Answer:
0.5
0.9545
0.68268
0.4986501
Step-by-step explanation:
The Canada Urban Transit Association has reported that the average revenue per passenger trip during a given year was $1.55. If we assume a normal distribution and a standard deviation of 5 $0.20, what proportion of passenger trips produced a revenue of Source: American Public Transit Association, APTA 2009 Transit Fact Book, p. 35.
a. less than $1.55?
b. between $1.15 and $1.95? c. between $1.35 and $1.75? d. between $0.95 and $1.55?
Given that :
Mean (m) = 1.55
Standard deviation (s) = 0.20
a. less than $1.55?
P(x < 1.55)
USing the relation to obtain the standardized score (Z) :
Z = (x - m) / s
Z = (1.55 - 1.55) / 0.20 = 0
p(Z < 0) = 0.5 ( Z probability calculator)
b. between $1.15 and $1.95?
P(x < 1.15)
USing the relation to obtain the standardized score (Z) :
Z = (x - m) / s
Z = (1.15 - 1.55) / 0.20 = - 2
p(Z < - 2) = 0.02275 ( Z probability calculator)
P(x < 1.95)
USing the relation to obtain the standardized score (Z) :
Z = (x - m) / s
Z = (1.95 - 1.55) / 0.20 = 2
p(Z < - 2) = 0.97725 ( Z probability calculator)
0.97725 - 0.02275 = 0.9545
c. between $1.35 and $1.75?
P(x < 1.35)
USing the relation to obtain the standardized score (Z) :
Z = (x - m) / s
Z = (1.35 - 1.55) / 0.20 = - 1
p(Z < - 2) = 0.15866 ( Z probability calculator)
P(x < 1.75)
USing the relation to obtain the standardized score (Z) :
Z = (x - m) / s
Z = (1.75 - 1.55) / 0.20 = 1
p(Z < - 2) = 0.84134 ( Z probability calculator)
0.84134 - 0.15866 = 0.68268
d. between $0.95 and $1.55?
P(x < 0.95)
USing the relation to obtain the standardized score (Z) :
Z = (x - m) / s
Z = (0.95 - 1.55) / 0.20 = - 3
p(Z < - 3) = 0.0013499 ( Z probability calculator)
P(x < 1.55)
USing the relation to obtain the standardized score (Z) :
Z = (x - m) / s
Z = (1.55 - 1.55) / 0.20 = 0
p(Z < 0) = 0.5 ( Z probability calculator)
0.5 - 0.0013499 = 0.4986501
