B. A family on a University campus has a 150-gallon tank that is filled at the beginning ofeach week. The weekly demand of the family shows a relative frequency behaviour thatincreases steadily up to 100 gallons and then levels off between 100 and 150 gallons. If Xdenotes weekly demand in hundreds of gallons, the relative frequency of demand can bemodelled by0sxs1)(x,f(x) = }1, 1

Given information: fx=x, 0≤x≤11, 1<x<1.50, elsewhere

Answered: B. A family on a University campus has…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

B. A family on a University campus has a 150-gallon tank that is filled at the beginning of
each week. The weekly demand of the family shows a relative frequency behaviour that
increases steadily up to 100 gallons and then levels off between 100 and 150 gallons. If X
denotes weekly demand in hundreds of gallons, the relative frequency of demand can be
modelled by
0sxs1)
(x,
f(x) = }1, 1<x < 1.5
(o, elsewhere)
i) Find F(x)
ii) Find P(0 < X < 0.5)
iii) Find P(0.5 < X S 1.2)
iv) Find P(X 2 1|X < 1.4)

Expert Solution

Step 1

Given information:

$f (x) = \{\begin{cases} x, 0 \leq x \leq 1 \\ 1, 1 < x < 1.5 \\ 0, e l s e w h e r e \end{cases}$

Step 2

(i)

The value of $F (x)$ will be calculated as,

When $x < 0$ , the value of $F (x)$ will be,

$F (x) = \int_{- \infty}^{x} f (t) d t = \int_{- \infty}^{x} 0 d t = 0$

When $0 \leq x \leq 1$ ,

$F (x) = \int_{- \infty}^{x} f (t) d t = \int_{- \infty}^{0} f (t) d t + \int_{0}^{x} f (t) d t = \int_{- \infty}^{0} 0 d t + \int_{0}^{x} t d t = 0 + {[\frac{t^{2}}{2}]}_{0}^{x} = \frac{x^{2}}{2}$

When $1 < x \leq 1.5$

$F (x) = \int_{- \infty}^{x} f (t) d t = \int_{- \infty}^{0} f (t) d t + \int_{0}^{1} f (t) d t + \int_{1}^{x} f (t) d t = \int_{- \infty}^{0} 0 d t + \int_{0}^{1} t d t + \int_{1}^{x} 1 d t = 0 + {[\frac{t^{2}}{2}]}_{0}^{1} + {[t]}_{1}^{x} = \frac{1}{2} + x - 1 = x - 0.5$

When $x > 1.5$

$F (x) = \int_{- \infty}^{x} f (t) d t = \int_{- \infty}^{0} f (t) d t + \int_{0}^{1} f (t) d t + \int_{1}^{1.5} f (t) d t + \int_{1.5}^{x} f (t) d t = \int_{- \infty}^{0} 0 d t + \int_{0}^{1} t d t + \int_{1}^{1.5} 1 d t + \int_{1.5}^{x} 0 d t = 0 + {[\frac{t^{2}}{2}]}_{0}^{1} + {[t]}_{1}^{1.5} + 0 = \frac{1}{2} + 1.5 - 1 = 1.5 - 0.5 = 1$