Answer:
The correct answer is "100".
Step-by-step explanation:
The given values are:
X (mean) = 5
X (standard deviation) = 25
Variance of X will be:
=
![25^(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/d1tedeuw1f46cu9w78aznvclepohko07ov.png)
=
![625](https://img.qammunity.org/2021/formulas/mathematics/high-school/lbihnvgbxsled26skf219n1l0ivg7b4028.png)
The solution of the part first is:
The given equation is :
![Y=2+4X](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/ng144d0rfe0oalma79i2ut670fxxg23jcy.png)
On putting the value of x in above equation we get ,
![Y=2+4* 5 \\Y=2+20\\Y=22](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/svjh7i8lffi8rrrbt9y3dhtbwusw1nby7d.png)
So we get the mean of Y is 22
For finding the S.D of Y
![Y = 2+4X\\](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/o8tl9d2zctfrw6n720w77gbzz64mlgkm1y.png)
So,
variance of Y= Var(2+4X)
As we know that the variance of constant is zero
So, variance (2) =0
⇒
![Variance(aX) = a^2Variance(X)](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/srxyki6091ehcn93469lcztig10jsmu2vm.png)
⇒
![Var(4X) = 42Var(X) = 16Var(X)](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/a6dpkqcd10obhvi3k422c4tjbd3a8kwbtc.png)
So,
⇒
![Variance(Y) = Variance (2) + Variance(4X)](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/pnjomxldtvfyv84ubpuan5v81iwfutjfpx.png)
We know the variance of constant is zero
So, var(2)=0
⇒
![Variance(Y)= 16* 625\\ Variance(Y)=10000](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/9yj1ywbxjf1mec31xt1y2ck6idwqz1ej7p.png)
Thus the standard deviation of Y is:
=
![(10000)^(0.5)](https://img.qammunity.org/2021/formulas/advanced-placement-ap/high-school/93xx4127v2ioygesjcv971e1df76ozzdht.png)
=
![100](https://img.qammunity.org/2021/formulas/mathematics/middle-school/fe77yyrb40zgk0muzj9eo8un7d99yye4xv.png)