Question

A variable has the density curve whose equation is ? = 2? for 0 < x < 1 and y = 0.

a. Show that the area under this density curve to the left of any number x between 0 and 1 equals ? 2

B. What percentage of all possible outcomes of the variable lie between 1 2 and 3 4?

C. What percentage of all possible outcomes are at least 1 4 ?

Answer 1

2 And 2 77 77 Graph of the density of the given solution It is given that the variable (a) has density function Y=21, OLULI and yzo. The Graph of the density function is given below. 1 0.5- 0.0 K ROM MULI .: 216 d (a) show that the area under the density Curve to the left of any number 'a' is equal to ar between 0 and I for the density function from above Graph, the base of the triangle is 'n' and the height is "24" Thus, the area of the triangle is = 1 { basen (height) Area = 4(x)(21) Hence, the of any area under number n iar the density Curve to the left is equal to 92 between 0 and 1

find the percentage of all observations lie between - 1 and 2 ...Area between 1 and 3 4 = (Area to the left of 3) - (Area to the left of 1) =@je[from part ] = ab ta rom (C) . = 0.3125 Thus the percentage of all observations that lie between and & ů [31:25% . Find the percentage of all observations are at least the (Area of the right of t1 = 1- (Area to the left ft) =1-6? : [from part a] =l-to = 0.9375 Thus, the percentage of all observations are at least to ů [930756]