A market maker posts 400 independent quotes, and each is filled with probability 1/2. Using the normal approximation without continuity correction, what is the z-score of exactly 230 fills?

Learn / Library / Statistics

Fills on posted quotes

StatisticsEasy

A market maker posts $400$ independent quotes, and each is filled with probability $\frac{1}{2}$ . Using the normal approximation without continuity correction, what is the $z$ -score of exactly $230$ fills?

Checking an answer needs JavaScript. The hint and the solution open without it.

Hint

Find the mean and standard deviation of a binomial count first.

Solution

The number of fills is binomial, with mean $400 \times \frac{1}{2} = 200$ .
The variance is $400 \times \frac{1}{2} \times \frac{1}{2} = 100$ , so the standard deviation is $10$ .
The distance of $230$ from the mean is $230 - 200 = 30$ .
Therefore the $z$ -score is $30 \div 10 = 3$ .

Answer: $3$

Next question Back to the library