Answer:
Step-by-step explanation:
Variance is the sum of squares of all items x from mean. Â i.e
Variance = [tex]\Sigma (x-xbar)^2[/tex]
Being the sum of squares variance can never be negative irrespective of mean being negative.
Std deviation is the square root of variances. Â This will thus be less than variance if variance >1 other wise std deviation will be bigger than variance.
Hence option d is right.
