(Solved): Anticipated consumer demand in a restaurant for free range steaks next month can be modeled by a n ...

Anticipated consumer demand in a restaurant for free range steaks next month can be modeled by a normal random variable with mean 1,400 pounds and standard deviation 90 pounds. a. What is the probability that demand will exceed 1,200 pounds? b. What is the probability that demand will be between 1,300 and 1,500 pounds? c. The probability is \( 0.15 \) that demand will be more than how many pounds? Click the icon to view the standard normal table of the cumulative distribution function. a. The probability that demand will exceed 1,200 pounds is (Round to four decimal places as needed.) b. The probability that demand will be between 1,300 and 1,500 pounds is (Round to four decimal places as needed.) c. The probability is \( 0.15 \) that demand will be more than pounds. (Round to one decimal place as needed.) The number of hits per day on the Web site of a tool company is normally distributed with a mean of 590 and a standard deviation of 130. a. What proportion of days have more than 710 hits? b. What proportion of days have between 610 and 710 hits? c. Find the number of hits such that only \( 10 \% \) of the days will have the number of hits below this number. Click the icon to view the standard normal table of the cumulative distribution function. a. The proportion of days with more than 710 hits is (Round to four decimal places as needed.) b. The proportion of days with between 610 and 710 hits is (Round to four decimal places as needed.) c. Only \( 10 \% \) of the days will have less than hits. (Round to one decimal place as needed.)