Chapter 7, Problem 108CE

a.

To determine

Compute the mean.

Answer to Problem 108CE

The mean is approximately 68.33.

Explanation of Solution

Calculation:

It is given that the distribution of scores on a statistics exam follows triangular distribution with lower limit 50, mode 60 and upper limit of 95, that is, $T\left(50,60,95\right)$.

Triangular distribution:

A continuous random variable X is said to follow triangular distribution if the probability density function of X is,

$f\left(x\right)=\{\begin{array}{l}\frac{2\left(x-a\right)}{\left(b-a\right)\left(c-a\right)}\text{ for}a\le x\le b\\ \frac{2\left(c-x\right)}{\left(c-a\right)\left(c-b\right)}\text{ for}b\le x\le c\end{array}$ ,

where a, b and c are the lower limit, mode and upper limit, respectively.

Mean of a Triangular distribution is defined as,

$\mu =\frac{a+b+c}{3}$.

It is given that, $a=50$, $b=60$ and $c=95$.

Thus, the mean of the Triangular distribution is,

$\begin{array}{c}\mu =\frac{50+60+95}{3}\\ =\frac{205}{3}\\ =\underset{\_}{68.33}.\end{array}$

Therefore, the mean is 68.33.

b.

To determine

Compute the standard deviation.

Answer to Problem 108CE

The standard deviation is 9.65.

Explanation of Solution

Calculation:

The standard deviation of a Triangular distribution is defined as,

$\sigma =\sqrt{\frac{a^2+b^2+c^2-ab-ac-bc}{18}}$.

It is given that, $a=50$, $b=60$ and $c=95$.

Thus, the standard deviation of the Triangular distribution is,

$\begin{array}{c}\sigma =\sqrt{\frac{a^2+b^2+c^2-ab-ac-bc}{18}}\\ =\sqrt{\frac{{50}^2+{60}^2+{95}^2-\left(50\right)\left(60\right)-\left(50\right)\left(95\right)-\left(60\right)\left(95\right)}{18}}\\ =\sqrt{\frac{2,500+3,600+9,025-3,000-4,750-5,700}{18}}\\ =\sqrt{\frac{1,675}{18}}\\ =\sqrt{93.056}\\ \approx \underset{\_}{9.65}.\end{array}$

Therefore, the standard deviation of the Triangular distribution is 9.65.

c.

To determine

Compute the probability that a score will be less than 75.

Answer to Problem 108CE

The probability that a score will be less than 75 is 0.746.

Explanation of Solution

Calculation:

The cumulative distribution function (CDF) of a Triangular distribution is defined as,

$P\left(X\le x\right)=\{\begin{array}{l}\frac{{\left(x-a\right)}^2}{\left(b-a\right)\left(c-a\right)}\text{for }a\le x\le b\\ 1-\frac{{\left(c-x\right)}^2}{\left(c-a\right)\left(c-b\right)}\text{for b}\le x\le c\end{array}$ .

Assume that the random variable X denotes the scores of Statistics exam.

It is given that, $a=50$, $b=60$, $c=95$ and $x=75$.

Hence, the CDF of the Triangular distribution will be,

$P\left(X\le x\right)=1-\frac{{\left(c-x\right)}^2}{\left(c-a\right)\left(c-b\right)}\text{for b}\le x\le c$.

Thus, the probability that a score will be less than 75 is,

$\begin{array}{c}P\left(X\le 75\right)=1-\frac{{\left(95-75\right)}^2}{\left(95-50\right)\left(95-60\right)}\\ =1-\frac{{20}^2}{\left(45\right)\left(35\right)}\\ =1-\frac{400}{1,575}\\ =1-0.2539\\ =\underset{\_}{0.746}.\end{array}$

Therefore, the probability that a score will be less than 75 is 0.746.

d.

To determine

Draw the distribution and shade the area for the event in part (c).

Answer to Problem 108CE

The shaded sketch is obtained as:

Explanation of Solution

Calculation:

It is given that, $a=50$, $b=60$ and $c=95$.

Graphical Procedure:

Step by step procedure to obtain the sketch is given below:

• Draw a left skewed triangle with base from the lower limit 50 to upper limit 95 and with the height at the mode of 60.
• Shade the left side area from the mode of 75.

The shaded sketch is obtained as:

The shaded area in the sketch is the event in part (c).