Probability

Random variables

1. A wheel of fortune is divided in 15 sections numbered with values 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5. The random variable $X$ is the number selected when the wheel is spun. Write the table of all possible results and their respective probabilities.
   Solutions:    $\begin{array}{|lccccc|}\hline \mathrm{Result}x& 1& 2& 3& 4& 5\\ \mathrm{Probability}P(X=x)& \frac{1}{3}& \frac{4}{15}& \frac{1}{5}& \frac{2}{15}& \frac{1}{15}\\ \hline\end{array}$
2. Three marbles are extracted (without replacement) out of a jar which includes 3 white and 4 red marbles. The random variable $X$ is the number of red marbles obtained. Write the table of probability distribution.
   Solutions:    $\begin{array}{|lcccc|}\hline x& 0& 1& 2& 3\\ P(X=x)& \frac{1}{35}& \frac{12}{35}& \frac{18}{35}& \frac{4}{35}\\ \hline\end{array}$
3. ?
   ?

   Expected value (in general):
   
   $E(X)=x_1{p}_1+x_2{p}_2+\cdots +x_n{p}_n$
   The random variable $X$ is the number of heads obtained when you toss a coin four times.
   

   (a)   Write the table of probability distribution.

   (b)   Find the expected value $E(X)$.

   Solutions:    (a)   $\begin{array}{|lccccc|}\hline x& 0& 1& 2& 3& 4\\ P(X=x)& \frac{1}{16}& \frac{1}{4}& \frac{3}{8}& \frac{1}{4}& \frac{1}{16}\\ \hline\end{array}$     (b)  $E(X)=2$
4. Math test is composed of three exercises: the first is on trigonometry, the second on logarithms and the third one on probability. An average student can solve 65% of exercises on trigonometry, 70% of exercises on logarithms and 80% of exercises on probability. The random variable $X$ represents the number of exercises (0, 1, 2, or all 3) solved by an average student.
   

   (a)   Write the table of probability distribution.

   (b)   Find the expected value $E(X)$.

   Solutions:    (a)   $\begin{array}{|lcccc|}\hline x& 0& 1& 2& 3\\ P(X=x)& 2.1\mathrm{\%}& 17.2\mathrm{\%}& 44.3\mathrm{\%}& 36.4\mathrm{\%}\\ \hline\end{array}$     (b)  $E(X)=2.15$
5. The table shows the probability distribution of a random variable $X$.
   $\begin{array}{|lcccccc|}\hline x& 2& 4& 6& 8& 10& 12\\ P(X=x)& \frac{5}{20}& \frac{7}{20}& \frac{3}{20}& \frac{3}{20}& p& \frac{1}{20}\\ \hline\end{array}$
   

   (a)   Find the unknown $p$.

   (b)   Find the expected value $E(X)$.

   Solutions:    (a)  $p=\frac{1}{20}$;    (b)  $E(X)=5.1$
6. Probability distribution of a random variable is given by the formula: ${\displaystyle P(X=x)=\frac{6-x}{12}}$, for $x\in \{1,2,3\}$.
   

   (a)   Write the table of probability distribution.

   (b)   Find the expected value $E(X)$.

   Solutions:    (a)   $\begin{array}{|lccc|}\hline x& 1& 2& 3\\ P(X=x)& \frac{5}{12}& \frac{1}{3}& \frac{1}{4}\\ \hline\end{array}$     (b)  $E(X)\approx 1.83$
7. Probability distribution of a random variable is given by the formula: $P(X=x)=\frac{1}{29}(x+5)$, for $x\in \{3,4,7\}$.
   

   (a)   Write the table of probability distribution.

   (b)   Find the expected value $E(X)$.

   Solutions:    (a)   $\begin{array}{|lccc|}\hline x& 3& 4& 7\\ P(X=x)& \frac{8}{29}& \frac{9}{29}& \frac{12}{29}\\ \hline\end{array}$     (b)  $E(X)\approx 4.97$

Binomial distribution

8. ?
   ?

   Binomial distribution $B(n,p)$:
   
   $P(X=r)=(\genfrac{}{}{0ex}{}{n}{r})p^r{(1-p)}^{n-r}$
   A standard fair playing die is rolled 10 times. The random variable $X$ represents the number of sixes. Calculate:
   

   (a)   $P(X=0)$.

