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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.

7 answers

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 \( \boxed{0.5541} \).

your math is incorrect

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!

The answer is 0.0850.
Knowing this, find the mean and round it to two decimal places.

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:

\[ \boxed{3.60} \]

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

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 \( \sigma \): \( \boxed{1.587} \)