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For a binomial distribution with n = 124 and p = 0.24, what is the mean, variance, and standard deviation?

(a) Mean: 29.76, Variance: 22.618, Standard deviation: 4.756
(b) Mean: 29.770000000000003, Variance: 22.618, Standard deviation: 4.756
(c) Mean: 29.76, Variance: 22.628, Standard deviation: 4.756
(d) Mean: 29.76, Variance: 22.618, Standard deviation: 4.766

1 Answer

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Final answer:

The mean, variance, and standard deviation of a binomial distribution with n = 124 and p = 0.24 are 29.76, 22.5984, and 4.7542, respectively. The correct answer (a) has a slight typo in the variance value.

Step-by-step explanation:

To find the mean, variance, and standard deviation for a binomial distribution with parameters n = 124 and p = 0.24, we can use the shortcut formulas: μ = np, σ² = npq, and σ = √(npq), where q is the probability of failure (1 - p).

In this case, q would be 1 - 0.24 = 0.76.

So,

  • Mean (µ) = np = 124 * 0.24 = 29.76
  • Variance (σ²) = npq = 124 * 0.24 * 0.76 = 22.5984
  • Standard Deviation (σ) = √(npq) = √(124 * 0.24 * 0.76) = 4.7542, which rounded to four decimal places becomes 4.7542.

Comparing the results to options (a) through (d), the closest correct answer is (a) Mean: 29.76, Variance: 22.618, Standard deviation: 4.756, albeit there is a slight discrepancy in the variance value due to rounding. The actual calculated variance is 22.5984, which rounds to 22.60, indicating a minor typo in all options presented.

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