Question

In​ 2008, the per capita consumption of coffee in Country A was reported to be 18.43 pounds. Assume that the per capita consumption of coffee in Country A is approximately normally​ distributed, with a mean of 18.43 pounds and a standard deviation of 4 pounds. Complete parts​ (a) through​ (d) below.

a. What is the probability that someone in Country A consumed more than 11 pounds of coffee in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

b. What is the probability that someone in Country A consumed between 7 and 9 pounds of coffee in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

c. What is the probability that someone in Country A consumed less than 9 pounds of coffee in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

d. 98​% of the people in Country A consumed less than how many pounds of​ coffee?

The probability is 98​% that someone in Country A consumed less than nothing pounds of coffee. ​(Round to two decimal places as​ needed.)

Answer 1

Define random variable X: The per capita consumption of coffee in Country A which is approximately normally​ distributed with mean = and standard deviation =

a) Here we have to find P(X > 11)

                        

                                   where z is standard normal variable

                        = P(z > -1.86)                (Round to 2 decimal)

                        = 1 - P(z < -1.86)

                        = 1 - 0.0314             (From statistical table of negative z values)

                        = 0.9686

Probability that someone in Country A consumed more than 11 pounds of coffee in​ 2008 is 0.9686

b) Here we have to find P(7 < X < 9)

                               

                                = P(-2.86 < z < -2.36) (Round to 2 decimal)

                                = P(z < -2.36) - P(z < -2.86)

                                = 0.0091 - 0.0021               (From statistical table of negative z values)

                                = 0.0070

Probability that someone in Country A consumed between 7 and 9 pounds of coffee in​ 2008 is 0.0070

c) Here we have to find P(X < 9)

                      

                       = P(z < -2.36)         (Round to 2 decimal)

                       = 0.0091    (From statistical table of negative z values)

Probability that someone in Country A consumed less than 9 pounds of coffee in​ 2008 is 0.0091

d) Here we have to find x such that P(X < x) = 0.98

z for left tail area = 0.98 is

z = 2.05      (From statistical table of positive z values, value close to 0.98 is 0.9798)

   = 18.43 + 2.05 * 4

   = 18.43 + 8.2

= 26.63

The probability is 98​% that someone in Country A consumed less than 26.63 pounds of coffee.