According to a study by Dr. John McDougall of his live-in weight

According to the given information, the weight loss of a randomly selected individual following the program for one month follows a uniform distribution. We can denote this distribution as $X~U(a,b)$, where $a$ and $b$ represent the minimum and maximum possible weight loss values, respectively.
Given that the weight loss is uniformly distributed, we can determine the values of $a$ and $b$ from the information provided. The study states that people lose between 6 and 15 pounds a month until they approach trim body weight. Therefore, we have $a=6$ and $b=15$.
To find the probability that the individual lost more than 8 pounds in a month, we need to calculate $P(X>8)$. Since the distribution is uniform, the probability density function (PDF) is constant within the range of the distribution. The formula for the PDF of a uniform distribution is:
$f(x)=\frac{1}{b-a}\text{for }a\le x\le b$
Substituting the values of $a$ and $b$, we have:
$f(x)=\frac{1}{15-6}\text{for }6\le x\le 15$
Now, to calculate the probability that the individual lost more than 8 pounds, we need to integrate the PDF from 8 to 15:
$P(X>8)={\int }_8^{15}f(x)dx$
Similarly, to find the probability that the individual lost less than 13 pounds given that they lost more than 9 pounds, we need to calculate $P(X<13|X>9)$. This can be found by dividing the probability of the intersection of the two events (losing more than 9 pounds and less than 13 pounds) by the probability of the event (losing more than 9 pounds).
The probability density function remains the same, but the range changes. In this case, we integrate from 9 to 13 to find the numerator:
${\int }_9^{13}f(x)dx$
The denominator is the probability of losing more than 9 pounds, which we calculated earlier.
To summarize:
- The distribution of $X$ is given by $X~U(6,15)$.
- The probability that the individual lost more than 8 pounds is $P(X>8)={\int }_8^{15}\frac{1}{15-6}dx$.
- The probability that the individual lost less than 13 pounds, given that they lost more than 9 pounds, is $P(X<13|X>9)=\frac{{\int }_9^{13}\frac{1}{15-6}dx}{P(X>9)}$, where $P(X>9)$ is the probability of losing more than 9 pounds.

Answer:
$\frac{2}{3}$
Explanation:
The distribution of X can be represented as:
$f(x)=\{\begin{array}{cc}\frac{1}{15-6}=\frac{1}{9}\hfill & \text{if }6\le x\le 15\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$
The mean (μ) and standard deviation (σ) of the distribution are both equal to:
$\mu =\sigma =\frac{6+15}{2}=10.5$
To find the probability that the individual lost more than 8 pounds in a month, we need to calculate the area under the probability density function (PDF) from x = 8 to x = 15:
$P(X>8)={\int }_8^{15}f(x)dx=\frac{1}{9}{\int }_8^{15}dx=\frac{1}{9}[x{]}_8^{15}=\frac{7}{9}$
Therefore, the probability that the individual lost more than 8 pounds in a month is $\frac{7}{9}$.
Now, given that the individual lost more than 9 pounds in a month, we want to find the probability that he lost less than 13 pounds. This can be calculated by finding the area under the PDF from x = 9 to x = 13, and dividing it by the probability that the individual lost more than 9 pounds:
$P(9<X<13|X>9)=\frac{P(9<X<13)}{P(X>9)}$
The probability that the individual lost between 9 and 13 pounds can be calculated as:
$P(9<X<13)={\int }_9^{13}f(x)dx=\frac{1}{9}{\int }_9^{13}dx=\frac{1}{9}[x{]}_9^{13}=\frac{4}{9}$
The probability that the individual lost more than 9 pounds is:
$P(X>9)=1-P(X\le 9)=1-{\int }_6^{9}f(x)dx=1-\frac{1}{9}{\int }_6^{9}dx=1-\frac{1}{9}[x{]}_6^9=\frac{2}{3}$
Therefore, the probability that the individual lost less than 13 pounds in the month, given that he lost more than 9 pounds, is:
$P(9<X<13|X>9)=\frac{P(9<X<13)}{P(X>9)}=\frac{\frac{4}{9}}{\frac{2}{3}}=\frac{2}{3}$
Hence, the probability that the individual lost less than 13 pounds in the month, given that he lost more than 9 pounds, is $\frac{2}{3}$.

Given that the weight loss is uniformly distributed between 6 and 15 pounds, we can determine the probability density function (PDF) for X, denoted as f(x).
The range of X is between 6 and 15, so the probability density is constant within this range. We can calculate the height of the uniform distribution by dividing 1 by the width of the range.
The width of the range is 15 - 6 = 9 pounds. Therefore, the height of the uniform distribution is 1/9.
The PDF for X can be defined as follows:
$f(x)=\{\begin{array}{cc}\frac{1}{9}\hfill & \text{if }6\le x\le 15\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$
Next, we need to find the probability that the individual lost more than 8 pounds in a month. This can be calculated by finding the area under the PDF curve from x = 8 to x = 15.
The probability can be expressed as follows:
$P(X>8)={\int }_8^{15}f(x)dx$
Since f(x) is a constant within the range [8, 15], we can simply multiply the height (1/9) by the width (15 - 8) to find the area:
$P(X>8)=\frac{1}{9}·(15-8)$
Simplifying this expression:
$P(X>8)=\frac{7}{9}$
Therefore, the probability that the individual lost more than 8 pounds in a month is $\frac{7}{9}$.
Now, let's find the probability that the individual lost less than 13 pounds in the month, given that we know they lost more than 9 pounds.
We can use conditional probability to calculate this. The probability that the individual lost less than 13 pounds given that they lost more than 9 pounds can be expressed as:
$P(X<13|X>9)=\frac{P(X<13\text{ and }X>9)}{P(X>9)}$
To find the numerator, we calculate the area under the PDF curve from x = 9 to x = 13:
$\text{Numerator}={\int }_9^{13}f(x)dx$
Since f(x) is a constant within the range [9, 13], we can again multiply the height (1/9) by the width (13 - 9) to find the area:
$\text{Numerator}=\frac{1}{9}·(13-9)$
Simplifying this expression:
$\text{Numerator}=\frac{4}{9}$
To find the denominator, we use the probability that X > 9, which we calculated earlier as 7/9.
Now we can substitute the numerator and denominator into the conditional probability formula:
$P(X<13|X>9)=\frac{\text{Numerator}}{P(X>9)}=\frac{\frac{4}{9}}{\frac{7}{9}}$
Simplifying this expression:
$P(X<13|X>9)=\frac{4}{7}$
Therefore, the probability that the individual lost less than 13 pounds in the month, given that they lost more than 9 pounds, is $\frac{4}{7}$.