Question

Let \({\rm{X}}\) denote the amount of space occupied by an article placed in a \({\rm{1 - f}}{{\rm{t}}^{\rm{3}}}\) packing container. The pdf of \({\rm{X}}\) is

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{90}}{{\rm{x}}^{\rm{8}}}{\rm{(1 - x)}}}&{{\rm{0 < x < 1}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

a. Graph the pdf. Then obtain the cdf of \({\rm{X}}\) and graph it. b. What is \({\rm{P(X}} \le {\rm{.5)}}\) (i.e., \({\rm{F(}}{\rm{.5)}}\))? c. Using the cdf from (a), what is \({\rm{P(}}{\rm{.25 < X}} \le {\rm{.5)}}\)? What is \({\rm{P(}}{\rm{.25}} \le {\rm{X}} \le {\rm{.5)}}\)? d. What is the \({\rm{75th}}\) percentile of the distribution? e. Compute \({\rm{E(X)}}\) and \({{\rm{\sigma }}_{\rm{X}}}\). f. What is the probability that \({\rm{X}}\) is more than \({\rm{1}}\) standard deviation from its mean value?

Short answer

(a) The value is \({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x}} \le {\rm{0}}}\\{{{\rm{x}}^{\rm{9}}}{\rm{(10 - 9x)}}}&{{\rm{0 < x < 1}}}\\{\rm{1}}&{{\rm{x}} \ge {\rm{1}}}\end{array}} \right.\).

(b)The value is\({\rm{0}}{\rm{.01074}}\).

(c)The value is\({\rm{0}}{\rm{.01071}}\).

(d)The value is\({\rm{0}}{\rm{.9036}}\).

(e)The values are\({\rm{0}}{\rm{.8182}}\)and\({\rm{0}}{\rm{.111}}\).

(f) The value is \({\rm{0}}{\rm{.3146}}\).

Step-by-step solution

Define mean

The average of a group of numbers is simply defined as the mean. In statistics, the mean is also one of the indicators of central tendency.

Explanation

(a)We are given the following pdf\({\rm{f(x)}}\):

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{90 \times }}{{\rm{x}}^{\rm{8}}}{\rm{(1 - x)}}}&{{\rm{0 < x < 1}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

The definition of cdf for a continuous random variable X is as follows:

For each number\({\rm{x}}\), the cumulative distribution function\({\rm{F(x)}}\)for a continuous\({\rm{rv}}\)\({\rm{X}}\)is defined by

\(\begin{array}{c}{\rm{F(x) = P(X}} \le {\rm{x)}}\\{\rm{ = }}\int_{{\rm{ - }}\infty }^{\rm{x}} {\rm{f}} {\rm{(y) \times dy}}\end{array}\)

If\({\rm{x}}\)is an integer between\({\rm{0}}\)and\({\rm{1}}\),

\(\begin{array}{c}{\rm{F(X) = }}\int_{\rm{0}}^{\rm{x}} {\rm{9}} {\rm{0 \times }}{{\rm{y}}^{\rm{8}}}{\rm{(1 - y) \times dy}}\\{\rm{ = 90}}\int_{\rm{0}}^{\rm{x}} {{{\rm{y}}^{\rm{8}}}} {\rm{dy - 90}}\int_{\rm{0}}^{\rm{x}} {{{\rm{y}}^{\rm{9}}}} {\rm{dy}}\\{\rm{ = 90}}\left( {\frac{{{{\rm{y}}^{\rm{9}}}}}{{\rm{9}}}} \right)_{\rm{0}}^{\rm{x}}{\rm{ - 90}}\left( {\frac{{{{\rm{y}}^{{\rm{10}}}}}}{{{\rm{10}}}}} \right)_{\rm{0}}^{\rm{x}}\\{\rm{ = }}\frac{{{\rm{90}}}}{{\rm{9}}}\left( {{{\rm{x}}^{\rm{9}}}{\rm{ - 0}}} \right){\rm{ - }}\frac{{{\rm{90}}}}{{\rm{9}}}\left( {{{\rm{x}}^{{\rm{10}}}}} \right)\\{\rm{F(X) = }}{{\rm{x}}^{\rm{9}}}{\rm{(10 - 9x)}}\end{array}\)

As a result,\({\rm{F(X)}}\)can be written as:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x}} \le {\rm{0}}}\\{{{\rm{x}}^{\rm{9}}}{\rm{(10 - 9x)}}}&{{\rm{0 < x < 1}}}\\{\rm{1}}&{{\rm{x}} \ge {\rm{1}}}\end{array}} \right.\)

Explanation

(b) We can write: Using the cdf provided in part(a),

\(\begin{aligned}{\rm{P(X}} \le {\rm{0}}.5) &= F(0{\rm{.5)}}\\ &= 0{\rm{.}}{{\rm{5}}^{\rm{9}}}{\rm{(10 - 9(0}}{\rm{.5))}}\\ &= 0 {\rm{.}}{{\rm{5}}^{\rm{9}}}{\rm{(5}}{\rm{.5)}}\\{\rm{P(X}} \le {\rm{0}}.5) &= 0 {\rm{.01074}}\end{aligned}\)

Therefore, the value is \({\rm{0}}{\rm{.01074}}\).