   (b)   $P(X=1)$.

   (c)   $P(X⩾1)$.

   (d)   $P(X⩾2)$.

   Solutions:    (a)  $P(X=0)\approx 16.2\mathrm{\%}$;    (b)  $P(X=1)\approx 32.3\mathrm{\%}$;    (c)  $P(X⩾1)\approx 83.8\mathrm{\%}$;    (d)  $P(X⩾2)\approx 51.5\mathrm{\%}$
9. ?
   ?

   For binomial distribution only:
   
   $E(X)=np$
   
   $Var(X)=np(1-p)$
   
   $\sigma =\sqrt{np(1-p)}$
   A standard fair playing die is rolled 120 times. The random variable $X$ represents the number of sixes. Calculate $E(X)$.
   Solutions:    $E(X)=20$
10. The random variable $X$ is distributed as $X\sim B(60,0.2)$. Find $E(X)$ and $\sigma$
    Solutions:    $E(X)=12,\sigma \approx 3.10$

11. A standard fair playing die is rolled 12 times. The random variable $X$ represents the number of sixes. Calculate:
    

    (a)   $P(X=1)$ and $P(X=2)$.

    (b)   $P(X⩽3)$ and $P(X>3)$.

    (c)   $E(X)$ and $\sigma$.

    Solutions:    (a)  $P(X=1)\approx 0.269,P(X=2)\approx 0.296$;    (b)  $P(X⩽3)\approx 0.875,P(X>3)\approx 0.125$;    (c)  $E(X)=np=2,\sigma =\sqrt{np(1-p)}\approx 1.29$
12. Random variable $X$ is the number of heads obtained when you toss a coin twenty times. Calculate:
    

    (a)   $P(X=7)$, $P(X⩽7)$ and $P(X<7)$.

    (b)   $P(7⩽X⩽13)$ and $P(7<X<13)$.

    (c)   $E(X)$ and $\sigma$.

    Solutions:    (a)  $P(X=7)\approx 0.0739,P(X⩽7)\approx 0.132,P(X<7)\approx 0.0577$;    (b)  $P(7⩽X⩽13)\approx 0.885,P(7<X<13)\approx 0.737$;    (c)  $E(X)=10,\sigma \approx 2.24$
13. The Singsing company produces LCD screens for personal computers. It's known that 1.25% of their LCD screens are defective. The Gertrude Cox Middle School buys 160 of their LCD screens.
    

    (a)   Find the expected number of defective LCD screens.

    (b)   Find the probability that there are 3 defective LCD screens.

    (c)   Find the probability that the number of defective LCD screens is less or equal 4.

    (d)   Find the probability that the number of defective LCD screens is greater or equal 5.

    Solutions:    (a)  $E(X)=2$;     (b)  $P\approx 0.182$;     (c)  $P\approx 0.948$;     (d)  $P\approx 0.0515$

Normal distribution

14. A random variable is normally distributed with the mean $\mu =70$ and the standard deviation $\sigma =10$.
    

    (a)   Calculate $P(65⩽X⩽75)$.

    (b)   Calculate $P(X⩽50)$.

    (c)   Calculate $P(X⩾85)$.

    Solutions:    (a)  $P(65⩽X⩽75)\approx 0.383$;     (b)  $P(X⩽50)\approx 0.0228$;     (c)  $P(X⩾85)\approx 0.0668$
15. A random variable $X$ is normally distributed: $X\sim N(20,25)$.
    

    (a)   Calculate $P(15⩽X⩽20)$.

    (b)   Calculate $P(X⩾23)$.

    Hint:    The notation $X\sim N(20,25)$ means that $X$ is normally distributed with the mean $\mu =20$ and variance ${\sigma }^2=25$ (which means that the standard deviation $\sigma =5$).
    Solutions:    (a)  $P(15⩽X⩽20)\approx 0.341$;     (b)  $P(X⩾23)\approx 0.274$

16. A random variable is normally distributed with the mean $\mu =70$ and the standard deviation $\sigma =10$. We will write this as $X\sim N(70,{10}^2)$ or $X\sim N(70,100)$.
    

    (a)   Find $a$ where $P(X⩽a)=0.75$.

    (b)   Find $b$ where $P(X⩽b)=0.31$.

