Questions: Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X ≤ 14) n = 18, p = 0.8

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places.

P(X ≤ 14) n = 18, p = 0.8
Transcript text: Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X ≤ 14) n = 18, p = 0.8
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Solution

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Solution Steps

Step 1: Calculate Cumulative Probability

To find the cumulative probability \( P(X \leq 14) \) for a binomial distribution with parameters \( n = 18 \) and \( p = 0.8 \), we compute:

\[ P(X \leq 14) \approx 0.499 \]

Step 2: Calculate Probability of Exactly 14 Successes

Using the formula for the probability of exactly \( x \) successes in a binomial distribution:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

we find:

\[ P(X = 14) \approx 0.2153 \]

Step 3: Calculate Statistical Measures

The mean \( \mu \), variance \( \sigma^2 \), and standard deviation \( \sigma \) of the binomial distribution are calculated as follows:

  • Mean: \[ \mu = n \cdot p = 18 \cdot 0.8 = 14.4 \]

  • Variance: \[ \sigma^2 = n \cdot p \cdot q = 18 \cdot 0.8 \cdot 0.2 = 2.88 \]

  • Standard Deviation: \[ \sigma = \sqrt{npq} = \sqrt{18 \cdot 0.8 \cdot 0.2} \approx 1.6971 \]

Final Answer

The results are summarized as follows:

  • \( P(X \leq 14) \approx 0.499 \)
  • \( P(X = 14) \approx 0.2153 \)
  • Mean \( \mu = 14.4 \)
  • Variance \( \sigma^2 = 2.88 \)
  • Standard Deviation \( \sigma \approx 1.6971 \)

Thus, the final boxed answer is:

\[ \boxed{P(X \leq 14) \approx 0.499, \; P(X = 14) \approx 0.2153, \; \mu = 14.4, \; \sigma^2 = 2.88, \; \sigma \approx 1.6971} \]

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