The lifespan of zebras in a particular zoo are normally distributed the average zebra lives 20.5 years the standard deviation is 3.9 years estimate the probability of a zebra living longer than32.2 years

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

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step-by-step explanation:

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry