# The question in the picture below..?

##### 1 Answer

**(i)**

**(ii)**

**(iii)**

#### Explanation:

**(i)** Let

#"P"(C>=200)="P"((C-mu)/sigma >= (200-mu)/sigma)#

#color(white)("P"(C>=200))="P"(Z >= (200-203)/2.5)#

#color(white)("P"(C>=200))="P"(Z >= –1.2)#

#color(white)("P"(C>=200))="P"(Z < 1.2)#

#color(white)("P"(C>=200))=Phi(1.2)#

#color(white)("P"(C>=200))=0.8849" "=88.49%#

**(ii)** Let

#"P"(400 <= D<= 405)#

#="P((400-406)/sqrt(12.5)<=(D-mu)/sigma <= (405-406)/sqrt(12.5))#

#~~ "P(–1.70 <= Z <= –0.28)#

#=Phi(–0.28)-Phi(–1.70)#

#=0.3897-0.0446#

#=0.3451" "=34.51%#

**(iii)**

#"P"(abs(barC - 203)< a)=0.95#

#=>2{"P"(0 < [barC - 203] < a)}=0.95#

#=>2{"P"(0 < [barC - 203]/0.3125 < a/0.3125)}=0.95#

#=>"P"(0 < Z < a/0.3125)=0.475#

#=>Phi(a/0.3125)-Phi(0)=0.475#

#=>Phi(a/0.3125)-0.5=0.475#

#=>Phi(a/0.3125)=0.975#

#=>a/0.3125 = Phi^(–1)(0.975)#

#=>a/0.3125=1.96#

#=>a=0.6125#

Note: The value of

Also, the *mean* of