Question

In​ 2008, the per capita consumption of soft drinks in Country A was reported to be 18.97 gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 18.97 gallons and a standard deviation of 5 gallons. Complete parts​ (a) through​ (d) below.

a. What is the probability that someone in Country A consumed more than 12 gallons of soft drinks in​ 2008?

The probability is

​(Round to four decimal places as​ needed.)

b. What is the probability that someone in Country A consumed between 7 and 9 gallons of soft drinks in​ 2008?

The probability is

​(Round to four decimal places as​ needed.)

c. What is the probability that someone in Country A consumed less than 9 gallons of soft drinks in​ 2008?

The probability is

​(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 _____ gallons of soft drinks.

​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 18.97

standard deviation = = 5

a.

P(x > 12) = 1 - P(x < 12)

= 1 - P[(x - ) / < (12 - 18.97) / 5)

= 1 - P(z < -1.39)

= 1 - 0.0823

= 0.9177

Probability = 0.9177

b.

P(7 < x < 9) = P[(7 - 18.97)/ 5) < (x - ) /  < (9 - 18.97) / 5) ]

= P(-2.39 < z < -1.99)

= P(z < -1.99) - P(z < -2.39)

= 0.0233 - 0.0084

= 0.0149

Probability = 0.0149

c.

P(x < 9) = P[(x - ) / < (9 - 18.97) / 5]

= P(z < -1.99)

= 0.0233

Probability = 0.0233

d.

Using standard normal table ,

P(Z < z) = 97%

P(Z < 1.88) = 0.97

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 5 + 18.97 = 28.37

The probability is 97% that someone in Country A consumed less than 28.37 gallons of soft drinks