Exercise · ZW-NORMAL-DISTRIBUTION-007 · ★★☆ Intermediate
If \(X\sim N(100,225)\), by the 68-95-99.7 rule, \(P(85\le X\le115)\) is approximately:
Choose one option.
Show answer
- C. \(0.68\)
Explanation
Here \(\sigma=15\), so the interval is \(\mu\pm\sigma\), which holds about 68% of the probability.
Review the concept: Normal distribution
A normal variable is a continuous variable with density \(f\) shaped like a symmetric bell, determined by its mean \(\mu\) and its standard deviation \(\sigma\).
\[f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}}\]
More practice problems: normal distribution
- ZW-NORMAL-DISTRIBUTION-001 If Xsim N(5,9), what is the standard deviation of X?
- ZW-NORMAL-DISTRIBUTION-002 If Xsim N(50,16), what is the standardized value z of x=58?
- ZW-NORMAL-DISTRIBUTION-003 If Z is standard normal, what is P(Z≤0)?
- ZW-NORMAL-DISTRIBUTION-004 If Xsim N(μ,1), at which point does the density reach its maximum?
- ZW-NORMAL-DISTRIBUTION-005 If Z is standard normal, what is f_Z(0), the height of the density at 0?
- ZW-NORMAL-DISTRIBUTION-006 If Xsim N(3,4), what distribution does Y=2X+1 have?
- ZW-NORMAL-DISTRIBUTION-008 If Z is standard normal, what is E[Z^2]?
- ZW-NORMAL-DISTRIBUTION-009 If Xsim N(μ,σ^2), approximately what is P(X≤μ+σ)?