Question: A binomial experiment has the given number of trials \( n \) and the given success probability…

A binomial experiment has the given number of trials \( n \) and the given success probability \( p \).

\( n = 15 \), \( p = 0.2 \)

Part: 0 / 3

Part 1 of 3

(a) Determine the probability \( P(7) \). Round the answer to at least three decimal places.

\( P(7) = \)

Part: 1 / 3

Part 2 of 3

(b) Find the mean. Round the answer to two decimal places.

The mean is

Part: 2 / 3

Part 3 of 3

(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 is

The standard deviation is

Solution

A binomial experiment is described with \( n = 15 \) and \( p = 0.2 \). (a) Determine the probability \( P(7) \). The probability mass function for a binomial distribution is given by: \[ P(x) = \binom{n}{x} p^x (1-p)^{n-x} \] Substitute \( n = 15 \), \( p = 0.2 \), and \( x = 7 \): \[ P(7) = \binom{15}{7} (0.2)^7 (0.8)^{8} \] Calculate \(\binom{15}{7}\): \[ \binom{15}{7} = \frac{15!}{7!(15-7)!} = 6435 \] Now calculate: \[ P(7) = 6435 \times (0.2)^7 \times (0.8)^8 \] \[ P(7) = 6435 \times 0.000128 \times 0.16777216 \] \[ P(7) \approx 0.03809 \] So the probability \( P(7) \approx 0.038 \). (b) Find the mean. The mean of a binomial distribution is given by: \[ \mu = n \cdot p \] Substituting the values: \[ \mu = 15 \cdot 0.2 = 3 \] The mean is 3.00. (c) Find the variance and standard deviation. The variance of a binomial distribution is given by: \[ \sigma^2 = n \cdot p \cdot (1-p) \] Substitute the values: \[ \sigma^2 = 15 \cdot 0.2 \cdot 0.8 = 2.4 \] The variance is 2.40. The standard deviation is the square root of the variance: \[ \sigma = \sqrt{2.4} \approx 1.54919 \] So the standard deviation is approximately 1.549.