Explanation

(c) As, \({\rm{P(0}}{\rm{.25 < X}} \le {\rm{0}}{\rm{.5)}}\) signifies the probability that an article placed in a \({\rm{1f}}{{\rm{t}}^{\rm{3}}}\) packing container will take up at least \({\rm{0}}{\rm{.25f}}{{\rm{t}}^{\rm{3}}}\) but less than or equal to \({\rm{0}}{\rm{.5f}}{{\rm{t}}^{\rm{3}}}\).

\(\begin{aligned}{\rm{P(0}}{\rm{.25 < X}} \le {\rm{0}} .5) &= F(0 {\rm{.5) - F(0}}{\rm{.25)}}\\ & = \left( {{\rm{0}}{\rm{.}}{{\rm{5}}^{\rm{9}}}{\rm{(10 - 9(0}}{\rm{.5))}}} \right){\rm{ - }}\left( {{\rm{0}}{\rm{.2}}{{\rm{5}}^{\rm{9}}}{\rm{(10 - 9(0}}{\rm{.25))}}} \right)\\& = (0{\rm{.01074) - (0}}{\rm{.00003)}}\\{\rm{P(0}}{\rm{.25 < X}} \le {\rm{0}} .5) &= 0{\rm{.01071}}\end{aligned}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as a proposition\({\rm{(x)}}\). Then, using\({\rm{a < b}}\), for any two numbers a and b,

\({\rm{P(a}} \le {\rm{X}} \le {\rm{b) = F(b) - F(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.01071}}\).

Explanation

(d)Recall that the percentile of any given distribution is defined as follows: Let p be a positive integer between\({\rm{0}}\)and\({\rm{1}}\). The\({{\rm{(100p)}}^{{\rm{th}}}}\)percentile of a continuous \({\rm{rv}}\)X's distribution (denoted by\({{\rm{\eta }}_{\rm{p}}}\)) is defined by

\(\begin{array}{c}{\rm{p = F}}\left( {{{\rm{\eta }}_{\rm{p}}}} \right)\\{\rm{ = }}\int_{{\rm{ - }}\infty }^{{{\rm{\eta }}_{\rm{p}}}} {\rm{f}} {\rm{(y)dy}}\end{array}\)

We need to find the\({\rm{7}}{{\rm{5}}^{\rm{t}}}{\rm{h}}\)percentile for the given distribution. Let's use\({{\rm{\eta }}_{{\rm{75}}}}\)as an example.

\(\begin{array}{c}{\rm{0}}{\rm{.75 = F}}\left( {{{\rm{\eta }}_{{\rm{75}}}}} \right)\\{\rm{0}}{\rm{.75 = }}{\left( {{{\rm{\eta }}_{{\rm{75}}}}} \right)^{\rm{9}}}\left( {{\rm{10 - 9}}\left( {{{\rm{\eta }}_{{\rm{75}}}}} \right)} \right)\end{array}\)

This problem can be overcome with the help of technology. As a result, we have:

\({{\rm{\eta }}_{{\rm{75}}}}{\rm{ = 0}}{\rm{.9036}}\)

Therefore, the value is \({\rm{0}}{\rm{.9036}}\).

Explanation

(e)The following is a formula for calculating the average depth value:

\(\begin{aligned} E(X) &= \int_{{\rm{ - }}\infty }^\infty {\rm{x}} {\rm{ \times f(x) \times dx}}\\ &= \int_{\rm{0}}^{\rm{1}} {\rm{x}} {\rm{ \times 90}}{{\rm{x}}^{\rm{8}}}{\rm{(1 - x) \times dx}}\\ &= 90\int_{\rm{0}}^{\rm{1}} {{{\rm{x}}^{\rm{9}}}} {\rm{(1 - x) \times dx}}\\ &= 90 \int_{\rm{0}}^{\rm{1}} {\left( {{{\rm{x}}^{\rm{9}}}{\rm{ - }}{{\rm{x}}^{{\rm{10}}}}} \right)} {\rm{ \times dx}}\\ &= 90\left( {\frac{{{{\rm{x}}^{{\rm{10}}}}}}{{{\rm{10}}}}{\rm{ - }}\frac{{{{\rm{x}}^{{\rm{11}}}}}}{{{\rm{11}}}}} \right)_{\rm{0}}^{\rm{1}}\\ &= 90\left( {\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{11}}}}} \right)\\ &= \frac{{\rm{9}}}{{{\rm{11}}}}\\ E(X) &= 0 {\rm{.8182}}\end{aligned}\)

The expected or mean value of a continuous\({\rm{rv}}\)X with pdf\({\rm{f(x)}}\)is defined as,

\(\begin{array}{c}{\rm{\mu = E(X)}}\\{\rm{ = }}\int_{{\rm{ - }}\infty }^\infty {\rm{x}} {\rm{ \times f(x) \times dx}}\end{array}\)

