the binomial.
Without any further calculations, I cannot provide you with an exact answer. So, for the estimation using the normal distribution as an approximation, you'll need to calculate the mean and standard deviation and then use the standard normal distribution table or a calculator to find the probability.
To find the binomial probability P(9) given n=13 and p=0.7, we can use a binomial probability table.
1. Binomial Probability Table Approach:
Using the binomial probability table, we need to find the probability associated with x = 9.
The binomial probability table provides the probabilities for a given value of x.
From the table, we find that when n=13 and p=0.7:
P(x=9) = 0.137
2. Normal Distribution Approximation Approach:
To determine whether we can use the normal distribution as an approximation to the binomial distribution, we need to check if both np>5 and nq>5.
Here,
n = 13
p = 0.7
q = 1 - p = 1 - 0.7 = 0.3
Calculating np and nq:
np = 13 * 0.7 = 9.1
nq = 13 * 0.3 = 3.9
Since np>5 (9.1 > 5) and nq>5 (3.9 > 5) are both false, the normal approximation cannot be used in this case.
Therefore, for this specific example, we would rely on the binomial probability table approach to find the probability P(9) accurately.
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.