    Solutions:    (a)  $a=\mathrm{invNorm}(0.75,70,10)\approx 76.7$;     (b)  $b=\mathrm{invNorm}(0.31,70,10)\approx 65.0$
17. A random variable $X$ is normally distributed: $X\sim N(20,25)$.
    

    (a)   Find $a$ where $P(X⩽a)=\frac{2}{3}$.

    (b)   Find $m$ where $P(X⩾m)=0.45$.

    Solutions:    (a)  $a\approx 22.2$;     (b)  $m\approx 20.6$
18. A random variable $X$ is normally distributed: $X\sim N(40,{15}^2)$.
    

    (a)   Write down $\mu$ and $\sigma$.

    (b)   Calculate $P(X⩽30)$.

    (c)   Find $m$ where $P(30⩽X⩽m)=0.6$.

    Solutions:    (a)  $\mu =40,\sigma =15$;     (b)  $P(X⩽30)\approx 0.252$;     (c)  $m\approx 55.7$
19. A factory produces soft drinks in 500 ml bottles. The quantity of drink in a bottle is normally distributed with standard deviation of 5 ml. According to EU standards a 500 ml bottle must not contain less then 500 ml of drink. For this reason the filling machine is set so that the quantity of drink has the mean of 512 ml. The factory produces 5000 bottles in one day.
    

    (a)   How many percent of the bottles contain less than 500 ml?

    (b)   How many bottles containing less than 500 ml are produced in one day?

    Solutions:    (a)  $P(X<500)\approx 0.820\mathrm{\%}$;     (b)  41 bottles
20. A farmer is growing tomatoes. The weights of his tomatoes are normally distributed with mean 140 g and standard deviation 15 g.
    

    (a)   How many percent of tomatoes weight from 120 to 150 g?

    (b)   How many percent of tomatoes weight 150 g or more?

    The 5% of heaviest tomatoes are considered "extra-large".

    (c)   Find the minimal weight of an extra-large tomato.

    Solutions:    (a)  $P(120⩽X⩽150)\approx 65.6\mathrm{\%}$;     (b)  $P(X⩾150)\approx 25.2\mathrm{\%}$;     (c)  minimal weight: 165 g
21. Results of a psychological test are normally distributed with the mean 85 and standard deviation 15.
    

    (a)   How many percent of the population have the score greater than 100?

    (b)   How many percent of the population have the score between 70 and 80?

    (c)   How many percent of the population have the score smaller than 60?

    (d)   Find the value $a$, given that 10% of the population have the score smaller than $a$.

    Solutions:    (a)  $P(X⩾100)\approx 0.159=15.9\mathrm{\%}$;     (b)  $P(70⩽X⩽80)\approx 0.211=21.1\mathrm{\%}$;     (c)  $P(X⩽60)\approx 0.0478=4.78\mathrm{\%}$;     (d)  $a\approx 65.8$
22. ?
    ?

    If  $Y=aX+b$  then:
    
    ${\mu }_Y=a\cdot {\mu }_X+b$
    
    ${\sigma }_Y=a\cdot {\sigma }_X$
    A random variable $X$ is normally distributed: $X\sim N(20,8^2)$. The values of the variable $Y$ are obtained by multiplying $X$ by 5 and then adding 50.
    

    (a)   Write down $\mu$ and $\sigma$ for the variable $X$.

    (b)   Write down $\mu$ and $\sigma$ for the variable $Y$.

    (c)   Calculate values of $Y$ corresponding to $X_1=16$ and $X_2=30$.

    (d)   Calculate $P(16⩽X⩽30)$.

    (e)   Find the probability for the corresponding interval of $Y$.

    Hint:    $Y=5X+50$ means that the variable $Y$ has the mean ${\mu }_Y=5\cdot 20+50=150$ and standard deviation ${\sigma }_Y=5\cdot 8=40$, so $Y\sim N(150,{40}^2)$.
    Solutions:    (a)  ${\mu }_X=20,{\sigma }_X=8$;     (b)  ${\mu }_Y=150,{\sigma }_Y=40$;     (c)  $Y_1=130,Y_2=200$;     (d)  $P(16⩽X⩽30)\approx 0.586$;     (e)  $P(130⩽Y⩽200)\approx 0.586$
23. A random variable $X$ is normally distributed: $X\sim N(10,4)$. The values of the variable $Y$ are obtained by multiplying $X$ by 3 and then adding 6.
    

