Given mean =122.3 and standard deviation =18.5
[tex]z=\frac{x-mean}{standard deviation} = \frac{168.4-122.3}{18.5}[/tex]
[tex]= \frac{46.1}{18.5}[/tex]
≈2.49
As a general rule z-scores lower than -1.96 or higher than 1.96 are considered unusual.
Since our z-score = 2.49>1.96 , it is unusual.
Hence correct option is 'C' = "2.49 ; unusual"
Answer-
Z- score of the sample is 2.49, Unusual.
Solution-
We know the formula for calculating Z score,
[tex]Z=\frac{X- \mu}{\sigma}[/tex]
where,
X = raw score = 168.4
μ = mean of the sample = 122.3
σ = standard deviation of the sample = 18.5,
Putting all the values in the equation,
[tex]Z = \frac{168.4-122.3}{18.5} = 2.49[/tex]
∵ According to the rule, if the Z-score is lower than -1.96 or higher than 1.96 (or its absolute value greater than 1.96 ) then it is considered as unusual and interesting.
Here, as the value of Z is more than 1.96 which is 2.49, so it's unusual.
