The question in the picture below..?

Question

Adrian Alexander · Accepted Answer

(i) #"P"(C>=200)=0.8849#
(ii) #"P"(400 <= D<= 405)=0.3451#
(iii) #a=0.6125# (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.