This Question: 12 eticipated consumer domand in a restaurant for free range slea

Question

This Question: 12 eticipated consumer domand in a restaurant for free range sleaks nexd month can be modeled by a normal random variabb a What ls the probability that demand will exceed 1,400 pounds? b What is the probability that demand wil be behweon 1,500 and 1,700 pounds? c The probability is 0.10 that demand wil be more than how many pounds? with mean 1,600 pounds and standarnd deviation 90 pounds Cick the lcon to view the standard normal table of the cumulative ibuion function The probabillty thait demand will xceed 1,400 pounds Round to four decimal places as needed) b The probability that demand will be behween 1,500 and 1,700 pounds ib Round to four dedimal places as needed ) c The probab y is 0 t0 ful demand wa be more than?ponds. (Round to one decinal place as needed ) Enter your answer in each of the answer boxes ave for La O Type here to search

Explanation / Answer

Ans:

a)

z=(1400-1600)/90=-2.22

P(z>-2.22)=0.9869

b)

z(1500)=(1500-1600)/90=-1.11

z(1700)=(1700-1600)/90=1.11

P(-1.11<z<1.11)=P(z<1.11)-P(z<-1.11)

=0.8667-0.1333=0.7335

c)

P(Z>z)=0.1

P(Z<=z)=1-0.1=0.9

z=normsinv(0.9)=1.282

x=1600+1.282*90=1715.3

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.