Question

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

Answer 1

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

Answer 2

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

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