Questions: Practice E5 Score: 4/10 Answered: 2/5 Question 3 Assume that a procedure yields a binomial distribution with a trial repeated n = 5 times. Use some form of the probability distribution given the probability p = 0.682 of success on a single trial. (Report answers accurate to 4 decimal places.) P(X ≤ 1)

Solution Steps

To solve this problem, we need to calculate the cumulative probability of getting at most 1 success in a binomial distribution with 5 trials and a success probability of 0.682. We will use the cumulative distribution function (CDF) for the binomial distribution to find this probability.

We are given a binomial distribution with parameters \( n = 5 \) (number of trials) and \( p = 0.682 \) (probability of success on a single trial). We need to find the probability that the number of successes \( X \) is less than or equal to 1, i.e., \( P(X \leq 1) \).

The cumulative distribution function (CDF) for a binomial distribution gives the probability that the random variable \( X \) is less than or equal to a certain value. We calculate \( P(X \leq 1) \) using the CDF:

\[ P(X \leq 1) = \sum_{k=0}^{1} \binom{n}{k} p^k (1-p)^{n-k} \]

Substituting the given values into the formula:

\[ P(X \leq 1) = \binom{5}{0} (0.682)^0 (1-0.682)^5 + \binom{5}{1} (0.682)^1 (1-0.682)^4 \]

Calculating each term:

• For \( k = 0 \): \[ \binom{5}{0} (0.682)^0 (0.318)^5 = 1 \times 1 \times 0.0032 = 0.0032 \]

• For \( k = 1 \): \[ \binom{5}{1} (0.682)^1 (0.318)^4 = 5 \times 0.682 \times 0.0102 = 0.0349 \]

Adding these probabilities gives:

\[ P(X \leq 1) = 0.0032 + 0.0349 = 0.0381 \]

Final Answer