Question

consider a binomial random variable with \\(n = 6\\) and \\(p = 0.8\\). let \\(x\\) be the number of successes in the sample. evaluate the probability. (round your answer to three decimal places.)

\\p(x \le 3)\\

you may need to use the appropriate appendix table to answer this question.

Explanation:

Identify the distribution parameters

Using the Discrete Random Variables knowledge point, we identify that \(x\) is a binomial random variable. The given parameters are:

• Number of trials: \(n = 6\)
• Probability of success: \(p = 0.8\)
• Probability of failure: \(q = 1 - p = 0.2\)

Set up the probability interval

Using the Probability Intervals knowledge point, we express the target cumulative probability \(P(x \le 3)\) as the sum of individual probability mass functions:

$$ P(x \le 3) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) $$

Calculate individual binomial probabilities

The binomial probability formula is:

$$ P(x = k) = \binom{n}{k} p^k (1-p)^{n-k} $$

We calculate each term for \(k = 0, 1, 2, 3\):

• For \(k = 0\):

$$ P(x = 0) = \binom{6}{0} (0.8)^0 (0.2)^6 = 1 \times 1 \times 0.000064 = 0.000064 $$

• For \(k = 1\):

$$ P(x = 1) = \binom{6}{1} (0.8)^1 (0.2)^5 = 6 \times 0.8 \times 0.00032 = 0.001536 $$

• For \(k = 2\):

$$ P(x = 2) = \binom{6}{2} (0.8)^2 (0.2)^4 = 15 \times 0.64 \times 0.0016 = 0.015360 $$

• For \(k = 3\):

$$ P(x = 3) = \binom{6}{3} (0.8)^3 (0.2)^3 = 20 \times 0.512 \times 0.008 = 0.081920 $$

Sum the probabilities

We add the calculated individual probabilities to find the cumulative probability:

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Round to three decimal places

Rounding \(0.09888\) to three decimal places gives:

$$ P(x \le 3) \approx 0.099 $$

Answer:

Consider a binomial random variable with \(n = 6\) and \(p = 0.8\). Let \(x\) be the number of successes in the sample. Evaluate the probability. (Round your answer to three decimal places.)

\(P(x \le 3) =\) <blank>0.099</blank>