Question

In​ 2008, the per capita consumption of soft drinks in Country A was reported to be 19.35 gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 19.35 gallons and a standard deviation of 4 gallons. Complete parts​ (a) through​ (d) below.

a. What is the probability that someone in Country A consumed more than 14 gallons of soft drinks 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 9 and 10 gallons of soft drinks 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 10 gallons of soft drinks in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

d. 97​% of the people in Country A consumed less than how many gallons of soft​ drinks? The probability is 97​% that someone in Country A consumed less than nothing gallons of soft drinks. ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 19.35

standard deviation = = 4

a) P(x > 14 ) = 1 - p( x< 14 )

=1- p P[(x - ) / < ( 14 - 19.35) / 4 ]

=1- P(z < -1.34)

Using z table,

= 1 - 0.0901

= 0.9099

b) P( 9 < x < 10 ) = P[(9 - 19.35 )/ 4) < (x - ) /  < (10 - 19.35) / 4) ]

= P( -2.59 < z < -2.34 )

= P(z < -2.34) - P(z < -2.59)

Using z table,

= 0.0096 - 0.0048

= 0.0048

c) P(x < 10 ) = P[(x - ) / < (10 - 19.35) / 4 ]

= P(z < -2.34)

Using z table,

= 0.0096

d) Using standard normal table,

P(Z < z) = 97%

= P(Z < z) = 0.97

= P(Z < 1.881) = 0.97

z = 1.881

Using z-score formula,

x = z * +

x = 1.881 * 4 + 19.35

x = 26.87 gallons.