Difference between revisions of "Probability Norm Exercises"
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<ol> | <ol> | ||
| − | <li><p> | + | <li><p><math>L1,L2</math> If <math>X\sim N(0,1)</math> evaluate</p> |
<ol> | <ol> | ||
| − | <li><p><math>\Pr (X\leq 0.23)</math></p></li> | + | <li><p><math>\Pr (X\leq 0.23)</math> Solution: 0.5910</p></li> |
| − | <li><p><math>\Pr (X\geq 0.23)</math></p></li> | + | <li><p><math>\Pr (X\geq 0.23)</math> 0.4090</p></li> |
| − | <li><p><math>\Pr (-0.5 \leq X \leq 1.84)</math></p></li></ol> | + | <li><p><math>\Pr (-0.5 \leq X \leq 1.84)</math> 0.6586</p></li></ol> |
</li> | </li> | ||
| − | <li><p> | + | <li><p><math>L1,L2</math> Find the number <math>z_{0}</math> such that if <math>Z\sim N(0,1)</math></p> |
<ol> | <ol> | ||
| − | <li><p><math>\Pr (Z\geq z_{0})=0.05</math></p></li> | + | <li><p><math>\Pr (Z\geq z_{0})=0.05</math> 1.645</p></li> |
| − | <li><p><math>\Pr (Z<-z_{0})=0.025</math></p></li> | + | <li><p><math>\Pr (Z<-z_{0})=0.025</math> 1.96</p></li> |
| − | <li><p><math>\Pr (-z_{0}<Z\leq z_{0})=0.95</math></p></li></ol> | + | <li><p><math>\Pr (-z_{0}<Z\leq z_{0})=0.95</math> 1.96</p></li></ol> |
<p>and check your answers using EXCEL.</p></li> | <p>and check your answers using EXCEL.</p></li> | ||
| − | <li><p> | + | <li><p><math>L1,L2</math> If <math>X\sim N(4,0.16)</math> evaluate</p> |
<ol> | <ol> | ||
| − | <li><p><math>\Pr (X\geq 4.2)</math></p></li> | + | <li><p><math>\Pr (X\geq 4.2)</math> 0.3085</p></li> |
| − | <li><p><math>\Pr (3.9<X\leq 4.3)</math></p></li> | + | <li><p><math>\Pr (3.9<X\leq 4.3)</math> 0.3721</p></li> |
| − | <li><p><math>\Pr \left( (X\leq 3.8)\cup (X\geq 4.2)\right) </math></p></li></ol> | + | <li><p><math>\Pr \left( (X\leq 3.8)\cup (X\geq 4.2)\right)</math> 0.6170</p></li></ol> |
<p>and check your answers using EXCEL. (Note for part (c), define the “events” <math>A=\left( X\leq 3.8\right) </math> and <math>B=\left( X\geq 4.2\right) </math> and calculate <math>\Pr \left( A\cup B\right)</math>.</p></li></ol> | <p>and check your answers using EXCEL. (Note for part (c), define the “events” <math>A=\left( X\leq 3.8\right) </math> and <math>B=\left( X\geq 4.2\right) </math> and calculate <math>\Pr \left( A\cup B\right)</math>.</p></li></ol> | ||
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= Footnotes = | = Footnotes = | ||
Revision as of 12:05, 5 September 2014
Exercises
[math]L1,L2[/math] If [math]X\sim N(0,1)[/math] evaluate
[math]\Pr (X\leq 0.23)[/math] Solution: 0.5910
[math]\Pr (X\geq 0.23)[/math] 0.4090
[math]\Pr (-0.5 \leq X \leq 1.84)[/math] 0.6586
[math]L1,L2[/math] Find the number [math]z_{0}[/math] such that if [math]Z\sim N(0,1)[/math]
[math]\Pr (Z\geq z_{0})=0.05[/math] 1.645
[math]\Pr (Z\lt -z_{0})=0.025[/math] 1.96
[math]\Pr (-z_{0}\lt Z\leq z_{0})=0.95[/math] 1.96
and check your answers using EXCEL.
[math]L1,L2[/math] If [math]X\sim N(4,0.16)[/math] evaluate
[math]\Pr (X\geq 4.2)[/math] 0.3085
[math]\Pr (3.9\lt X\leq 4.3)[/math] 0.3721
[math]\Pr \left( (X\leq 3.8)\cup (X\geq 4.2)\right)[/math] 0.6170
and check your answers using EXCEL. (Note for part (c), define the “events” [math]A=\left( X\leq 3.8\right) [/math] and [math]B=\left( X\geq 4.2\right) [/math] and calculate [math]\Pr \left( A\cup B\right)[/math].
