Answer to Question #219514 in Statistics and Probability for mimi

Question

The amount of time it takes for a steak order to be prepared at an expensive restaurant is uniformly distributed between 10 and 60 minutes.

(a) What is the probability that a steak order is prepared in less than 20 minutes? (5)

(b) What percentage of steak orders take more than 55 minutes to prepare? (5)

(c) What is the standard deviation of the time taken for a steak order to be prepared?

Answer 1

$f(x)=\frac{1}{50}, x\in[10,60]$

(a) $P(x<20)$

$P(x<20)=\int_{10}^{20}\frac{1}{50}dx$

$=\frac{x}{50}|_{10}^{20}$

$=\frac{20}{50}-\frac{10}{50}$

$=\frac{3}{5}$

(b)$P(x>55)$

$P(x>55)=\int_{55}^{60}\frac{1}{50}dx$

$=\frac{x}{50}|_{55}^{60}$

$=\frac{60}{50}-\frac{55}{50}$

$=\frac{1}{10}$

$= 10\%$

(c) $\sigma$

$\sigma=\sqrt{E(X^2)-(E(x))^2}$

$E(x)=\int_{10}^{60}\frac{x}{50}dx$

$=\frac{x^2}{100}|_{10}^{60}$

$=\frac{60^2-10^2}{100}$

$=35$

$E(x^2)=\int_{10}^{60}\frac{x^2}{50}dx$

$=\frac{x^3}{150}|_{10}^{60}$

$=\frac{60^3-10^3}{150}$

$=\frac{4300}{3}$

$\sigma=\sqrt{\frac{4300}{3}-35^2}$

$=14.434$