A binomial experiment has the given number of trials N and the given success probability P.

Question

n= 10, p= 0.3
Determine the probability P(5). Round the answer to at least 3 decimal places.

Answer 1

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 2

Incorrect

Answer 3

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 4

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

Answer 5

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 6

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 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:

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