Question

Anticipated consumer demand in a restaurant for free range steaks next month can be modeled by a normal random variable with mean pounds and standard deviation pounds. a. What is the probability that demand will exceed ​pounds? b. What is the probability that demand will be between and ​pounds? c. The probability is that demand will be more than how many​ pounds?

a. The probability that demand will exceed pounds is ??????? ​(Round to four decimal places as​ needed.)

b. The probability that demand will be between and pounds is ?????? . ​(Round to four decimal places as​ needed.)

c. The probability is that demand will be more than ??????? pounds.

Answer 1

A) the probability in which the demand will exceed more than 1000 pounds is given by

P(X>1000)

Which can be written in the form

=P[(X-1200)/100 > (1000-1200)/100)

Which is equal to P(Z>-2)

P(Z>-2)=P(|Z|<2) + P(Z>2)

=0.95+0.025=0.975

Here it shows that with 2 standard deviation of the mean about 95 percentage of normal distribution falls within tge S.D of mean.which further leaves about 5% lie without and remaining 2.5% lying to the other side

B) next we want to find the probability to which demand will be in between 1100 and 1300

The equation for further findings is given as follows

P(1100<X<1300)

=P[(1100-1200)/100 < (X-1200)/100 < (1300-1200)/100]

Which is equals to P(-1<Z<1)

This probability is about 0.68 because 68% of normal distribution lies in one standard devaition within the mean

C) given the probability 0.10 for demand exceeding a certain pounds

Hence the eq is as follows

P(X>k)=0.10( given)

P(X>k)=P[(X-1200)/100 > (k-1200)/100]

=P(Z>k*)=0.10

(k-1200)/100=1.2816

(k-1200)=1.2816*100

k-1200=128.16

k=128.16+1200

=1328.16

Hence for a probability of 0.10 there exist a demand exceeding 1328 pounds