Question

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

Answer 1

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.

Answer:

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

Answer 2

An integral of a probability distribution function is being used to find probabilities for a continuous random variable, and further calculation can be defined as follows:

Probability density function:

Following is the graph of the pdf, please find the CDF graph file.

$\to F(X) =\int^{x}_{0} \ f(x) \ dx= \int^{x}_{0} \frac{2(x+2)}{5}\ dx=\frac{2}{5}[\frac{x^2}{2}+2x]^{x}_{0}=\frac{2}{5} [\frac{x^2}{2}+2x]$

Following is the CDF of the graph:

Calculating the mean of X:

$\to E(X) =\int^{1}_{0} x \ f(x) \ dx= \int^{1}_{0} \frac{2(x^2+2x)}{5}\ dx=\frac{2}{5}[\frac{x^3}{3}+x^2]^{1}_{0}=\frac{2}{5} [\frac{1}{3}+1]=\frac{8}{15}$

For point c)

Using the X  that shows the 40th percentile. So

$\to \frac{2}{5}\left [ \frac{x^{2}}{2}+2x \right ]=0.40\\\\\to \frac{x^{2}}{5}+\frac{4x}{5}=0.40\\\\\to x^{2}+4x-2=0$

The one solution of the above equation is 0.45. So 40th percentile is 0.45.

For point d)

$\to P(\frac{1}{4} < X < \frac{1}{2})=F(\frac{1}{2})- F(\frac{1}{4})=\frac{2}{5}\left [ \frac{1}{8}+1-\frac{1}{32}-\frac{1}{2} \right ]=0.2375$

Find out more information about the density function here:

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