Question

In​ 2008, the per capita consumption of soft drinks in Country A was reported to be 19.31 gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 19.31 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?

b. What is the probability that someone in Country A consumed between 5 and 6 gallons of soft drinks in​ 2008?

c.What is the probability that someone in Country A consumed less than 6 gallons of soft drinks in​ 2008?

d.98​%of the people in Country A consumed less than how many gallons of soft​ drinks? The probability is 98% that someone in Country A consumed less than ____ gallons of soft drinks.

Answer 1

Part a)
X ~ N ( µ = 19.31 , σ = 4 )
P ( X > 14 ) = 1 - P ( X < 14 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 14 - 19.31 ) / 4
Z = -1.3275
P ( ( X - µ ) / σ ) > ( 14 - 19.31 ) / 4 )
P ( Z > -1.3275 )
P ( X > 14 ) = 1 - P ( Z < -1.3275 )
P ( X > 14 ) = 1 - 0.0922
P ( X > 14 ) = 0.9078

Part b)
X ~ N ( µ = 19.31 , σ = 4 )
P ( 5 < X < 6 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 5 - 19.31 ) / 4
Z = -3.5775
Z = ( 6 - 19.31 ) / 4
Z = -3.3275
P ( -3.58 < Z < -3.33 )
P ( 5 < X < 6 ) = P ( Z < -3.33 ) - P ( Z < -3.58 )
P ( 5 < X < 6 ) = 0.0004 - 0.0002
P ( 5 < X < 6 ) = 0.0003

Part c)
X ~ N ( µ = 19.31 , σ = 4 )
P ( X < 6 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 6 - 19.31 ) / 4
Z = -3.3275
P ( ( X - µ ) / σ ) < ( 6 - 19.31 ) / 4 )
P ( X < 6 ) = P ( Z < -3.3275 )
P ( X < 6 ) = 0.0004

Part d)
X ~ N ( µ = 19.31 , σ = 4 )
P ( X < x ) = 98% = 0.98
To find the value of x
Looking for the probability 0.98 in standard normal table to calculate Z score = 2.0537
Z = ( X - µ ) / σ
2.0537 = ( X - 19.31 ) / 4
X = 27.5248 ≈ 28
P ( X < 28 ) = 0.98