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Digiron [165]
3 years ago
11

Let x be the amount of time (in minutes) that a particular San Francisco commuter must wait for a BART train. Suppose that the d

ensity curve is as pictured below (a uniform distribution):
(a) What is the probability that x is less than 11 min? more than 14 min? P (x is less than 11 minutes) = P (x is more than 14 minutes) =

(b) What is the probability that x is between 8 and 13 min? P (x is between 8 and 13 minutes) =

(c) Find the value c for which P(x < c) = .8. c = mins You may need to use the appropriate table in Appendix A to answer this question.

Mathematics
2 answers:
Finger [1]3 years ago
8 0

Your question is not complete, as you have not provided the uniform distribution, please see attached to view the distribution, as (0, 0.05) and (20, 0.05).

Answer:

A. I) 0.45

II) 0.3

B) 0.25

C) 4

Step-by-step explanation:

The probability will be calculated using (base)(height). Since the height is static at 0.05, which is the density and the base is the X values. Therefore the probability will be a function of minutes which is the X values.

                    A.

I) The probability that X is less than 11 minutes (X<11)

(0.05)(20-11)

0.05 × 9= 0.45

II) The probability that X is greater than 14 minutes is (X>14)

(0.05)(20-14)

0.05 × 6 = 0.3

                    B.

The probability that X is between 8 and 13 minutes (8<X<13)

(0.05)(13-8)

0.05 × 5 = 0.25

                    C.

(X<c)= 0.8 find the value of c

Since the probability is (base)(height)

Therefore;

(0.05)(20-c) = 0.8

Multiply bracket

1-0.05c = 0.8

1 - 0.8 = 0.05c

0.2 = 0.05c

C = 0.2 ÷ 0.05

C = 4

babymother [125]3 years ago
3 0

Answer:

a) P (x is less than 11 minutes) =  0.55

P (x is more than 14 minutes) = 0.3

b) P (x is between 8 and 13 minutes) = 0.25

c)  P(x < c) =0.8 is 0.05 x c = 0.8

Step-by-step explanation:

a) P (x is less than 11 minutes) = 11 x 0.05

                                                 = 0.55

P (x is more than 14 minutes) = 0.05 x (20 - 14)

                                                = 0.05 x 6

                                                = 0.3

b) P (x is between 8 and 13 minutes) = 0.05 x (13 -8)

                                                            = 0.05 x 5

                                                             = 0.25

c)  P(x < c) =0.8

The area between 0 and c is 0.8

Hence,

0.05 x c = 0.8

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34% of the scores lie between 433 and 523.

Solution:

Given data:

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$Z(X)=\frac{x-\mu}{\sigma}

$Z(433)=\frac{433-\ 433}{90}=0

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P(433 < x < 523) = P(0 < Z < 1)

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Here μ to μ + σ = \frac{68\%}{2} =34\%

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Answer:

y=-3/7x+2

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