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

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places