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Q. 5.1

Expert-verifiedFound in: Page 217

Book edition
9th

Author(s)
Sheldon M. Ross

Pages
432 pages

ISBN
9780321794772

The number of minutes of playing time of a certain high school basketball player in a randomly chosen game is a random variable whose probability density function is given in the following figure:

Find the probability that the player plays

(a) more than $15$ minutes;

(b) between $20and35$ minutes;

(c) less than $30$ minutes;

(d) more than $36$ minutes

(a) The probability that the player plays more than $15$ minutes is $0.875$

(b) The probability that the player plays between $20and35$ is $0.625$

(c) The probability that the player plays less than $30$minutes is $0.75$

(d) The probability that the player plays more than $36$ minutes is $0.1$

Formalize the given probability function. As may be observed from the graph,

$f\left(x\right)=0.025,x\in [10,20)\cup (30,40]$

$f\left(x\right)=0.05,x\in [20,30]$

$f\left(x\right)=0$, otherwise

$X$ is the random variable with the density function defined $P$. The needed probabilities are calculated as the integrals of the density function $f$ over the relevant intervals.

$P(X>15)=1-P(X\le 15)\phantom{\rule{0ex}{0ex}}=1-{\int}_{10}^{15}f\left(x\right)dx\phantom{\rule{0ex}{0ex}}=1-{\int}_{10}^{15}0.025dx$

$=1-0.025\times 5\phantom{\rule{0ex}{0ex}}=0.875$

$P(X\in (20,35\left)\right)=P(X\in (20,30\left)\right)+P(X\in (30,35\left)\right)$

$={\int}_{20}^{30}f\left(x\right)dx+{\int}_{30}^{35}f\left(x\right)dx$

$=0.05\times 10+0.025\times 5\phantom{\rule{0ex}{0ex}}=0.625$

$P(X<30)=1-P(X\ge 30)\phantom{\rule{0ex}{0ex}}=1-{\int}_{30}^{40}f\left(x\right)dx\phantom{\rule{0ex}{0ex}}=1-{\int}_{30}^{40}0.025dx$

$=1-0.025\times 10\phantom{\rule{0ex}{0ex}}=0.75$

$P(X>36)={\int}_{36}^{40}f\left(x\right)dx\phantom{\rule{0ex}{0ex}}={\int}_{36}^{40}0.025dx\phantom{\rule{0ex}{0ex}}=0.025\times 4\phantom{\rule{0ex}{0ex}}=0.1$

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