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2

$begingroup$

$P(c le |Z|) = 0.016$

where Z is normally distributed .

I know that this means that either $Z ge c$ or $Z le -c$, but I'm not sure how to use this to find the value of c.

probability statistics

share|cite|improve this question

edited Mar 12 at 19:31

Elena Torre

asked Mar 12 at 19:03

Elena TorreElena Torre

596

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  where $Z$ is standard normal $mathcalN(0, 1)$ ?
  $endgroup$
  – ippiki-ookami
  Mar 12 at 19:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ippiki-ookami correct
  $endgroup$
  – Elena Torre
  Mar 12 at 19:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You want $c$ to be quantile 0.992 of the std normal dist'n. In R statistical software qnorm (without extra arguments) is the inverse CDF or quantile function of std. norm. In R, code qnorm(.992) returns $c = 2.408916.$ Then as in Answ (+1) of @callculus, the R code 2 - 2*pdf(c) returns 0.016, where pnorm is std norm CDF. You can get $c$ correct to about two decimal places (i.e, 2.41) using printed normal tables. I think this is a drill problem on use of such tables; try it.
  $endgroup$
  – BruceET
  Mar 12 at 22:52
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$P(c le |Z|) = 0.016$

where Z is normally distributed .

I know that this means that either $Z ge c$ or $Z le -c$, but I'm not sure how to use this to find the value of c.

probability statistics

share|cite|improve this question

edited Mar 12 at 19:31

Elena Torre

asked Mar 12 at 19:03

Elena TorreElena Torre

596

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  where $Z$ is standard normal $mathcalN(0, 1)$ ?
  $endgroup$
  – ippiki-ookami
  Mar 12 at 19:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ippiki-ookami correct
  $endgroup$
  – Elena Torre
  Mar 12 at 19:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You want $c$ to be quantile 0.992 of the std normal dist'n. In R statistical software qnorm (without extra arguments) is the inverse CDF or quantile function of std. norm. In R, code qnorm(.992) returns $c = 2.408916.$ Then as in Answ (+1) of @callculus, the R code 2 - 2*pdf(c) returns 0.016, where pnorm is std norm CDF. You can get $c$ correct to about two decimal places (i.e, 2.41) using printed normal tables. I think this is a drill problem on use of such tables; try it.
  $endgroup$
  – BruceET
  Mar 12 at 22:52
  
  

  
  
  
  

add a comment | 

$begingroup$

$P(c le |Z|) = 0.016$

where Z is normally distributed .

I know that this means that either $Z ge c$ or $Z le -c$, but I'm not sure how to use this to find the value of c.

probability statistics

share|cite|improve this question

edited Mar 12 at 19:31

Elena Torre

asked Mar 12 at 19:03

Elena TorreElena Torre

596

$endgroup$

share|cite|improve this question

edited Mar 12 at 19:31

Elena Torre

asked Mar 12 at 19:03

Elena TorreElena Torre

596

share|cite|improve this question

edited Mar 12 at 19:31

Elena Torre

asked Mar 12 at 19:03

Elena TorreElena Torre

596

asked Mar 12 at 19:03

Elena TorreElena Torre

596

asked Mar 12 at 19:03

Elena TorreElena Torre

596

2 Answers
2

active

oldest

votes

2

$begingroup$

Hint:

If $Z$ is $textsymmetric$ around $0$, then $P(Z>c)=P(Zleq -c)$, where $c>0$. These are the areas denoted by $4$ and $1$.

So you have to calculate $2cdot P(Z>c)$. Using the converse probability it becomes

$$P(c le |Z|))=2cdot (1-P(Zleq c))=2-2P(Zleq c)=0.016$$

Since $Z$ is standard normal distributed we have $2-2Phi( c)=0.016$

Now just solve for $c$.

share|cite|improve this answer

answered Mar 12 at 19:35

callculuscallculus

18.4k31428

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice presentation, except I think regions 1 and 4 should be smaller to fit this particular problem.
  $endgroup$
  – BruceET
  Mar 12 at 22:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @BruceET It´s just a shematic illustration, obviously.
  $endgroup$
  – callculus
  Mar 13 at 0:44
  

  
  
  
  

add a comment | 

2

$begingroup$

Comment continued:

In the figure below, each of the two areas under the standard normal density curve
outside the vertical dotted lines (at about $pm 2.41$) corresponds to probability 0.008.
So the total 'tail probability' is 0.016 as required.

share|cite|improve this answer

answered Mar 12 at 23:08

BruceETBruceET

36k71540

$endgroup$

add a comment | 

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2

$begingroup$

Hint:

If $Z$ is $textsymmetric$ around $0$, then $P(Z>c)=P(Zleq -c)$, where $c>0$. These are the areas denoted by $4$ and $1$.

