Given a long algebraic equation, what are some strategies that you can use to make simplifying and evaluating the equation more
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Mean = 12000
Standard Deviation = 2000
We have to find how many standard deviations is 14,500 away from the mean.
This can be achieved by calculating the z-score
Z-score tells us how many standard deviations above or below is a sample value from the mean. A positive z value shows, sample value is above the mean.
z score can be calculated as = (Sample Value - Mean )/Standard Deviation)
So,
Z-score =
This means, 14,500 is 1.25 standard deviations above the mean value 12,000.
Standard Deviation = 2000
We have to find how many standard deviations is 14,500 away from the mean.
This can be achieved by calculating the z-score
Z-score tells us how many standard deviations above or below is a sample value from the mean. A positive z value shows, sample value is above the mean.
z score can be calculated as = (Sample Value - Mean )/Standard Deviation)
So,
Z-score =
This means, 14,500 is 1.25 standard deviations above the mean value 12,000.