-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 \(\displaystyle X\), with mean \(\displaystyle \mu\) and standard deviation \(\displaystyle \sigma\)

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

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

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

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

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

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

 

[HallsofIvy]

View attachment 1020

still having trouble figuring this out!

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

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

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

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

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

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

in that \(\displaystyle \mu\) is in between 8 and 12 which would be \(\displaystyle \mu=10\).

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 \(\displaystyle \sigma = 1.56\) to an accuracy of three significant figures.
If a normal distribution has mean 10 and standard deviation \(\displaystyle \sigma\), then \(\displaystyle \frac{x- \mu}{\sigma}\) 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 \(\displaystyle \frac{12- 0}{\sigma}= z\)

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

Knowing both \(\displaystyle \sigma\) and \(\displaystyle \mu= 10\) you can find the "standard" variable \(\displaystyle z= \frac{11- 10}{\sigma}\) 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 \(\displaystyle X\), with mean \(\displaystyle \mu\) and standard deviation \(\displaystyle \sigma\)

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

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

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

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

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

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

(a) You are given that:

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

What relationship is there between this and the shaded region?

(b) Using the given:

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

We may state:

\(\displaystyle 8<\mu<12\)

and by symmetry:

\(\displaystyle \mu-8=12-\mu\)

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

(c) We know that:

\(\displaystyle z=\frac{\mu-x}{\sigma}\)

or:

\(\displaystyle \sigma=\frac{\mu-x}{z}\)

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 \(\displaystyle P(X \leq 11)\).

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