Question

According to a study by Dr. John McDougall of his live-in weight loss program, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. a. Define the random variable. X = _________ b. X ~ _________ c. Graph the probability distribution. d. f(x) = _________ e. μ = _________ f. σ = _________ g. Find the probability that the individual lost more than ten pounds in a month. h. Suppose it is known that the individual lost more than ten pounds in a month. Find the probability that he lost less than 12 pounds in the month. i. P(7 < x < 13|x > 9) = __________. State this in a probability question, similarly to parts g and h, draw the picture, and find the probability.

Short answer

All data has been given below.

Step-by-step solution

Measurement of variables

a.
$x=\left(\begin{array}{c}arandomselctedweightloss\\ onemonthindivualprogram\end{array}\right)$

b.

Uniform distribution of X is

$X~U(6,15)$

c. The probability distribution is

localid="1648231021239" $\begin{array}{rcl}f\left(x\right)& =& \frac{1}{b-a}\\ & =& \frac{1}{15-6}\\ & =& \frac{1}{9}\end{array}$

So, the graph is

d.

The calculation of part c is

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{l}\frac{1}{9}\\ 0\end{array}\right.6<x<15 \\ oisforotherwise \end{gathered}$$

e.

The mean value is

localid="1648231128407" $\begin{array}{rcl}\mu & =& \frac{a+b}{2}\\ & =& \frac{6+15}{2}\\ & =& 21/2\\ & =& 10.5\end{array}$

f.

The value of standard deviation

$\begin{array}{rcl}\sigma & =& \frac{\sqrt{b-a^2}}{12}\\ \sigma & =& \frac{\sqrt{15-6^2}}{12}\\ & =& 2.60\end{array}$

g.

$$\begin{gathered} P(x<10)=base\times height \\ =(15-10)\times \frac{1}{9} \\ =0.56 \end{gathered}$$

Calculation of variables

h.

The calculation of $P(7<x<13|x>9)$

$\begin{array}{rcl}P(7& <& x<13|x>9)=base\times height\\ & =& (12-10)\times \frac{1}{15-10}\\ & =& 0.40\end{array}$

i. The calculation of probability

localid="1648231805153" $\begin{array}{rcl}P(7& <& x<13|x>9)=base\times height\\ & =& (13-9)\times \frac{1}{(15-9)}\\ & =& 0.67\end{array}$

The curve of the probability