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میتونی لایو بذاری!

a) The diagrams show the graphs of two functions, g and h. For each of the functions g and h, give a reason why it cannot be a probability density function.

b) The distance, in kilometres, travelled in a given time by a cyclist is represented by the continuous random variable $X$ with probability density function given by

$f\left( x \right) = \left\{ \begin{gathered}
  \frac{{30}}{{{x^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10 \leqslant x \leqslant 15, \hfill \\
  0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise. \hfill \\ 
\end{gathered}  \right.$

(i) Show that $E\left( X \right) = 30{\text{ }}ln{\text{ }}1.5.$

(ii) Find the median of $X$. Find also the probability that $X$ lies between the median and the mean. 

پاسخ تشریحی :
نمایش پاسخ

a) $g:Area \ne 1$ or $ \gt 1$

h: pdf cannot be neg

b)(i) $\int\limits_{10}^{15} {\frac{{30}}{x}dx} $

$ = \left[ {15\,ln\,x} \right]_{10}^{15}$

$ = 30\left( {ln\,15 - ln\,10} \right)$

$\left( { = 30ln\,1.5} \right)$

(ii) $\int\limits_{10}^m {\frac{{30}}{{{x^2}}}\,dx}  = 0.5$

$\left[ { - 30{x^{ - 1}}} \right]_{10}^m = 0.5$

$ - \frac{{30}}{m} - ( - \frac{{30}}{{10}}) = 0.5$

$m = 12$

$\int\limits_{'12'}^{30ln1.5} {\frac{{30}}{{{x^2}}}\,dx} $

$ = 0.0337$ (3 sfs)

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