Tyra

2020-12-01

According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between 6 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. Give the distribution of X. Enter an exact number as an integer, fraction, or decimal.$f\left(x\right){=}_{}$ where $\le X\le .\mu =\sigma =$. Find the probability that the individual lost more than 8 pounds in a month.Suppose it is known that the individual lost more than 9 pounds in a month. Find the probability that he lost less than 13 pounds in the month.

unett

Step 1 Given, According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between 6 and 15 pounds a month until they approach trim body weight. Let's suppose that the weight loss is uniformly distributed. Step 2 The random variable $X=$ Weight loss in pounds Given $a=6,b=15$
$X\sim U\left(a,b\right)⇒X\sim U\left(6,15\right)$
$f\left(x\right)=\frac{1}{b-a}$
$=\frac{1}{15-6}$
$=\frac{1}{9}$
$\mu =\frac{a+b}{2}$
$=\left(6+\frac{15}{2}\right)$
$=\frac{21}{2}$
$=10.5$
$\sigma =\frac{b-a}{\sqrt{12}}$
$=\frac{15-6}{\sqrt{12}}$
$=\frac{9}{\sqrt{12}}$
$=2.5980$ Step 3 The probability that the individual lost more than 8 pounds in a month: $P\left(x>8\right)=\frac{b-x}{b-a}$
$=\frac{15-8}{15-6}$
$=0.7778$ (rounded off to 4 decimals) Step 4 Suppose it is known that the individual lost more than 9 pounds in a month. The probability that he lost less than 13 pounds in the month: $P\left(x<13\mu dx>9\right)=P\frac{99\right)$

$=\frac{\frac{13-9}{15-6}}{\frac{15-9}{15-6}}$
$=\frac{\frac{4}{9}}{\frac{6}{9}}$
$=\frac{2}{3}$
$=0.6667$ (rounded off to 4 decimals)

alenahelenash

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 $X~U\left(a,b\right)$, 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 $P\left(X>8\right)$. 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:

Substituting the values of $a$ and $b$, we have:

Now, to calculate the probability that the individual lost more than 8 pounds, we need to integrate the PDF from 8 to 15:
$P\left(X>8\right)={\int }_{8}^{15}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$
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 $P\left(X<13|X>9\right)$. 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:
${\int }_{9}^{13}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$
The denominator is the probability of losing more than 9 pounds, which we calculated earlier.
To summarize:
- The distribution of $X$ is given by $X~U\left(6,15\right)$.
- The probability that the individual lost more than 8 pounds is $P\left(X>8\right)={\int }_{8}^{15}\frac{1}{15-6}\phantom{\rule{0.167em}{0ex}}dx$.
- The probability that the individual lost less than 13 pounds, given that they lost more than 9 pounds, is $P\left(X<13|X>9\right)=\frac{{\int }_{9}^{13}\frac{1}{15-6}\phantom{\rule{0.167em}{0ex}}dx}{P\left(X>9\right)}$, where $P\left(X>9\right)$ is the probability of losing more than 9 pounds.

star233

$\frac{2}{3}$
Explanation:
The distribution of X can be represented as:

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:
$P\left(X>8\right)={\int }_{8}^{15}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=\frac{1}{9}{\int }_{8}^{15}dx=\frac{1}{9}\left[x{\right]}_{8}^{15}=\frac{7}{9}$
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:
$P\left(99\right)=\frac{P\left(99\right)}$
The probability that the individual lost between 9 and 13 pounds can be calculated as:
$P\left(9
The probability that the individual lost more than 9 pounds is:
$P\left(X>9\right)=1-P\left(X\le 9\right)=1-{\int }_{6}^{9}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=1-\frac{1}{9}{\int }_{6}^{9}dx=1-\frac{1}{9}\left[x{\right]}_{6}^{9}=\frac{2}{3}$
Therefore, the probability that the individual lost less than 13 pounds in the month, given that he lost more than 9 pounds, is:
$P\left(99\right)=\frac{P\left(99\right)}=\frac{\frac{4}{9}}{\frac{2}{3}}=\frac{2}{3}$
Hence, the probability that the individual lost less than 13 pounds in the month, given that he lost more than 9 pounds, is $\frac{2}{3}$.

karton

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:

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:
$P\left(X>8\right)={\int }_{8}^{15}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$
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:
$P\left(X>8\right)=\frac{1}{9}·\left(15-8\right)$
Simplifying this expression:
$P\left(X>8\right)=\frac{7}{9}$
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:

To find the numerator, we calculate the area under the PDF curve from x = 9 to x = 13:
$\text{Numerator}={\int }_{9}^{13}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$
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:
$\text{Numerator}=\frac{1}{9}·\left(13-9\right)$
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:
$P\left(X<13\phantom{\rule{0.167em}{0ex}}|\phantom{\rule{0.167em}{0ex}}X>9\right)=\frac{\text{Numerator}}{P\left(X>9\right)}=\frac{\frac{4}{9}}{\frac{7}{9}}$
Simplifying this expression:
$P\left(X<13\phantom{\rule{0.167em}{0ex}}|\phantom{\rule{0.167em}{0ex}}X>9\right)=\frac{4}{7}$
Therefore, the probability that the individual lost less than 13 pounds in the month, given that they lost more than 9 pounds, is $\frac{4}{7}$.

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