So you have to calculate $2cdot P(Z>c)$. Using the converse probability it becomes

$$P(c le |Z|))=2cdot (1-P(Zleq c))=2-2P(Zleq c)=0.016$$

Since $Z$ is standard normal distributed we have $2-2Phi( c)=0.016$

Now just solve for $c$.

share|cite|improve this answer

answered Mar 12 at 19:35

callculuscallculus

18.4k31428

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice presentation, except I think regions 1 and 4 should be smaller to fit this particular problem.
  $endgroup$
  – BruceET
  Mar 12 at 22:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @BruceET It´s just a shematic illustration, obviously.
  $endgroup$
  – callculus
  Mar 13 at 0:44
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint:

If $Z$ is $textsymmetric$ around $0$, then $P(Z>c)=P(Zleq -c)$, where $c>0$. These are the areas denoted by $4$ and $1$.

So you have to calculate $2cdot P(Z>c)$. Using the converse probability it becomes

$$P(c le |Z|))=2cdot (1-P(Zleq c))=2-2P(Zleq c)=0.016$$

Since $Z$ is standard normal distributed we have $2-2Phi( c)=0.016$

Now just solve for $c$.

share|cite|improve this answer

answered Mar 12 at 19:35

callculuscallculus

18.4k31428

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice presentation, except I think regions 1 and 4 should be smaller to fit this particular problem.
  $endgroup$
  – BruceET
  Mar 12 at 22:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @BruceET It´s just a shematic illustration, obviously.
  $endgroup$
  – callculus
  Mar 13 at 0:44
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

If $Z$ is $textsymmetric$ around $0$, then $P(Z>c)=P(Zleq -c)$, where $c>0$. These are the areas denoted by $4$ and $1$.

So you have to calculate $2cdot P(Z>c)$. Using the converse probability it becomes

$$P(c le |Z|))=2cdot (1-P(Zleq c))=2-2P(Zleq c)=0.016$$

Since $Z$ is standard normal distributed we have $2-2Phi( c)=0.016$

Now just solve for $c$.

share|cite|improve this answer

answered Mar 12 at 19:35

callculuscallculus

18.4k31428

$endgroup$

share|cite|improve this answer

answered Mar 12 at 19:35

callculuscallculus

18.4k31428

answered Mar 12 at 19:35

callculuscallculus

18.4k31428

answered Mar 12 at 19:35

callculuscallculus

18.4k31428

2

$begingroup$

Comment continued:

In the figure below, each of the two areas under the standard normal density curve
outside the vertical dotted lines (at about $pm 2.41$) corresponds to probability 0.008.
So the total 'tail probability' is 0.016 as required.

share|cite|improve this answer

answered Mar 12 at 23:08

BruceETBruceET

36k71540

$endgroup$

add a comment | 

2

$begingroup$

Comment continued:

In the figure below, each of the two areas under the standard normal density curve
outside the vertical dotted lines (at about $pm 2.41$) corresponds to probability 0.008.
So the total 'tail probability' is 0.016 as required.

share|cite|improve this answer

answered Mar 12 at 23:08

BruceETBruceET

36k71540

$endgroup$

add a comment | 

$begingroup$

Comment continued:

In the figure below, each of the two areas under the standard normal density curve
outside the vertical dotted lines (at about $pm 2.41$) corresponds to probability 0.008.
So the total 'tail probability' is 0.016 as required.

share|cite|improve this answer

answered Mar 12 at 23:08

BruceETBruceET

36k71540

$endgroup$

share|cite|improve this answer

answered Mar 12 at 23:08

BruceETBruceET

36k71540

answered Mar 12 at 23:08

BruceETBruceET

36k71540

answered Mar 12 at 23:08

BruceETBruceET

36k71540