The question in the picture below..?

(i) Let `C` be the mass of coffee in one random jar. Then `C" ~ N"(mu = 203, sigma^2 = 2.5^2).`

`"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 `C_1` and `C_2` be the masses of coffee in two independently chosen random jars, and let `D=C_1+C_2`. Then
`D" ~ N"(mu = 2xx203, sigma^2 = 2xx2.5^2).`
`color(white)(D)="N"(406, 12.5)`

`"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) `barC" ~ N"(mu=203, sigma^2=(2.5^2)/20)="N"(203,0.3125)`

`"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 `a` is equal to `z_0.025xxsigma,` where `z_(alpha//2)` is the `z`-coordinate of the standard normal curve `Z` that has an area of `alpha//2` to its right. Here, `alpha=1-0.95,` so `alpha//2 = 0.025.`

Also, the `sigma` used here is the standard deviation of `barC`. You may see a similar term used to help calculate confidence intervals: `z_(alpha//2)xxsigma/sqrtn.` In this form, `sigma` is the standard deviation of a single observation, thus `sigma/sqrtn` is the standard deviation of the mean of `n` observations.