    (a)   Calculate $P(X⩽11)$ and $P(Y⩽39)$.

    (b)   Calculate $P(Y⩾30)$ and find the appropriate interval for $X$.

    (c)   Find $a$ where $P(Y⩽a)=\frac{1}{4}$.

    (d)   Find $b$ where $P(33⩽Y⩽b)=0.3$.

    Solutions:    (a)  $P(X⩽11)=P(Y⩽39)\approx 0.691$;     (b)  $P(X⩾8)=P(Y⩾30)\approx 0.841$;     (c)  $a\approx 32.0$;     (d)  $b\approx 37.7$
24. A random variable $X$ is normally distributed: $X\sim N(30,9)$.
    

    (a)   Write down $\mu$ and $\sigma$.

    (b)   Calculate $P(X⩽\mu )$ and $P(X⩾\mu )$.

    (c)   Calculate $P(\mu ⩽X⩽\mu +\sigma )$ and $P(X⩾\mu +\sigma )$.

    (d)   Calculate $P(\mu -\sigma ⩽X⩽\mu +\sigma )$.

    Solutions:    (a)  $\mu =30,\sigma =3$;     (b)  $P(X⩽\mu )=P(X⩾\mu )=0.5=50\mathrm{\%}$;     (c)  $P(\mu ⩽X⩽\mu +\sigma )\approx 0.341$,  $P(X⩾\mu +\sigma )\approx 0.159$;     (d)   $P(\mu -\sigma ⩽X⩽\mu +\sigma )\approx 0.683$
25. A random variable $Z$ is normally distributed: $Z\sim N(0,1)$.
    

    (a)   Write down $\mu$ and $\sigma$.

    (b)   Calculate $P(Z⩽\mu )$ and $P(Z⩾\mu )$.

    (c)   Calculate $P(\mu ⩽Z⩽\mu +\sigma )$ and $P(Z⩾\mu +\sigma )$.

    (d)   Calculate $P(\mu -\sigma ⩽Z⩽\mu +\sigma )$.

    (e)   Calculate $P(\mu -2\sigma ⩽Z⩽\mu +2\sigma )$.

    (f)   Calculate $P(\mu -3\sigma ⩽Z⩽\mu +3\sigma )$.

    Solutions:    (a)  $\mu =0,\sigma =1$;     (b)  $P(Z⩽0)=P(Z⩾0)=0.5=50\mathrm{\%}$;     (c)  $P(0⩽Z⩽1)\approx 0.341$,  $P(Z⩾1)\approx 0.159$;     (d)   $P(-1⩽Z⩽1)\approx 0.683$;     (e)   $P(-2⩽Z⩽2)\approx 0.954$;     (f)   $P(-3⩽Z⩽3)\approx 0.997$
26. ?
    ?

    Standard score or Z-score is a $N(0,1)$ distributed variable:
    
    $Z=\frac{X-\mu }{\sigma }$
    
    
    A random variable $X$ is normally distributed with $\mu =50$ and $\sigma =15$.
    

    (a)   Find the standard score (the $Z$-score) for values: $X_1=55$,  $X_2=60$ and $X_3=20$.

    Hence calculate:

    (b)   $P(55⩽X⩽60)$

    (c)   $P(X⩽20)$

    Solutions:    (a)  $Z_1=\frac{1}{3},Z_2=\frac{2}{3},Z_3=-2$;     (b)  $P(55⩽X⩽60)=P(\frac{1}{3}⩽Z⩽\frac{2}{3})\approx 0.117$;     (c)  $P(X⩽20)=P(Z⩽-2)\approx 0.0228$
27. Results of a psychological test are normally distributed with the mean $\mu =90$.
    

    (a)   How many percent of the population have the score between $\mu -\frac{1}{2}\sigma$ and $\mu +\frac{1}{2}\sigma$?

    (b)   Find $\sigma$, given that only 5% of the population achieved score greater than 120.

    Hint:    Use $Z$-score in part (a). In part (b), first find the value $z_0$ so that $P(Z⩾z_0)=0.05$ and then use $Z=\frac{X-\mu }{\sigma }$ to find $\sigma$.
    Solutions:    (a)  $P(-\frac{1}{2}⩽Z⩽\frac{1}{2})\approx 0.383=38.3\mathrm{\%}$;     (b)  $\sigma \approx 18.2$

 Index