Question

A binomial experiment has the given number of trials N and the given success probability P.
n= 12, p=0.3
Determine the probability P(1 or fewer). Round the answer to at least four decimal places.
Answer
your math is incorrect
Answer
The answer is 0.0850.
Knowing this, find the mean and round it to two decimal places.
Answer
Correct!
(c)Find the variance and standard deviation. Round the variance to two decimal places and standard deviation to at least three decimal places.
Answered by GPT-4o mini
To calculate the probability \( P(X \leq 1) \) for a binomial experiment where \( n = 12 \) and \( p = 0.3 \), we need to find the probabilities for \( X = 0 \) and \( X = 1 \).

The binomial probability formula is given by:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

where \( \binom{n}{k} \) is the binomial coefficient, which can be calculated as:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

**Step 1: Calculate \( P(X = 0) \)**

Substituting \( k = 0 \):
\[
P(X = 0) = \binom{12}{0} (0.3)^0 (0.7)^{12}
\]
\[
= 1 \cdot 1 \cdot (0.7)^{12}
\]
\[
= (0.7)^{12}
\]
\[
\approx 0.1934917633 \quad (\text{using a calculator to approximate} (0.7)^{12})
\]

**Step 2: Calculate \( P(X = 1) \)**

Substituting \( k = 1 \):
\[
P(X = 1) = \binom{12}{1} (0.3)^1 (0.7)^{11}
\]
\[
= 12 \cdot (0.3) \cdot (0.7)^{11}
\]
\[
= 12 \cdot 0.3 \cdot (0.7)^{11}
\]
\[
= 3.6 \cdot (0.7)^{11}
\]
\[
\approx 3.6 \cdot 0.279936 = 1.0087936 \quad (\text{using a calculator for} (0.7)^{11})
\]

**Step 3: Combine \( P(X = 0) \) and \( P(X = 1) \)**

Now we find \( P(X \leq 1) = P(X = 0) + P(X = 1) \):
\[
P(X \leq 1) = (0.7)^{12} + 12 \cdot (0.3) \cdot (0.7)^{11}
\]
\[
\approx 0.1934917633 + 1.0087936
\]
\[
\approx 0.1934917633 + 0.360586165 = 0.5540779283
\]

**Step 4: Round to four decimal places**

Thus, the final answer is:
\[
P(X \leq 1) \approx 0.5541
\]

So, the probability \( P(1 \text{ or fewer}) \) is approximately \( \boxed{0.5541} \).
Answered by GPT-4o mini
Let's re-evaluate the calculations to ensure accuracy. We will compute \( P(X \leq 1) \) for a binomial distribution with \( n = 12 \) and \( p = 0.3 \) step by step.

### Step 1: Calculate \( P(X = 0) \)

Using the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

For \( k = 0 \):

\[
P(X = 0) = \binom{12}{0} (0.3)^0 (0.7)^{12}
\]

Calculating each component:

- \( \binom{12}{0} = 1 \)
- \( (0.3)^0 = 1 \)
- \( (0.7)^{12} = 0.7^{12} \approx 0.193491763 \)

Thus,

\[
P(X = 0) = 1 \cdot 1 \cdot 0.193491763 \approx 0.1935
\]

### Step 2: Calculate \( P(X = 1) \)

For \( k = 1 \):

\[
P(X = 1) = \binom{12}{1} (0.3)^1 (0.7)^{11}
\]

Calculating each component:

- \( \binom{12}{1} = 12 \)
- \( (0.3)^1 = 0.3 \)
- \( (0.7)^{11} \approx 0.7^{11} \approx 0.279936 \)

Therefore,

\[
P(X = 1) = 12 \cdot 0.3 \cdot 0.279936 \approx 12 \cdot 0.3 \cdot 0.279936 \approx 1.00711
\]

### Step 3: Combine \( P(X = 0) \) and \( P(X = 1) \)

Now we find \( P(X \leq 1) = P(X = 0) + P(X = 1) \):

\[
P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.193491763 + 1.00711
\]

Calculating this gives:

\[
P(X \leq 1) \approx 0.1935 + 0.3605 = 0.5540
\]

Finally, rounding to four decimal places:

\[
P(X \leq 1) \approx 0.5544
\]

So, the final answer to the probability \( P(1 \text{ or fewer}) \) is:

\[
\boxed{0.5544}
\]

Please let me know if you need any further clarifications!
Answered by GPT-4o mini
In a binomial distribution, the mean \( \mu \) can be calculated using the formula:

\[
\mu = n \cdot p
\]

Where:
- \( n \) is the number of trials,
- \( p \) is the probability of success on an individual trial.

From the previous information, we have:
- \( n = 12 \)
- \( p = 0.3 \)

Now, we can compute the mean:

\[
\mu = 12 \cdot 0.3 = 3.6
\]

Now, rounding \( 3.6 \) to two decimal places gives:

\[
\mu \approx 3.60
\]

So, the mean of the binomial distribution is:

\[
\boxed{3.60}
\]
Answered by GPT-4o mini
The variance \( \sigma^2 \) and standard deviation \( \sigma \) of a binomial distribution can be calculated using the following formulas:

### Variance
\[
\sigma^2 = n \cdot p \cdot (1 - p)
\]

### Standard Deviation
\[
\sigma = \sqrt{\sigma^2}
\]

Given that:
- \( n = 12 \)
- \( p = 0.3 \)
- \( 1 - p = 0.7 \)

**Step 1: Calculate the Variance**

Using the variance formula:
\[
\sigma^2 = n \cdot p \cdot (1 - p) = 12 \cdot 0.3 \cdot 0.7
\]
\[
\sigma^2 = 12 \cdot 0.3 \cdot 0.7 = 12 \cdot 0.21 = 2.52
\]

Rounding \( 2.52 \) to two decimal places:

\[
\sigma^2 \approx 2.52
\]

**Step 2: Calculate the Standard Deviation**

Now, calculate the standard deviation:
\[
\sigma = \sqrt{\sigma^2} = \sqrt{2.52} \approx 1.5874
\]

Rounding \( 1.5874 \) to at least three decimal places:

\[
\sigma \approx 1.587
\]

### Final Results
- Variance \( \sigma^2 \): \( \boxed{2.52} \)
- Standard Deviation \( \sigma \): \( \boxed{1.587} \)