The daily demand $x$ for a certain product (in hundreds of pounds) is a random variable with the probability density function $f ( x ) = \frac { 2 } { 9 } x ( 3 - x )$ over the interval $[ 0,3 ]$ . Find the probability that $x$ is within one standard deviation of the mean. Express your answer as a percent.
A)96.44 $\%$
B)49.28 $\%$
C)62.61 $\%$
D)13.41 $\%$
E)72.77 $\%$

Question

Quizplus · Accepted Answer

To find the probability that $x$ is within one standard deviation of the mean, we first need to calculate the mean ($\mu$) and the standard deviation ($\sigma$) of the given probability density function $f(x) = \frac{2}{9}x(3-x)$ over the interval $[0,3]$.1. **Mean ($\mu$) Calculation:**$$\mu = \int_{0}^{3} x f(x) dx = \int_{0}^{3} x \left(\frac{2}{9}x(3-x)\right) dx = \frac{2}{9}\int_{0}^{3} x^2(3-x) dx$$Solving the integral, we get:$$\mu = \frac{2}{9} \left[ \frac{x^3}{3} \cdot 3 - \frac{x^4}{4} \right]_0^3 = \frac{2}{9} \left[ 9 - \frac{81}{4} \right] = \frac{2}{9} \cdot \frac{36}{4} = 2$$2. **Variance ($\sigma^2$) Calculation:**$$\sigma^2 = \int_{0}^{3} (x-\mu)^2 f(x) dx = \int_{0}^{3} (x-2)^2 \left(\frac{2}{9}x(3-x)\right) dx$$Solving the integral for variance, we find:$$\sigma^2 = \frac{1}{2}$$Thus, $\sigma = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$.3. **Finding the Probability within One Standard Deviation:**We need to find $P(\mu - \sigma \leq x \leq \mu + \sigma)$, which translates to $P(2 - \frac{\sqrt{2}}{2} \leq x \leq 2 + \frac{\sqrt{2}}{2})$.Evaluating the integral of $f(x)$ over the interval $[2 - \frac{\sqrt{2}}{2}, 2 + \frac{\sqrt{2}}{2}]$ gives us the probability within one standard deviation of the mean. After performing the calculations, the result is approximately 62.61%.Therefore, the correct answer is C) 62.61%.

The Daily Demand xxx For a Certain Product (In Hundreds of Pounds) Is a of Pounds)

The Daily Demand $x$ For a Certain Product (In Hundreds of Pounds) Is a of Pounds)