Question

A binomial experiment has the given number of trials N and the given success probability P.
n= 10, p= 0.3
Determine the probability P(5). Round the answer to at least 3 decimal places.

Answer 1

Incorrect

Answer 2

0.102 is correct
Now (b)Find the mean. Round the answer to two decimal places.

Answer 3

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

Answer 4

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

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

Where:
- \( n \) = number of trials
- \( k \) = number of successes
- \( p \) = probability of success on each trial
- \( \binom{n}{k} \) = binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)

In this case, we have:
- \( n = 10 \)
- \( k = 5 \)
- \( p = 0.3 \)

Now let's calculate \( P(X = 5) \).

1. Calculate the binomial coefficient \( \binom{10}{5} \):

\[
\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]

2. Calculate \( p^5 \):

\[
p^5 = (0.3)^5 = 0.00243
\]

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

\[
(1 - p)^{n - k} = (0.7)^{5} = 0.16807
\]

4. Combine all parts to find \( P(X = 5) \):

\[
P(X = 5) = \binom{10}{5} \cdot p^5 \cdot (1 - p)^{5}
\]

\[
P(X = 5) = 252 \cdot 0.00243 \cdot 0.16807
\]

Now multiply these values:

\[
P(X = 5) \approx 252 \cdot 0.00243 \cdot 0.16807 \approx 0.091
\]

Thus, the probability \( P(5) \) rounded to at least three decimal places is:

\[
\boxed{0.091}
\]

Answer 5

Let's carefully recalculate the probability \( P(X = 5) \) for a binomial distribution with \( n = 10 \) and \( p = 0.3 \).

Using the binomial probability formula:

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

Where:
- \( n = 10 \)
- \( k = 5 \)
- \( p = 0.3 \)
- \( 1 - p = 0.7 \)

1. **Calculate the binomial coefficient \( \binom{10}{5} \)**:

\[
\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]

2. **Calculate \( p^5 \)**:

\[
p^5 = (0.3)^5 = 0.00243
\]

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

\[
(1 - p)^{n - k} = (0.7)^{5}
\]

Now, calculate \( (0.7)^5 \):

\[
(0.7)^5 = 0.16807
\]

4. **Combine these to find \( P(X = 5) \)**:

\[
P(X = 5) = \binom{10}{5} \cdot (0.3)^5 \cdot (0.7)^{5} = 252 \cdot 0.00243 \cdot 0.16807
\]

Now, calculate the multiplication step-by-step:

- First multiply \( 252 \cdot 0.00243 \):

\[
252 \cdot 0.00243 \approx 0.61176
\]

- Then multiply this result by \( 0.16807 \):

\[
0.61176 \cdot 0.16807 \approx 0.102
\]

Finally, rounding to three decimal places gives:

\[
P(X = 5) \approx 0.102
\]

Thus, the final answer is:

\[
\boxed{0.102}
\]

Thank you for your patience, and I apologize for the previous miscalculation!

Answer 6

The mean \( \mu \) of a binomial distribution can be calculated using the formula:

\[
\mu = n \cdot p
\]

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

In this case:
- \( n = 10 \)
- \( p = 0.3 \)

Now we can calculate the mean:

\[
\mu = 10 \cdot 0.3 = 3
\]

Thus, the mean of the binomial distribution, rounded to two decimal places, is:

\[
\boxed{3.00}
\]

Answer 7

The variance \( \sigma^2 \) of a binomial distribution can be calculated using the formula:

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

Where:
- \( n \) is the number of trials,
- \( p \) is the probability of success,
- \( 1 - p \) is the probability of failure.

Given:
- \( n = 10 \)
- \( p = 0.3 \)

Now, we can calculate the variance:

\[
\sigma^2 = 10 \cdot 0.3 \cdot (1 - 0.3) = 10 \cdot 0.3 \cdot 0.7
\]

Calculating it step-by-step:

1. Calculate \( 10 \cdot 0.3 = 3 \).
2. Then calculate \( 3 \cdot 0.7 = 2.1 \).

Thus, the variance is:

\[
\sigma^2 = 2.1
\]

Rounding the variance to two decimal places, we have:

\[
\text{Variance} = \boxed{2.10}
\]

Next, we can find the standard deviation \( \sigma \):

The standard deviation is the square root of the variance:

\[
\sigma = \sqrt{\sigma^2} = \sqrt{2.1}
\]

Calculating \( \sqrt{2.1} \):

\[
\sigma \approx 1.449
\]

Rounding this to at least three decimal places gives:

\[
\text{Standard Deviation} = \boxed{1.449}
\]