Answer to Question #224447 in Statistics and Probability for Qasim

Question

Question 2:

The Project Manager found that distribution of the amount of gravel (in tons) sold by a particular construction supply company in a given week is a continuous random variable X with PDF as 

f(x)={█(3/2(1-x^2)     0≤x ≤1@0             elsewhere)┤

Find the expected value and standard deviation of the amount of gravel sold in a week.

Consider a normally distributed random variable X with a mean of 85 and standard deviation of 10. Find P(90≤X≤100)?

Answer 1

The probability density function of X

$f(x)=\left\{\begin{matrix} (3/2)(1-x^2)\; 0 \leqslant x\leqslant 1 & \\ 0 \; elsewhere & \end{matrix}\right.$

Obtain the expected value of amount of gravel sold in a week:

$E(X)= \int^{1}_{0}xf(x)dx \\ = \int^{1}_{0} \frac{3}{2 }x(1-x^2)dx \\ = \frac{3}{2}\int^{1}_{0}(x-x^3)dx \\ = \frac{3}{2}(\frac{x^2}{2} -\frac{x^4}{4})^1_0 \\ = \frac{3}{2}(\frac{1}{2} -\frac{1}{4}) \\ = \frac{3}{2} \times \frac{2}{8} \\ = \frac{3}{8}$

Obtain the Standard deviation of amount of gravel sold in a week:

$E(X^2) = \int^{1}_{0}x^2f(x)dx \\ =\int^{1}_{0} \frac{3}{2}x^2(1-x^2)dx \\ = \frac{3}{2}\int^{1}_{0} x^2 -x^4 dx \\ = \frac{3}{2}(\frac{x^3}{3} -\frac{x^5}{5})^1_0 \\ = \frac{3}{2}(\frac{1}{3}-\frac{2}{15}) \\ = \frac{3}{2} \times \frac{2}{15} \\ = \frac{3}{15} \\ \sigma^2= E(X^2) -[E(X)]^2 \\ = \frac{3}{15}-(\frac{3}{8})^2 \\ = \frac{3}{15} -\frac{9}{64} \\ = 0.2-0.140625 \\ = 0.0594 \\ \sigma = \sqrt{0.0594}= 0.243$

The Standard deviation of amount of gravel sold in a week is 0.243.

Obtain the value of P(90≤X<100):

$P(90≤X≤100) = P(X≤100) -P(X≤90) \\ = P(Z≤ \frac{100-85}{10}) -P(Z≤\frac{90-85}{10}) \\ = P(Z≤1.5) -P(Z≤0.5) \\ = 0.9332-0.6915 \\ = 0.2417$

The indicated probability is 0.2417