MHB -aux.13.Normal Distribution area

[karush]

View attachment 1020

still having trouble figuring this out!

The graph shows a normal curve for the random variable $$X$$, with mean $$ \mu$$ and standard deviation $$\sigma$$

It is known that $$P \left(X \geq12 \right) = 0.1$$.

(a) The shaded region $$A$$ is the region under the curve where $$x \geq 12$$. Write down the area of the shaded region $$A$$.

It is also known that $$P(X \leq 8) = 0.1$$.

(b) Find the value of $$\mu$$, explaining your method in full.

in that $$\mu$$ is in between 8 and 12 which would be $$\mu=10$$(c) Show that $$\sigma = 1.56$$ to an accuracy of three significant figures.

(d) Find $$ P(X \leq 11)$$.

 

[HallsofIvy]

View attachment 1020

still having trouble figuring this out!

The graph shows a normal curve for the random variable $$X$$, with mean $$ \mu$$ and standard deviation $$\sigma$$

It is known that $$P \left(X \geq12 \right) = 0.1$$.

(a) The shaded region $$A$$ is the region under the curve where $$x \geq 12$$. Write down the area of the shaded region $$A$$.

Seriously? You were just told that this area is 0.1!

It is also known that $$P(X \leq 8) = 0.1$$.

(b) Find the value of $$\mu$$, explaining your method in full.

in that $$\mu$$ is in between 8 and 12 which would be $$\mu=10$$.

Okay, Since the graph of the normal distribution is symmetric about mu, and you are told that the probabilities that x is less than 8 and larger than 12 are equal, mu is exactly half way between them

(c) Show that $$\sigma = 1.56$$ to an accuracy of three significant figures.
If a normal distribution has mean 10 and standard deviation [math]\sigma[/math], then [math]\frac{x- \mu}{\sigma}[/math] has standard normal distribution. Look up the "z" that has probability .1 in a table of the standard distribution (a good one online is at Standard Normal Distribution Table) then solve [math]\frac{12- 0}{\sigma}= z[/math]

(d) Find $$ P(X \leq 11)$$.

Knowing both [math]\sigma[/math] and [math]\mu= 10[/math] you can find the "standard" variable [math]z= \frac{11- 10}{\sigma}[/math] using the table of the normal distribution.

 

[MarkFL]

View attachment 1020

still having trouble figuring this out!

The graph shows a normal curve for the random variable $$X$$, with mean $$ \mu$$ and standard deviation $$\sigma$$

It is known that $$P \left(X \geq12 \right) = 0.1$$.

(a) The shaded region $$A$$ is the region under the curve where $$x \geq 12$$. Write down the area of the shaded region $$A$$.

It is also known that $$P(X \leq 8) = 0.1$$.

(b) Find the value of $$\mu$$, explaining your method in full.

in that $$\mu$$ is in between 8 and 12 which would be $$\mu=10$$(c) Show that $$\sigma = 1.56$$ to an accuracy of three significant figures.

(d) Find $$ P(X \leq 11)$$.

(a) You are given that:

$$P \left(X \geq12 \right) = 0.1$$

What relationship is there between this and the shaded region?

(b) Using the given:

$$P(X \leq 8) = 0.1$$

We may state:

$$8<\mu<12$$

and by symmetry:

$$\mu-8=12-\mu$$

Note: we are simply stating mathematically that the mean is midway between $X=8$ and $X=12$.

(c) We know that:

$$z=\frac{\mu-x}{\sigma}$$

or:

$$\sigma=\frac{\mu-x}{z}$$

Once we know $\mu$, and we use $x=12$, what $z$-value should we use? What area is to the left of $x$ but to the right of $\mu$?

(d) Once we have $\sigma$, we may standardize $X=11$ (convert it to a $z$-score) and then use our table to determine $$ P(X \leq 11)$$.

So, what do you find? :D

I see, before I post, that another has posted, but I figure we are saying the same thing in slightly different ways, and I am not giving anything further away. :D

 

[karush]

(a) You are given that:

Once we know $\mu$, and we use $x=12$, what $z$-value should we use? What area is to the left of $x$ but to the right of $\mu$?

well from the table I found $0.5-0.1=0.4$ so $0.4$ on table is $\approx 0.3997$ or a $z$ value of $1.28$

so $\frac{|10-12|}{1.28}=1.56 = \sigma$
I assume the numerator has to be a abs value

(d) Once we have $\sigma$, we may standardize $X=11$ (convert it to a $z$-score) and then use our table to determine $P(X \leq 11)$.

So, what do you find? :D

so $\frac{10-11}{1.56} = z =.64$ from table is $.2389 = P(X \leq 11)$

however don't we include what is left of $\mu$ which would add $.5$ which would give us $\approx .7389$

View attachment 1025

 

[MarkFL]

Correct, and correct. :D