Calculating Probability with Binomial Distribution

[karush]

4000
not sure what this means

If $$X \sim B(8,0.3)$$, find $$P(X=5)$$

looks like we working off of a probability distribution table

 

[MarkFL]

Re: notation question

This is not an area in which I am very knowledgeable, but I would say that the notation:

$$B(n,p)$$

implies:

$$P(X=k)={n \choose k}p^k(1-p)^{n-k}$$

Binomial distribution - Wikipedia, the free encyclopedia

 

[Nono713]

Yes, $\displaystyle X \sim B(8,0.3)$ simply means $X$ is a random variable following the distribution $B(8,0.3)$ ;)

 

[chisigma]

In general $X \sim B (\mu,\sigma)$ means that X is distributed according to the 'B' distribution law with mean $\mu$ and variance $\sigma$. If 'B' means 'Binomial' then is...

$\displaystyle P \{X= x\} = \binom{n}{k} p^{k}\ (1-p)^{n-k}\ k=0,1,...,n\ (1)$

... and is...

$\mu = n\ p$

$\sigma= n\ p\ (1-p)\ (2)$

The problem is that if is $\mu=8$ and $\sigma = .3$ is doesn't exist any combination of n and p satisfying (2) so that may be that 'B' doesn't mean 'Binomial' in this case...

Kind regards

$\chi$ $\sigma$

 

[karush]

Re: notation question

This is not an area in which I am very knowledgeable, but I would say that the notation:

$$B(n,p)$$

implies:

$$P(X=k)={n \choose k}p^k(1-p)^{n-k}$$

let me try this if $$X \sim B(8,0.3) $$ and $$P(X=5)$$
then plug in would be

$${8 \choose 5}0.3^5(1-0.3)^{1-0.3}=0.047$$

 

[karush]

In general $X \sim B (\mu,\sigma)$ means that X is distributed according to the 'B' distribution law with mean $\mu$ and variance $\sigma$. If 'B' means 'Binomial' then is...

$\displaystyle P \{X= x\} = \binom{n}{k} p^{k}\ (1-p)^{n-k}\ k=0,1,...,n\ (1)$

... and is...

$\mu = n\ p$

$\sigma= n\ p\ (1-p)\ (2)$

The problem is that if is $\mu=8$ and $\sigma = .3$ is doesn't exist any combination of n and p satisfying (2) so that may be that 'B' doesn't mean 'Binomial' in this case...

Kind regards

$\chi$ $\sigma$

well, the next problem is $$X \sim N(15,9)$$, find $$P(X < 16)$$ so assume B and N are just function names.

 

[MarkFL]

let me try this if $$X \sim B(8,0.3) $$ and $$P(X=5)$$
then plug in would be

$${8 \choose 5}0.3^5(1-0.3)^{1-0.3}=0.047$$

Yes, I get:

$$P(X=5)={8 \choose 5}\left(\frac{3}{10} \right)^5\left(\frac{7}{10} \right)^3=\frac{583443}{12500000}\approx0.04667544$$

well, the next problem is $$X \sim N(15,9)$$, find $$P(X < 16)$$ so assume B and N are just function names.

The normal distribution can be denoted by $N\left(\mu,\sigma^2 \right)$.

Do you know how to standardize the data and use your table to find $P(x<16)$?

 

[karush]

The normal distribution can be denoted by $N\left(\mu,\sigma^2 \right)$.

Do you know how to standardize the data and use your table to find $P(x<16)$?

no but assume the first step is solve for just $$x=16$$

thanks for all your help btw...

 

[I like Serena]

no but assume the first step is solve for just $$x=16$$

thanks for all your help btw...

Standardizing the data means calculating the so called $z$-value.
$$z=\frac{\text{boundary value of x} - \mu}{\sigma}$$
You can look up that value in a table of the standard normal distribution.
Do you have such a table?

