Actually when data is skewed, interquartile range and median are better parameters of descriptive stats than mean and variation. IMO, this question is missing a point about normal distribution. Hence, the assumption about normal distribution is the only way to solve it. Otherwise, there is no way to establish a link between mean and standard deviation. For normal distribution though, when we divide the difference between a mean and any variable by a standard deviation we get the standard normal distribution with mean=0 and standard deviation=1
KarunMendiratta wrote:
greenmonomer wrote:
why did you consider it as normal distribution?
Because that is the only Distribution asked by ETS in GRE