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Chapter 5: Q14E (page 316)

Suppose that on a certain examination in advanced mathematics, students from university A achieve scores normally distributed with a mean of 625 and a variance of 100, and students from university B achieve scores normally distributed with a mean of 600 and a variance of 150. If two students from university A and three students from university B take this examination, what is the probability that the average of the scores of the two students from university A will be greater than the average of the scores of the three students from university B? Hint: Determine the distribution of the difference between the two averages

Short Answer

Expert verified

The probability that the average of the scores of the two students from university A will be greater than the average of the scores of the three students from university B is 0.9938.

Step by step solution

01

Given information

Let\(\bar X\)the average scores of two students from university A be the average of three students from university B.

It is given that scores of students from university A are normally distributed with a mean of 625 and a variance of 100.i.e \(\bar X \sim N\left( {625,100} \right)\).

Also,university B students' scores are normally distributed with a mean of 600 and a variance of 150.i.e.,\(\bar Y \sim N\left( {600,150} \right)\)

02

Compute the Probability

Since\(\bar X\)it has a normal distribution with mean =625 and variance =\(\frac{{100}}{2}\)

i.e.,\(\bar X\)has a normal distribution with \(E\left( {\bar X} \right) = 625\)and\(V\left( {\bar X} \right) = 50\).

Also,\[\bar Y\]has a normal distribution with mean =600 and variance =\(\frac{{150}}{3}\)

i.e.,\[\bar Y\]has a normal distribution with \(E\left( {\bar Y} \right) = 600\)and\(V\left( {\bar Y} \right) = 50\).

Therefore, by the property of Normal distribution, if X and Y are independent normal variates, then\(E\left( {X - Y} \right) = E\left( X \right) - E\left( Y \right)\)and\(V\left( {X - Y} \right) = V\left( X \right) + V\left( Y \right)\)

It implies that \(E\left( {\bar X - \bar Y} \right) = E\left( {\bar X} \right) - E\left( {\bar Y} \right)\) and \(V\left( {\bar X - \bar Y} \right) = V\left( {\bar X} \right) - V\left( {\bar Y} \right)\)

Therefore,

\(\begin{array}{c}E\left( {\bar X - \bar Y} \right) = 625 - 600\\ = 25\end{array}\)

\[\begin{array}{c}V\left( {\bar X - \bar Y} \right) = 50 + 50\\ = 100\end{array}\]

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Most popular questions from this chapter

Suppose that X has the normal distribution for which the mean is 1 and the variance is 4. Find the value of each of the following probabilities:

(a). \({\rm P}\left( {X \le 3} \right)\)

(b). \({\rm P}\left( {X > 1.5} \right)\)

(c). \({\rm P}\left( {X = 1} \right)\)

(d). \({\rm P}\left( {2 < X < 5} \right)\)

(e). \({\rm P}\left( {X \ge 0} \right)\)

(f). \({\rm P}\left( { - 1 < X < 0.5} \right)\)

(g). \({\rm P}\left( {\left| X \right| \le 2} \right)\)

(h). \({\rm P}\left( {1 \le - 2X + 3 \le 8} \right)\)

Suppose that in a large lot containingTmanufactured items, 30 percent of the items are defective, and 70 percent are non-defective. Also, suppose that ten items are selected randomly without replacement from the lot.

Determine (a) an exact expression for the probability that not more than one defective item will be obtained and (b) an approximate expression for this probability based on the binomial distribution.

It is said that a random variable has the Weibull distribution with parameters a and b (a > 0 and b > 0) if X has a continuous distribution for which the p.d.f. f (x|a, b) is as follows:

\({\bf{f}}\left( {{\bf{x|a,b}}} \right){\bf{ = }}\frac{{\bf{b}}}{{{{\bf{a}}^{\bf{b}}}}}{{\bf{x}}^{{\bf{b - 1}}}}{{\bf{e}}^{{\bf{ - }}{{\left( {\frac{{\bf{x}}}{{\bf{a}}}} \right)}^{\bf{b}}}}}\,{\bf{,x > 0}}\)

Show that if X has this Weibull distribution, then the random variable \({{\bf{X}}^{\bf{b}}}\) has the exponential distribution with parameter \({\bf{\beta = }}{{\bf{a}}^{{\bf{ - b}}}}\)

Let \({{\bf{X}}_{{\bf{1,}}}}{\bf{ }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\)be i.i.d. random variables having the normal distribution with mean \({\bf{\mu }}\) and variance\({{\bf{\sigma }}^{\bf{2}}}\). Define\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\frac{{\bf{1}}}{{\bf{n}}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}_{\bf{i}}}} \) , the sample mean. In this problem, we shall find the conditional distribution of each \({{\bf{X}}_{\bf{i}}}\)given\(\overline {{{\bf{X}}_{\bf{n}}}} \).

a.Show that \({{\bf{X}}_{\bf{i}}}\)and\(\overline {{{\bf{X}}_{\bf{n}}}} \) have the bivariate normal distribution with both means \({\bf{\mu }}\), variances\({{\bf{\sigma }}^{\bf{2}}}{\rm{ }}{\bf{and}}\,\,\frac{{{{\bf{\sigma }}^{\bf{2}}}}}{{\bf{n}}}\),and correlation\(\frac{{\bf{1}}}{{\sqrt {\bf{n}} }}\).

Hint: Let\({\bf{Y = }}\sum\limits_{{\bf{j}} \ne {\bf{i}}} {{{\bf{X}}_{\bf{j}}}} \).

Now show that Y and \({{\bf{X}}_{\bf{i}}}\) are independent normal and \({{\bf{X}}_{\bf{n}}}\)and \({{\bf{X}}_{\bf{i}}}\) are linear combinations of Y and \({{\bf{X}}_{\bf{i}}}\) .

b.Show that the conditional distribution of \({{\bf{X}}_{\bf{i}}}\) given\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\overline {{{\bf{x}}_{\bf{n}}}} \)\(\) is normal with mean \(\overline {{{\bf{x}}_{\bf{n}}}} \) and variance \({{\bf{\sigma }}^{\bf{2}}}\left( {{\bf{1 - }}\frac{{\bf{1}}}{{\bf{n}}}} \right)\).

Consider a random variableXhaving the lognormal distribution with parameters渭 补苍诲蟽2. Determine the p.d.f. ofX.
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