Question

In​ 2008, the per capita consumption of coffee in Country A was reported to be 19.37 pounds. Assume that the per capita consumption of coffee in Country A is approximately normally​ distributed, with a mean of 19.37 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 13 pounds of coffee in​ 2008?

b. What is the probability that someone in Country A consumed between 6 and 10 pounds of coffee in​ 2008?

c. What is the probability that someone in Country A consumed less than 10 pounds of coffee in​ 2008?

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 ?? pounds of coffee.

Answer 1

Solution :

Given that ,

mean = = 19.37

standard deviation = = 4

a)P(x > 13 ) = 1 - p( x<13 )

=1- p [(x - ) / < (13-19.37) /4 ]

=1- P(z <-1.59 )

= 1 - 0.0559 = 0.9441

probability = 0.9441

b)

P( 6< x <10) = P[(6 - 19.37)/4 ) < (x - ) /  < (10 - 19.37) /4 ) ]

= P( -3.34< z < -2.34 )

= P(z < -2.34 ) - P(z < -3.34 )

Using standard normal table

= 0.0096 - 0.0004 = 0.0092

Probability = 0.0092

c)

P(x < 10 ) = P[(x - ) / < (10-19.37) /4 ]

= P(z < -2.34 )

= 0.0096

probability =0.0096

d)

P(Z < z) = 0.98

z = 2.054

Using z-score formula,

x = z * +

x = 2.054 * 4+19.37

x = 27.59

Answer = 27.59 pounds