Chapter 6: Q. 6.30 (page 260)

A variable has a density curve whose equation is $y = 1$ for $0 < x < 1$ and $y = 0$ and otherwisse,
(a) Graph the density curve of this variable.
(b) Show that the area under this density curve to the left of any number x between 0 and 1equals x.
What percentage of all possible observations of the variable
(c) lie between $\frac{1}{4}$ and $\frac{3}{4}$ .
(d) exceed $\frac{7}{8}$ .

Short Answer

Expert verified

(a) The density curve of this variable is:

(b) Thus, the area under this density curve to the left of any number x between 0 and 1equals x.

(d) The percentage of all possible observations that exceed $\frac{7}{8} i s 1.25 %$

Step by step solution

Part (a) Step 1. Given Information.

A variable has a density curve whose equation is $y = 1, 0 < x < 1, y = 0$ .

Part (a) Step 2. Graph the density function.

Graph the density function $y = 1, 0 < x < 1, y = 0$

Part (b) Step 1. The area between 0, 1.

From part (a),

Length, $l = x$

Breadth, $b = 1$

Thus the area of the rectangle is given by,

$A = l b = x (1) = x$

Part (c) Step 1. Find the percentage.

Area between $\frac{1}{4}, \frac{3}{4} =$ [Area to the left of $\frac{3}{4}] -$ [Area to the left of $\frac{1}{4}]$

$= \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = 0.5$

The percentage of all possible observations that lie between $\frac{1}{4}, \frac{3}{4} i s 50 %$

Part (d) Step 1. Find the percentage.

[Area to the right of $\frac{7}{8}] = 1 -$ [Area to the left of $\frac{7}{8}]$

$= 1 - \frac{7}{8} = \frac{1}{8} = 0.125$

The percentage of all possible observations that exceeds $\frac{7}{8} i s 1.25 %$ .

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Short Answer

Step by step solution

Part (a) Step 1. Given Information.

Part (a) Step 2. Graph the density function.

Part (b) Step 1. The area between 0, 1.

Part (c) Step 1. Find the percentage.

Part (d) Step 1. Find the percentage.

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