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Chapter 4: Q127E (page 197)

An individual鈥檚 credit score is a number calculated based on that person鈥檚 credit history that helps a lender determine how much he/she should be loaned or what credit limit should be established for a credit card. An article in the Los Angeles Times gave data which suggested that a \({\rm{\beta }}\) distribution with parameters \({\rm{A = 150,B = 850,\alpha = 8,\beta = 2}}\)would provide a reasonable approximation to the distribution of American credit scores. (Note: credit scores are integer-valued).

a. Let \({\rm{X}}\)represent a randomly selected American credit score. What are the mean value and standard deviation of this random variable? What is the probability that \({\rm{X}}\) is within 1 standard deviation of its mean value?

b. What is the approximate probability that a randomly selected score will exceed \({\rm{750}}\) (which lenders consider a very good score)?

Short Answer

Expert verified

(a) The average is 710, while the standard deviation is \(\frac{{{\rm{280}}\sqrt {{\rm{11}}} }}{{{\rm{11}}}}{\rm{\gg 84}}{\rm{.8232}}\)\({\rm{P(\mu - \sigma < x < \mu + \sigma ) = 0}}{\rm{.6845 = 68}}{\rm{.45\% }}\)

(b) \({\rm{P(X > 750) = 0}}{\rm{.3757 = 37}}{\rm{.57\% }}\)

Step by step solution

01

Definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Calculating the mean value and standard deviation of this random variable

Given:

\begin{aligned}A&= 150,\hfill\\B&= 850,\hfill\\\alpha&= 8,\hfill\\\beta&= 2\hfill\\\end{aligned}

(a) The mean and variance formula

\(\begin{aligned}{}{\rm{\mu }}&{{\rm{ = A + (B - A)}}\frac{{\rm{\alpha }}}{{{\rm{\alpha + \beta }}}}}\\{{{\rm{\sigma }}^{\rm{2}}}}&{{\rm{ = }}\frac{{{{{\rm{(B - A)}}}^{\rm{2}}}{\rm{\alpha \beta }}}}{{{{{\rm{(\alpha + \beta )}}}^{\rm{2}}}{\rm{(\alpha + \beta + 1)}}}}}\end{aligned}\)

Substitute known values for the variables and evaluate:

\(\begin{aligned}{{}{}}{\rm{\mu }}&{{\rm{ = A + (B - A)}}\frac{{\rm{\alpha }}}{{{\rm{\alpha + \beta }}}}{\rm{ = 150 + (850 - 150)}}\frac{{\rm{8}}}{{{\rm{8 + 2}}}}{\rm{ = 710}}}\\{{{\rm{\sigma }}^{\rm{2}}}}&{{\rm{ = }}\frac{{{{{\rm{(B - A)}}}^{\rm{2}}}{\rm{\alpha \beta }}}}{{{{{\rm{(\alpha + \beta )}}}^{\rm{2}}}{\rm{(\alpha + \beta + 1)}}}}{\rm{ = }}\frac{{{{{\rm{(850 - 150)}}}^{\rm{2}}}{\rm{(8)(2)}}}}{{{{{\rm{(8 + 2)}}}^{\rm{2}}}{\rm{(8 + 2 + 1)}}}}{\rm{ = }}\frac{{{\rm{78400}}}}{{{\rm{11}}}}}\end{aligned}\)

The square root of the variance is the standard deviation:

\({\rm{\sigma = }}\sqrt {{{\rm{\sigma }}^{\rm{2}}}} {\rm{ = }}\sqrt {\frac{{{\rm{78400}}}}{{{\rm{11}}}}} {\rm{ = }}\frac{{{\rm{280}}\sqrt {{\rm{11}}} }}{{{\rm{11}}}}{\rm{\gg 84}}{\rm{.8232}}\)

Calculate the values that differ by one standard deviation from the mean:

\(\begin{aligned}{}{}&{{\rm{\mu - \sigma = 710 - 84}}{\rm{.8232 = 625}}{\rm{.1768}}}\\{}&{{\rm{\mu + \sigma = 710 + 84}}{\rm{.8232 = 794}}{\rm{.8232}}}\end{aligned}\)

03

Calculating the probability that \({\rm{X}}\) is within \({\rm{1}}\) standard deviation of its mean value

A Beta distribution's probability density function is:

\({\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{B - A}}}}\frac{{{\rm{\Gamma (\alpha + \beta )}}}}{{{\rm{\Gamma (\alpha )\Gamma (\beta )}}}}{\left( {\frac{{{\rm{x - A}}}}{{{\rm{B - A}}}}} \right)^{{\rm{\alpha - 1}}}}{\left( {\frac{{{\rm{B - x}}}}{{{\rm{B - A}}}}} \right)^{{\rm{\beta - 1}}}}\)

The integral of the probability density function over the appropriate period is then the probability that \({\rm{X}}\) is within \({\rm{1}}\) standard deviation of its mean value

\begin{aligned}{}{P(\mu - \sigma < x < \mu + \sigma )= `o _{625.1768}^{794.8232}n\frac{1}{{850 - 150}}\frac{{\Gamma (8 + 2)}}{{\Gamma (8)\Gamma (2)}}{{\left( {\frac{{x - 150}}{{850 - 150}}} \right)}^{8 - 1}}{{\left( {\frac{{850 - x}}{{850 - 150}}}\right)}^{2 - 1}}dx} \\{= `o _{625.1768}^{794.8232}n\frac{1}{{700}}\frac{{362880}}{{5040}}{{\left( {\frac{{x - 150}}{{700}}} \right)}^7}{{\left( {\frac{{850 - x}}{{700}}} \right)}^1}dx} \\{ = \frac{1}{{{{700}^9}}}\frac{{362880}}{{5040}}`o _{625.1768}^{794.8232}n{{(x - 150)}^7}(850 - x)dx} \\{\gg 0.6845 = 68.45\% }\end{aligned}

04

Calculating the approximate probability that a randomly selected score will exceed \({\rm{750}}\)

(b) The probability density function is only specified on 150拢x拢850

The integral of the probability density function over the corresponding interval is the probability:

\begin{aligned}{}{P(X > 750)= `o _{750}^{850}n\frac{1}{{850 - 150}}\frac{{\Gamma (8 + 2)}}{{\Gamma (8)\Gamma (2)}}{{\left( {\frac{{x - 150}}{{850 - 150}}} \right)}^{8 - 1}}{{\left( {\frac{{850 - x}}{{850 - 150}}} \right)}^{2 - 1}}dx} \\{ = `o _{750}^{850}n\frac{1}{{700}}\frac{{362880}}{{5040}}{{\left( {\frac{{x - 150}}{{700}}} \right)}^7}{{\left( {\frac{{850 - x}}{{700}}} \right)}^1}dx} \\{ = \frac{1}{{{{700}^9}}}\frac{{362880}}{{5040}}`o _{750}^{850}n{{(x - 150)}^7}(850 - x)dx} \\{ = \frac{{15159367}}{{40353607}}\gg 0.3757 = 37.57\% }\end{aligned}

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