 

[karush]

Standardizing the data means calculating the so called $z$-value.
$$z=\frac{\text{boundary value of x} - \mu}{\sigma}$$
You can look up that value in a table of the standard normal distribution.
Do you have such a table?

this is what was given:

If $$X \sim N (15,9),$$ Find
a) $$ P(X < 16)$$
b) $$P(X \geq 17)$$
c) $$P(14 \leq X \leq 15.5)$$
d) $$P(X > 6)$$
e) $$P(X>k)=0.34$$, then $$k=?$$

so from $$N\left(\mu,\sigma^2 \right)$$ and $$N (15,9)$$

$$z=\frac{\text{boundary value of x} - 15}{3}$$

so how do we get boundary value of x?

 

[I like Serena]

Let's start with:

If $$X \sim N (15,9),$$ Find
a) $$ P(X < 16)$$

so from $$N\left(\mu,\sigma^2 \right)$$ and $$N (15,9)$$

$$z=\frac{\text{boundary value of x} - 15}{3}$$

so how do we get boundary value of x?

For (a) the boundary value of x is 16.

So $$z = \frac{16 - 15}{3} = \frac 1 3$$.

Btw, I'm assuming your $\sigma^2=9$. There is a little ambiguity here, since it's also possible that your $\sigma=9$, which would of course give different results.

For instance from the wiki z-table, we can find that the corresponding lower tail probability is approximately 0.6293.

 

[karush]

Let's start with:
For (a) the boundary value of x is 16.

So $$z = \frac{16 - 15}{3} = \frac 1 3$$.

Btw, I'm assuming your $\sigma^2=9$. There is a little ambiguity here, since it's also possible that your $\sigma=9$, which would of course give different results.

For instance from the wiki z-table, we can find that the corresponding lower tail probability is approximately 0.6293.

from the given $$ X-N(15,9)$$ and from $$N(\mu,\sigma^2)$$ looks like $$\sigma$$ is $$3$$let me try the next one
b) $$P(X \geq 17)$$

so $$z = \frac{17 - 15}{3} = \frac {2}{3} \approx .6667$$

so from z table $$.7454$$

 

[I like Serena]

from the given $$ X-N(15,9)$$ and from $$N(\mu,\sigma^2)$$ looks like $$\sigma$$ is $$3$$

If your course is using $$N(\mu,\sigma^2)$$, which is also on wiki, that is is correct.
However, if your course is using $$N(\mu,\sigma)$$, it becomes a different matter.
Can you tell from your course notes?

let me try the next one
b) $$P(X \geq 17)$$

so $$z = \frac{17 - 15}{3} = \frac {2}{3} \approx .6667$$

so from z table $$.7454$$

Close.
However, in (a) we were talking about the lower tail probability.
In (b) we're talking about the upper tail probability.
The z-table you used gives the lower tail probability (see the graph next to it).
But you can calculate the upper tail probability by taking 1 minus the lower tail probability.

 

[karush]

If your course is using $$N(\mu,\sigma^2)$$, which is also on wiki, that is is correct.
However, if your course is using $$N(\mu,\sigma)$$, it becomes a different matter.
Can you tell from your course notes?

OK but, I am not taking a course in this nor do I have a text or notes just trying to do a work sheet from a "summer 2014 math SL IB Questions" will be studying the subject more formerly this fall just want to get head start on it.

Close.
However, in (a) we were talking about the lower tail probability.
In (b) we're talking about the upper tail probability.
The z-table you used gives the lower tail probability (see the graph next to it).
But you can calculate the upper tail probability by taking 1 minus the lower tail probability.

so for b) $$1-.7454 = .2546$$ are you referring to the bell shaped graph

for c) we have $$P(14 \leq X \leq 15.5)$$

$$z_u = \frac{14 - 15}{3} = - \frac {1}{3} \approx -.3334$$
I presume this is upper since $$14 \leq X$$

from z table $$-0.6293$$ however this was from a neg number

and $$z_l=\frac{15.5 - 15}{3} = \frac {.5}{3} \approx .16667$$

from z table $$0.5636$$

so $$-0.6293 \leq X \leq 0.5636 $$

this was a shot in the dark...