To calculate variance for the pdf\({\rm{f(x)}}\), we first calculate\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\):

\(\begin{aligned}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right) &= \int_{{\rm{ - }}\infty }^\infty {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x) \times dx}}\\ &= \int_{\rm{0}}^{\rm{1}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times 90}}{{\rm{x}}^{\rm{8}}}{\rm{(1 - x) \times dx}}\\ &= 90 \int_{\rm{0}}^{\rm{1}} {{{\rm{x}}^{{\rm{10}}}}} {\rm{(1 - x) \times dx}}\\ &= 90 \int_{\rm{0}}^{\rm{1}} {\left( {{{\rm{x}}^{{\rm{10}}}}{\rm{ - }}{{\rm{x}}^{{\rm{11}}}}} \right)} {\rm{ \times dx}}\\ &= 90 \left( {\frac{{{{\rm{x}}^{{\rm{11}}}}}}{{{\rm{11}}}}{\rm{ - }}\frac{{{{\rm{x}}^{{\rm{12}}}}}}{{{\rm{12}}}}} \right)_{\rm{0}}^{\rm{1}}\\ &= 90\left( {\frac{{\rm{1}}}{{{\rm{11}}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{12}}}}} \right)\\ &= \frac{{{\rm{15}}}}{{{\rm{22}}}}\\ E(X) &= 0{\rm{.6818}}\end{aligned}\)

We use the following proposition because we already calculated\({\rm{E(X)}}\)in the previous section:\({\rm{E(X) = 0}}{\rm{.8182}}\).

\({\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - E(X}}{{\rm{)}}^{\rm{2}}}\)

This allows us to write:

\(\begin{array}{c}{\rm{V(X) = 0}}{\rm{.6818 - (0}}{\rm{.8182}}{{\rm{)}}^{\rm{2}}}\\{\rm{V(X) = 0}}{\rm{.0123}}\end{array}\)

Standard deviation\(\left( {{{\rm{\sigma }}_{\rm{X}}}} \right)\)can now be represented as follows:

\(\begin{array}{c}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} \\{\rm{ = }}\sqrt {{\rm{0}}{\rm{.0123}}} \\{{\rm{\sigma }}_{\rm{X}}}{\rm{ = 0}}{\rm{.111}}\end{array}\)

\({\rm{h(X)}}\)is any function of X if X is a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\), then,

\({\rm{E(h(x)) = }}\int_{{\rm{ - }}\infty }^\infty {\rm{h}} {\rm{(x) \times f(x) \times dx}}\)

Therefore, the values are \({\rm{0}}{\rm{.8182}}\) and \({\rm{0}}{\rm{.111}}\).

Explanation

(f) We may deduce the mean and standard deviation of the provided pdf from part(a):

\(\begin{aligned}E(X) &= \mu \\ &= 0 {\rm{.8182}}\\{{\rm{\sigma }}_{\rm{X}}} &= 0 {\rm{.1111}}\end{aligned}\)

\({\rm{P}}\left( {{\rm{\mu - }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + }}{{\rm{\sigma }}_{\rm{X}}}} \right)\) stands for the chance that the observed depth is within one standard deviation of the mean value.

\(\begin{aligned}{\rm{P}}\left( {{\rm{\mu - }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + }}{{\rm{\sigma }}_{\rm{X}}}} \right) &= P(0 {\rm{.7071 < X < 0}}{\rm{.9293)}}\\ &= F(0 {\rm{.9293) - F(0}}{\rm{.7071)}}\\ &= \left( {{{{\rm{(0}}{\rm{.9293)}}}^{\rm{9}}}{\rm{(10 - 9(0}}{\rm{.9293)) - }}\left( {{{{\rm{(0}}{\rm{.7071)}}}^{\rm{9}}}{\rm{(10 - 9(0}}{\rm{.7071))}}} \right.} \right.\\ &= (0{\rm{.8458) - (0}}{\rm{.1604)}}\\{\rm{P}}\left( {{\rm{\mu - }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + }}{{\rm{\sigma }}_{\rm{X}}}} \right) &= 0 {\rm{.6854}}\end{aligned}\)

We need to figure out the likelihood that X is more than one standard deviation away from its mean value. Let p be the symbol for it.

Then

\(\begin{aligned} p &= 1 - P\left( {{\rm{\mu - }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + }}{{\rm{\sigma }}_{\rm{X}}}} \right)\\ &= 1 - 0 {\rm{.6854}}\\ p &= 0 {\rm{.3146}}\end{aligned}\)

As a result, the chance that\({\rm{X}}\)is more than one standard deviation away from its mean value is\({\rm{0}}{\rm{.3149}}\).

Definition: If\({\rm{\bar A}}\)is the complement of\({\rm{A}}\), then

\({\rm{P(\bar A) = 1 - P(A)}}\)

Therefore, the value is \({\rm{0}}{\rm{.3146}}\).