Quizzes and exercises of distribution
These quizzes and exercises help us review and practice what we obtained in this chapter. They are presented in "show-hide" form similar to the sections of a web page as we are already familiar. Quizzes are multiple choice questions. After choosing an option, an announcement of result appears. To return back to this page, we click "OK" on the announcement. Of course we can choose again.
In exercises, we are required to calculate a value then fill it in an empty rectangle. Note that in this rectangle, only value is acceptable, its unit is not required. After filling the result, we click on "Answer". If the answer is correct, the border of the rectangle is green and there is green "V" symbol in the adjacent square. If the answer is wrong, the border of the rectangle is red and there is red "X" symbol in the adjacent square. We can erase the answer and retry by clicking on "Retry".
For each exercise, there is a hidden solution. To show this solution, we click on "Solution" tab. But try to solve exercises by ourself, don't abuse these solutions.
Quiz 1
Which statement concerns cumulative distribution function?
Quiz 2
Expected value of a discrete random variable is
Quiz 3
In filling machine of yoghurt, we want each package having 4 pieces of fruits. Due to the imperfect of this machine, the number of pieces of fruit varies from package to package. To study this phenomenon, which distribution we have to apply?
Quiz 4
To investigate the participation of students in sportive activities, a survey was realized with 100 students. The result shows that there are `p` students participating in sportive activities. What is the appropriate distribution of `p`?
Quiz 5
`X` is a continuous random variables conforms to standard normal distribution. `a` is the probability that `X` less than - 1, `b` is the probability that `X` greater than 1,2. Then
Quiz 6
Distribution of mean of samples approaches normal distribution when
Exercise 1
15% of people in the world is left-handed. Group A consists of 8 persons.
a. What is the probability that there are 3 left-handed persons in group A?
b. What is the probability that there are less than 3 left-handed persons in group A?
c. What is the probability that there are more than 3 left-handed persons in group A?
Solution
We can apply binomial distribution for these cases. We have `p=0,15` and `n=8`, so probability that there are `x` left-handed persons in group A is :
`p(x)=P(X=x)=(8!)/(x!(8-x)!)\ 0,15^x\ 0,85^(8-x)`
a. Probability that there are 3 left-handed persons in group A is p(3) :
`p(3)=(8!)/(3!xx5!)\ 0,15^3xx0,85^5=0,08386`
b. Because there are less than 3 left-handed persons in group A, so this group can have 0, 1 or 2 left-handed persons.
Then probability that there are less than 3 left-handed persons in group A is :
`P(X< 3)=p(0)+p(1)+p(2)`
`P(X< 3)=(8!)/(0!xx8!)0,15^0xx0,85^8+(8!)/(1!xx7!)0,15^1xx0,85^7+(8!)/(2!xx6!)0,15^2xx0,85^6`
`P(X< 3)=0,27249+0,38469+0,23760=0,89478`
c. Event "more than 3 left-handed person in group A" and event "less than or equal to 3 left-handed persons in group A" are two complementary events. So :
`P(X> 3)=1-P(X<=3)=1-[P(X< 3)+p(3)]`
`P(X> 3)=1-(0,89478+0,08386)=0,02136`
Exercise 2
For quality management in a factory, they count the defective products in one hour. The result after 50 hours of observation is shown in Table 1.
| 0 | 1 | 2 | 1 | 0 | 2 | 3 | 1 | 0 | 1 |
| 1 | 2 | 0 | 1 | 4 | 0 | 2 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 | 2 | 0 | 1 | 3 | 0 |
| 0 | 0 | 0 | 1 | 0 | 2 | 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 2 | 1 | 0 | 0 | 2 | 0 | 0 |
a. What is the mean of defective products in one hour?
b. If we consider the numbers of defective products in one hour, defined as `X`, conforms to Poisson distribution. Then how many hours correspond to `X` from 0 to 4 in theory?
| • `X = 0` |
| • `X=1` |
| • `X=2` |
| • `X=3` |
| • `X=4` |
c. Compare with the actual data.
Solution
To facilitate the calculation and comparison, we summarize the data in Table 1 and we obtain Table 2.
| Defective products `X` (product/hour) |
Hours `T(X)` |
Defective products `XxxT(X)` |
Time ratio `T(X)//sum T(X)` |
|---|---|---|---|
| 0 | 21 | 0 | 0,42 |
| 1 | 18 | 18 | 0,36 |
| 2 | 8 | 16 | 0,16 |
| 3 | 2 | 6 | 0,04 |
| 4 | 1 | 1 | 0,02 |
| Total | 50 | 44 | 1,00 |
a. From Table 2, we recognize that the observation was realized in 50 hours. In this time, there are 44 defective products. So the mean of defective product per hour is 0,88.
