Questions: Based on a survey, assume that 36% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones. Identify the values of n, x, p, and q. The value of n is

Based on a survey, assume that 36% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones. Identify the values of n, x, p, and q.

The value of n is
Transcript text: Based on a survey, assume that $36 \%$ of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones. Identify the values of $n, x, p$, and $q$. The value of $n$ is $\square$
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Solution

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Solution Steps

Step 1: Identify Parameters

We are given the following parameters for our binomial distribution analysis:

  • Number of trials \( n = 5 \)
  • Number of successes \( x = 3 \)
  • Probability of success \( p = 0.36 \)
  • Probability of failure \( q = 1 - p = 0.64 \)
Step 2: Calculate Probability

Using the binomial probability formula:

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

we find that the probability of exactly 3 consumers being comfortable with drone delivery is:

\[ P(X = 3) = \binom{5}{3} \cdot (0.36)^3 \cdot (0.64)^{5-3} = 0.1911 \]

Step 3: Calculate Statistical Measures

Next, we calculate the key statistical measures for the binomial distribution:

  • Mean (expected value) \( \mu \):

\[ \mu = n \cdot p = 5 \cdot 0.36 = 1.8 \]

  • Variance \( \sigma^2 \):

\[ \sigma^2 = n \cdot p \cdot q = 5 \cdot 0.36 \cdot 0.64 = 1.152 \]

  • Standard Deviation \( \sigma \):

\[ \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{5 \cdot 0.36 \cdot 0.64} = 1.0733 \]

Final Answer

The results of the analysis are as follows:

  • Probability of exactly 3 successes: \( P(X = 3) = 0.1911 \)
  • Mean (expected value): \( \mu = 1.8 \)
  • Variance: \( \sigma^2 = 1.152 \)
  • Standard Deviation: \( \sigma = 1.0733 \)

Thus, the final boxed answer is:

\[ \boxed{P(X = 3) = 0.1911, \mu = 1.8, \sigma^2 = 1.152, \sigma = 1.0733} \]

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