A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ($\pi$=0.25). To apply the normal distribution to approximate the binomial, what are the mean and standard deviation?
A) µ = 100 and $\sigma$= 0.25
B) µ =25 and $\sigma$ = 100
C) µ =25 and $\sigma$= 4.33
D) µ =100 and $\sigma$ = 2500

Question

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ($\pi$=0.25). To apply the normal distribution to approximate the binomial, what are the mean and standard deviation?
A) µ = 100 and  $\sigma$= 0.25
B) µ =25 and  $\sigma$ = 100
C) µ =25 and  $\sigma$= 4.33
D) µ =100 and  $\sigma$ = 2500

Quizplus · Accepted Answer

To approximate a binomial distribution using normal distribution, we need to calculate the mean (µ) and standard deviation (σ) using the following formulas:
- µ = n * p
- σ = sqrt(n * p * (1 - p))

In this case, n = 100 and p = 0.25, so:
- µ = 100 * 0.25 = 25
- σ = sqrt(100 * 0.25 * (1 - 0.25)) = 4.33 (rounded to 2 decimal places)

Therefore, the best choice is C, which has µ = 25 and σ = 4.33.

A Binomial Distribution Has 100 Trials (N = 100) with a Probability