1- 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?

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

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?