The Quartile Deviation of the normal distribution f(x)=\frac{1}\sqrt{18 \pi} e \frac{-(x-10)^{2}}{18}, -\infty

Theoretical Distribution

Question : 112 of 118

Marks: +1, -0

The Quartile Deviation of the normal distribution

f(x)=\;{1}/{√{18 π}} e \;{-(x-10)^{2}}/{18}

-∞ < x < ∞

Go to Question:

Prev Question

Next Question