What is the probability that a randomly selected day from 2013 will have a Group Trades total within 57,000 of the mean?

Question

the mean is 3421439 and the standard deviation is 508638.08

Answer 1

Since this is an individual day and you know the population mean µ=3,421,439 and the population standard deviation σ=508,638.08 you will compute the z-score:

z=(µ ± 57,000) - µ) / σ

=((3,421,439 ± 57,000) - 3,421,439) / 508,638.08

= ± 57,000 / 508,638

= ± 0.112063982635981

z is a standard normal random variable.

this means we require the probability that a standard normal random variable has value’s between

[-0.112063982635981 , 0.112063982635981]

Pr(-0.1121 <= z <= 0.1121)=0.089176

The probability that group trading volume for a randomly selected day within the time frame of a year is 0.089176.