3.9 The proportion of people who respond to a certain mail-order solicitation is a continuous random variable X that has the density function f(x) = 2(x+2) 5 , 0

Question

The proportion of people who respond to a certain mail-order solicitation is a continuous random variable X that has the density function f(x) = 2(x+2)/5 , 0 <x< 1, 0, elsewhere.

(a) Show that P(0 <X< 1) = 1.

(b) Find the probability that more than ¼ but fewer than ½ of the people contacted will respond to this type of solicitation.

a. See Explanation Below

b. P(¼<X< ½) = 0.2375

Step-by-step explanation:

Given

f(x) = 2(x+2)/5 , 0 <x< 1

Calculating P(0 <X< 1)

P(0 <X< 1) = ∫ f(x) dx {0.1}

Substitute 2(x+2)/5 for f(x)

P(0 <X< 1) = ∫ 2(x+2)/5 dx {0.1}

P(0 <X< 1) = 2/5 ∫(x+2) dx {0.1}

Integrate with respect to x

P(0 <X< 1) = 2/5 (x²/2+2x) {0.1}

P(0 <X< 1) = 2/5 (1²/2+2(1))

P(0 <X< 1) = 2/5 (½+2)

P(0 <X< 1) = 2/5 * 5/2

P(0 <X< 1) = 1 Proved

b.

Here we're to solve for P(¼<X< ½)

P(¼<X< ½) = ∫ f(x) dx {¼,½}

Substitute 2(x+2)/5 for f(x)

P(¼<X< ½) = ∫ 2(x+2)/5 dx {¼,½}

P(¼<X< ½) = 2/5 ∫(x+2) dx {¼,½}

P(¼<X< ½) = 2/5 (x²/2+2x) {¼,½}

P(¼<X< ½) = 2/5 [ (½²/2+2(½)) - (¼²/2+2(¼))]

P(¼<X< ½) = 2/5 [(⅛+1)-(1/32 + ½)]

P(¼<X< ½) = 0.2375

The probability that more than ¼ but fewer than ½ of the people contacted will respond to this type of solicitation is 0.2375

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

where are the images of the functions i can only see the question you typed.

step-by-step explanation: