Suppose X ~N (3,5) . Between what x values does about 68% of the data lie?...

Suppose X ~N (3,5) .

Between what x values does about 68% of the data lie? (The range of x values is centered at the mean of the distribution, i.e. 3.)

Expert Solution

According to empirical rule, about 68% of the area lies between one standard deviation from the mean.

We are given

Mean = 3

Variance = 5

Standard deviation = SD = sqrt(5) = 2.236068

Mean - 1*SD = 3 - 1*2.236068 = 0.763932

Mean + 1*SD = 3 + 1*2.236068 = 5.236068

Lower X = 0.763932

Upper X = 5.236068

orchestra answered 2 years ago

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