Subjective Type

How do you find the z-score for which 98% of the distribution's area lies between -z and z?

Solution

z = 2.33

Explanation:
You need to look this up from a z-score table or use a numerical implementation of the inverse normal distribution cumulative density function ( e.g. normsinv in Excel). Since you desire the 98% percent interval you desire 1% on each side of $$ \pm z,$$ look up 99% (0.99) for z to obtain this.

The closet value for 0.99 on the table gives z = 2.32 on the table (2.33 in Excel) this is your z score.


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