Question

A binomial experiment has the given number of trials N and the given success probability P.
n= 11, p= 0.1

Determine the probability P(4). Round the answer to at least 3 decimal places.

Answer 1

Find the mean .Round the answer to 2 decimal places.

Answer 2

(c)Find the variance and standard deviation. Round the variance to two decimal places and standard deviation to at least three decimal places.

Answer 3

To determine the probability \( P(X = k) \) in a binomial experiment, we can use the binomial probability formula:

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

Where:
- \( n \) is the number of trials,
- \( k \) is the number of successes (in this case, \( k = 4 \)),
- \( p \) is the probability of success (in this case, \( p = 0.1 \)),
- \( \binom{n}{k} \) is the binomial coefficient, which can be calculated as \( \frac{n!}{k!(n-k)!} \).

Given:
- \( n = 11 \)
- \( k = 4 \)
- \( p = 0.1 \)

Let's calculate \( P(X = 4) \):

1. Calculate the binomial coefficient \( \binom{11}{4} \):

\[
\binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4! \cdot 7!}
\]

Calculating \( 11! \), \( 4! \), and \( 7! \):

\[
= \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \frac{7920}{24} = 330
\]

2. Calculate \( p^k \):

\[
p^4 = (0.1)^4 = 0.0001
\]

3. Calculate \( (1-p)^{n-k} \):

\[
(1-p)^{7} = (0.9)^7
\]

Now, calculating \( (0.9)^7 \):

\[
(0.9)^7 \approx 0.4782969
\]

4. Now, combine these results into the binomial probability formula:

\[
P(X = 4) = \binom{11}{4} (0.1)^4 (0.9)^7
\]
\[
P(X = 4) = 330 \times 0.0001 \times 0.4782969
\]

Calculating \( 330 \times 0.0001 \):

\[
= 0.033
\]

Then, finally multiply:

\[
P(X = 4) = 0.033 \times 0.4782969 \approx 0.0158
\]

5. Round to at least three decimal places:

\[
P(X = 4) \approx 0.016
\]

Thus, the probability \( P(X = 4) \) is approximately **0.016**.

Answer 4

In a binomial experiment, the mean (or expected value) \( \mu \) can be calculated using the formula:

\[
\mu = n \cdot p
\]

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

Given:
- \( n = 11 \)
- \( p = 0.1 \)

Now we can calculate the mean:

\[
\mu = 11 \cdot 0.1 = 1.1
\]

Rounding to two decimal places, the mean is:

\[
\mu \approx 1.10
\]

Thus, the mean of the binomial experiment is **1.10**.

Answer 5

In a binomial experiment, the variance \( \sigma^2 \) and standard deviation \( \sigma \) can be calculated using the following formulas:

1. **Variance** \( \sigma^2 \):

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

2. **Standard Deviation** \( \sigma \):

\[
\sigma = \sqrt{\sigma^2}
\]

### Given:
- \( n = 11 \)
- \( p = 0.1 \)

### Step 1: Calculate Variance

\[
\sigma^2 = n \cdot p \cdot (1 - p) = 11 \cdot 0.1 \cdot (1 - 0.1)
\]

Calculate \( 1 - p \):

\[
1 - 0.1 = 0.9
\]

Now substitute back:

\[
\sigma^2 = 11 \cdot 0.1 \cdot 0.9
\]

Calculating:

\[
\sigma^2 = 11 \cdot 0.1 \cdot 0.9 = 11 \cdot 0.09 = 0.99
\]

Round the variance to two decimal places:

\[
\sigma^2 \approx 0.99
\]

### Step 2: Calculate Standard Deviation

Now calculate the standard deviation:

\[
\sigma = \sqrt{\sigma^2} = \sqrt{0.99}
\]

Calculating the square root:

\[
\sigma \approx 0.995
\]

Round the standard deviation to at least three decimal places:

\[
\sigma \approx 0.995
\]

### Final Results
- Variance \( \sigma^2 \) is approximately **0.99** (rounded to two decimal places).
- Standard deviation \( \sigma \) is approximately **0.995** (rounded to at least three decimal places).