Question

Given a mean of 61.2 and a standard deviation of 21.4, what is the z-score of the value 45 rounded to the nearest tenth?0.8−0.90.9−0.8

Answer 1

Given: Mean μ = 61.2 and standard deviation σ =21.4

The Z score of the value x=45 is given by

Z = $\frac{x - mean}{standard deviation}$

Z = $\frac{45 - 61.2 }{21.4}$

Z = $\frac{-16.2}{21.4}$

Z = -0.757

The z-score value for 45 rounded to nearest tenth is -0.8