 

[karush]

so the last 2 questions

If $$X \sim N(15,9)$$ find

d) $$P(X > 6)$$
$$
z_u = \frac{6 - 15}{3} = 3$$
z table $$= 0.9987$$

e) $$P(X>k)=0.34$$, then $$k=?$$

$$\frac{k-15}{3}=0.34$$ then $$k=16.02$$

 

[MarkFL]

d) Correct.

e) Because 0.34 < 0.5, we know $k>\mu$. We want to find the $z$-score associated with an area of $0.5-0.34=0.16$. Once you find this $z$-score, add this many standard deviations to the mean:

$$k=\mu+z_k\sigma$$

What do you find?

 

[karush]

d) Correct.

e) Because 0.34 < 0.5, we know $k>\mu$. We want to find the $z$-score associated with an area of $0.5-0.34=0.16$. Once you find this $z$-score, add this many standard deviations to the mean:

$$k=\mu+z_k\sigma$$

What do you find?

where does the $$0.5$$ come from?

like this?

$$k=15+0.8554(3)=17.5662 $$

 

[MarkFL]

Please refer to the following diagram:

View attachment 979

We know the red area plus the green area is $0.5$. We want the green area to be $0.34$, thus the red area must be $0.5-0.34=0.16$.

What $z$-score is associated with an area of 0.16?

 

[karush]

Please refer to the following diagram:

View attachment 979

We know the red area plus the green area is $0.5$. We want the green area to be $0.34$, thus the red area must be $0.5-0.34=0.16$.

What $z$-score is associated with an area of 0.16?

from the Cumulative from mean (0 to Z) table I find $$0.06356$$

 

[MarkFL]

from the Cumulative from mean (0 to Z) table I find $$0.06356$$

You are finding the area associated with $z=0.16$. You want to use the table the other way around. (Happy)

 

[I like Serena]

so for b) $$1-.7454 = .2546$$ are you referring to the bell shaped graph

Good! And yes.

for c) we have $$P(14 \leq X \leq 15.5)$$

$$z_u = \frac{14 - 15}{3} = - \frac {1}{3} \approx -.3334$$
I presume this is upper since $$14 \leq X$$

from z table $$-0.6293$$ however this was from a neg number

and $$z_l=\frac{15.5 - 15}{3} = \frac {.5}{3} \approx .16667$$

from z table $$0.5636$$

so $$-0.6293 \leq X \leq 0.5636 $$

this was a shot in the dark...

You're getting there...

You found that $P(X < 14) \approx -0.6293$ and that $P(X \leq 15.5) \approx .16667$.
Both are lower tail probabilities and you actually want the area in between in the bell curve.
To get it, you can simply subtract them from each other, effectively removing the lower tail from the leftmost one from the lower tail of the rightmost one.
That is, $P(14 \leq X \leq 15.5) = P(X \leq 15.5) - P(X < 14) \approx .1667 - -0.6293 = 0.7960$.

 

[karush]

You are finding the area associated with $z=0.16$. You want to use the table the other way around. (Happy)

$$0.15910$$ is $$z=.41$$ at least if you mean by the other way around...

 

[MarkFL]

$$0.15910$$ is $$z=.41$$ at least if you mean by the other way around...

Yes, if you want a bit more accuracy, I found:

$$z_k\approx0.412463129441$$

 

[karush]

Yes, if you want a bit more accuracy, I found:

$$z_k\approx0.412463129441$$

so then $$k=15+(0.412463129441)(3)=16.2374$$ or round down to $$16$$ since has been in integers

 

[MarkFL]

I would round to the hundredths since the probability is given to this many places.

 

[karush]

I would round to the hundredths since the probability is given to this many places.

ok $$16.24$$

thanks for all help I learned a lot...:)