Question

While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m. What is the probability that a randomly selected depth is within 1 standard deviation of the mean?

Answer 1

The probability that a randomly selected depth is within one standard deviation of the mean $\boxed{57.6\%}$.

Further Explanation:

Given:

A marine biologist selects water depths from a uniformly distributed collection between $2.00 m$ and $7.00 m$.

Explanation:

The depth follows uniform distribution with upper limit 7 and lower limit 2.

$\boxed{\overline X\sim U\left({a,b}\right)}$

Here,$\overline X$ is the mean, $a$ is the lower limit and $b$ is the upper limit.

The mean can be calculated as follows,

$\begin{aligned}{\text{Mean}}&=\frac{1}{2}\left({a+b}\right)\\&=\frac{1}{2}\left({2+7}\right)\\&=\frac{9}{2}\\&=4.5\\\end{aligned}$

The standard deviation of the depth can be calculated as follows,

$\begin{aligned}{\text{Standard deviation}}&=\sqrt{\frac{{{{\left({b-a}\right)}^2}}}{{12}}}\\&=\sqrt{\frac{{{{\left({7-2}\right)}^2}}}{{12}}}\\&=\sqrt{\frac{{{{\left(5\right)}^2}}}{{12}}}\\&=\sqrt{\frac{{25}}{{12}}}\\&=1.44\\\end{aligned}$

The probability density function can be obtained as follows,

$\begin{aligned}{\text{Pdf}}&=\frac{1}{{b-a}}\\&=\frac{1}{{7-2}}\\&=\frac{1}{5}\\\end{aligned}$

The probability that a randomly selected depth is within 1 standard deviation of the mean can be expressed as,

$\begin{aligned}{\text{Probability}}&=P\left({\overline X-\sigma$

Here,$\overline X$ represents the mean and $\sigma$ represents the standard deviation.

The probability that a randomly selected depth is within 1 standard deviation of the mean can be calculated as follows,

$\begin{aligned}{\text{Probability}}&=\int\limits_{3.06}^{5.94}{\frac{1}{5}dx}\\&=\frac{1}{5}\cdot\left.x\right|_{3.06}^{5.94}\\&=\frac{1}{5}\left[{5.94-3.06}\right]\\&=\frac{{2.88}}{5}\\&=0.576\\\end{aligned}$

The probability that a randomly selected depth is within one standard deviation of the mean $\boxed{57.6\%}$.

Learn more:

1. Learn more about normal distribution brainly.com/question/12698949

2. Learn more about standard normal distribution brainly.com/question/13006989

3. Learn more about confidence interval of mean brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter: Confidence Interval

Keywords: Z-score, Z-value, marine biologist, experiments, binomial distribution, standard normal distribution, standard deviation, test, measure, probability, low score, mean, repeating, indicated, normal distribution, percentile, percentage, uniformly distributed.

Answer 2

The answer is 10 and 15%