Question

Use the Normal model ​N(1133​,78​) for the weights of steers.

Answer 1

Answer:

Step-by-step explanation:

Let $X\sim N(1133, 78)$.

Here, the mean is 1133 and standard deviation is 78.

Since we don't have the table for N(1133, 78), so we use the standard table for N(0,1),

a)

All we have to do is find, within the table the specific percentile.

now to find (42nd percentile) 0.42 in the normal table N(0,1) and extract the number on the row and column.

$P(X\leq x)=0.42$

Using: $Z=\frac{X-\mu}{\sigma}$ where $\mu$ is the mean and $\sigma$ is the standard deviation.

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.42$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.42$

Now, using Standard Normal table to find the value of z- score;

Usually, tables are set up as the probability that a number, z  is less than or equal to Z.

i.e, $z= -0.20$

Now, putting this value in  $z=\frac{x-1133}{78}$, to find x;  

$\frac{x-1133}{78}=-0.20$

On Simplify,  we get;

$x=1,117.4$ pounds

b)

Similarly, find the weight for 91st percentile.

Follow the same steps that we have done in part (a),

$P(X\leq x)=0.91$ or

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.91$

⇒ $P(Z \leq \frac{x-1133}{78})=0.91$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.91$

Now, use normal table value to find the z-score;

Usually, tables are set up as the probability that a number, z is less than or equal to Z

we have, z= 1.34

putting the z value in  $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=1.34$

On simplify, we get

x= 1,237.52 pounds

c)

the interquartile range (IQR)=3rd Quartile - 1st quartile

First find the 1st quartile:

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.25$

⇒ $P(Z \leq \frac{x-1133}{78})=0.25$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.25$

Now, we use normal table to find that z=-0.675

putting the z value in  $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=-0.675$

On simplify, we get

x= 1,080.35 pounds

Similarly, for third quartile

By symmetry z=0.675

putting the z value in  $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=0.675$

On simplify, we get

x= 1,185.65 pounds

then, IQR = 1185.65 -1080.35=105.3 pounds