### Select your language

Suggested languages for you:

Americas

Europe

Q. 5.1

Expert-verified
Found in: Page 217

### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772

### Answers without the blur.

Just sign up for free and you're in.

# 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 $0.1$

See the step by step solution

## Step1: Find the probability that the player plays more than 15 minutes (part a)

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

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

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

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

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

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

## Step2: Find the probability that the player plays between 20 and 35 minutes (part b)

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

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

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

## Step3: Find the probability that the player plays less than 30 minutes (part c)

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

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

## step4: Find the probability that the player plays more than 36 minutes (part d)

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

### Want to see more solutions like these?

Sign up for free to discover our expert answers

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.