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swat32
3 years ago
5

A certain brand of dinnerware set comes in three colors: red, white, and blue. Twenty percent of customers order the red set, 45

% order the white, and 35% order the blue. Let X = 1 if a randomly chosen order is for a red set, let X = 0 otherwise; let Y = 1 if the order is for a white set, let Y = 0 otherwise; let Z = 1 if it is for either a red or white set, and let Z = 0 otherwise.
a. Let pX denote the success probability for X. Find pX.
b. Let pY denote the success probability for Y. Find pY.
c. Let pZ denote the success probability for Z. Find pZ.
d. Is it possible for both X and Y to equal 1?
e. Does pZ = pX + py ?
f. Does Z = X + Y ? Explain.
Mathematics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

a) 0.20

b) 0.45

c) 0.65

d) Yes

e) Yes

f) Z = X + Y (except when X = 1 and Y = 1)

This is because the successes of X and Y are mutually exclusive events but their failures aren't. X and Y cannot both be 1.

Step-by-step explanation:

Probability of a red set = 20% = 0.20

Probability of a white set = 45% = 0.45

Probability of a blue set = 35% = 0.35

Probability of the single set being a red or white set = 20% + 45% = 65% = 0.65

P(X=1) = 0.20, P(X=0) = 1 - 0.2 = 0.80

P(Y=1) = 0.45, P(Y=0) = 1 - 0.45 = 0.55

P(Z=1) = 0.65, P(Z=0) = 1 - 0.65 = 0.35

a) pX = P(X=1) = 0.20

b) pY = P(Y=1) = 0.45

c) pZ = P(Z=1) = 0.65

d) Since only one order is being considered at a time, it isn't possible to order red & white set in a single set, hence, both X and Y cannot both be successes (equal to 1) at the same time. But they can both be failures (both equal to 0) if a blue set is ordered. The successes of X and Y are mutually exclusive events but their failures aren't

e) Is pZ = pX + pY

pX = 0.2, pY = 0.45, pZ = 0.65

Hence, this statement is correct!

f) Z = X + Y

Let's check all the probabilities

when X = 1 and Y = 1, Z = 1

1 ≠ 1 + 1

when X = 0 and Y = 1, Z = 1

1 = 0 + 1

when X = 1 and Y = 0, Z = 1

1 = 1 + 0

when X = 0 and Y = 0, Z = 0

0 = 0 + 0

Hence, Z = X + Y (except when X = 1 and Y = 1)

This is because the success of X and Y are mutually exclusive events but their failures aren't.

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1- The Canada Urban Transit Association has reported that the average revenue per passenger trip during a given year was $1.55.
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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

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