When using the normal approximation to the binomial, what is the standard deviation used in the standardization formula when approximating a binomial distribution with 226 trials and p=0.25? Please submit your answer with 4 decimal places

To use the normal approximation to the binomial distribution, you need to calculate the standard deviation using the formula:

standard deviation (σ) = √(n * p * (1 - p))

where:
n = number of trials
p = probability of success in each trial

In this case, n = 226 and p = 0.25. Plugging the values into the formula, we get:

σ = √(226 * 0.25 * (1 - 0.25))

σ = √(226 * 0.25 * 0.75)

σ = √(42.75)

σ ≈ 6.5345

Therefore, the standard deviation used in the standardization formula is approximately 6.5345 (rounded to 4 decimal places).

To approximate a binomial distribution using the normal approximation, we need to find the mean (μ) and the standard deviation (σ) of the corresponding normal distribution.

The mean of a binomial distribution is calculated as the product of the number of trials (n) and the probability of success (p): μ = n * p.
In this case, n = 226 trials and p = 0.25, so the mean would be μ = 226 * 0.25 = 56.5.

The standard deviation of a binomial distribution is calculated using the formula: σ = √(n * p * (1 - p)).
Using the given values of n = 226 trials and p = 0.25, we can calculate the standard deviation as follows:
σ = √(226 * 0.25 * (1 - 0.25))
= √(226 * 0.25 * 0.75)
≈ √42.3750
≈ 6.5141 (rounded to four decimal places).

Therefore, the standard deviation used in the standardization formula when approximating a binomial distribution with 226 trials and p=0.25 is approximately 6.5141.