why did you consider it as normal distribution?

Because that is the only Distribution asked by ETS in GRE

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

True, but if this were a QC question, we'd couldn't assume a normal distribution.
On test day, if we're not told a distribution is normal, we can't assume it to be so.

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