b. If we consider defective products per hour `X` conforms to Poisson distribution, then `lambda` of this distribution is 0,88 and the probability density function is :
`p(x)=P(X=x)=(e^(-0,88)\ 0,88^x)/(x!)`
Therefore :
`p(0)=(e^(-0,88)\ 0,88^0)/(0!)=0,41478`
`p(1)=(e^(-0,88)\ 0,88^1)/(1!)=0,36501`
`p(2)=(e^(-0,88)\ 0,88^2)/(2!)=0,16060`
`p(3)=(e^(-0,88)\ 0,88^3)/(3!)=0,04711`
`p(4)=(e^(-0,88)\ 0,88^4)/(4!)=0,01036`
So if observation time is 50 hours, theoretical hours correspond to `X` defective products are :
`X=0` : `T_"th"(0)=p(0)xx50=0,41478xx50=20,739`
`X=1` : `T_"th"(1)=p(1)xx50=0,36501xx50=18,250`
`X=2` : `T_"th"(2)=p(2)xx50=0,16060xx50=8,030`
`X=3` : `T_"th"(3)=p(3)xx50=0,04711xx50=2,355`
`X=4` : `T_"th"(4)=p(4)xx50=0,01036xx50=0,518`
c. Comparison of theoretical results and reality is shown in Table 3.
| Defective products per hour `X` |
Observed hours `T(X)` |
Theoretical hours Theory(`X`) |
|---|---|---|
| 0 | 21 | 20,74 |
| 1 | 18 | 18,25 |
| 2 | 8 | 8,03 |
| 3 | 2 | 2,36 |
| 4 | 1 | 0,52 |
We recognize that the difference between observed values and theoretical ones are small. So we can conclude that observed values conform well to Poisson distribution.
Exercise 3
`Z` is a continuous random variable conforming to standard normal distribution. By using Table 1 or an appropriate table, determine :
| a. `P(Z<=1,11)` = |
| b. `P(Z<=-0,66)` = |
| c. `P(Z>1,44)` = |
| d. `P(Z>-0,55`) = |
| e. `P(-0,77< Z<=1,77)` = |
Solution
a. `P(Z<=1,11)=0,8665`
b. `P(Z<=-0,66)=1-P(Z<=0,66)=1-0,7454=0,2546`
c. `P(Z >1,44)=1-P(Z<=1,44)=1-0,9251=0,0749`
d. `P(Z>-0,55)=1-P(Z<=-0,55)=1-[1-P(Z<=0,55)]=P(Z<=0,55)=0,7088`
e. `P(-0,77< Z<=1,77)=P(Z<=1,77)-P(Z<=-0,77)`
`=P(Z<=1,77)-[1-P(Z<=0,77)]=P(Z<=1,77)+P(Z<=0,77)-1`
`=0,9616+0,7794-1=0,7410`
Exercise 4
`Z` is a continuous random variable conforming to standard normal distribution. By using Table 2 or a suitable table, determine :
| a. `P(Z<=a)=0,72` ⇒ `a` = |
| b. `P(Z<= b)=0,24` ⇒ `b` = |
| c. `P(Z>c)=0,66` ⇒ `c` = |
| d. `P(-d< Z< d)=0,84` ⇒ `d` = |
Solution
a. `a=0,5828`
b. `P(Z<=- b)=1-P(Z<=b)=1-0,24=0,76` ⇒ `- b = 0,7063`
⇒ `b=- 0,7063`
c. `P(Z>c)=P(Z<=-c)=0,66` ⇒ `-c=0,4125` ⇒ `c=-0,4125`
d. `P(Z<=-d)=[1-P(-d< Z< d)]/2=(1-0,84)/2=0,08=1-P(Z <= d)`
⇒ `P(Z<=d)=1-0,08=0,92` ⇒ `d=1,405`
Exercise 5
In high school A, `IQ` of students conforms to normal distribution with mean of 106 and standard deviation of 14.
a. How many percent of students of school A whose `IQ` are lower than 100?
b. How many percent of students of school A whose `IQ` are in the interval 80 - 120?
c. This school selects a group consisting of 10% of highest `IQ`. What is the minimal `IQ` of this group?
Solution
In order to use tables of normal distribution, we have to standardize `IQ` by formula :
`Z=(IQ-mu)/sigma=(IQ-106)/14`
a. `P(IQ< 100)=P(Z< (100-106)/14)=P(Z< -0,4268)=1-P(Z< 0,4268)`
`=1-0,666=0,334`
b. `P(80<=IQ<=120)=P((80-106)/14<=Z<=(120-106)/14)=P(-1,857<=Z<=1)`
`=P(Z< 1)-P(Z< -1,857)=0,8413-(1-0,969)=0,810`
c. `P(IQ<=a)=P(Z<=(a-106)/14)=0,90\ \ rArr\ \ (a-106)/14=1,2816`
So `a=(14xx1,2816)+106=124`
OK
This web page was last updated on 02 December 2018.