Tutorial 1

Example 1. The following date is given.

Make a steam-leaf plot

Find median using steam-leaf plot

Find weighted mean using steam-leaf plot

Find CV (coefficient of variation) of the data set.

Find CS (coefficient of skewness).
1.17 1.61 1.16 1.38 3.53
1.23 3.76 1.94 0.96 4.75
0.15 2.41 0.71 0.02 1.59
0.19 0.82 0.47 2.16 2.01
0.92 0.75 2.59 3.07 1.40

a)

Steam

Leaves

Frequency
0 02 15 19 47 71 75 82 92 96 9
1 16 17 23 38 40 59 61 94 8
2 01 16 41 59 4
3 07 53 76 3
4 75 1

b) median=1.38

 

 

EXAMPLE 2. A data set has mean 3.99 and median 4.99. Is it

Symmetric?

Positively skewed?

Has long left tail?

 

EXAMPLE 3. We have two data sets A and B. Both have the same sample standard deviation and means =2 and =3. Which data has more variability?

 

EXAMPLE 4. If the mean and variance of a data set of 100 observations are 7.8 cm and 0.36 cm square, find an interval

where 95% of the data is located

all date is located.

 

EXAMPLE 5. Find the upper and lower quartiles for the following data:

3, 8, 7, 1, 1, 12, 13, 9, 3, 2, 10.

Step 1. Ordered observations:

1, 1, 2, 3, 3, 7, 8, 9, 10, 12, 13.

Tutorial 2

 

Example 1. Suppose that the population of a certain city is 40% male and 60% female. It is known that 50% of the males and 30% of the females smoke.

a)   Find the probability that randomly selected person is smoker.

b)   Find the probability that a smoker is male.

 

Solution:

We consider the following events:

C= "selected person is smoker"

 M= "selected person is male"

 F= "selected person is female"

a)

%What we need to find in part a)?

P( C )=?

%Which formula we use?

P( C )=P( C|M)P(M)+P(C|F)P(F)

%How it can be shown by a tree?

b)

% What we need to find in part b)?

P(M|C)=?

%Which formula must be used?

P(M|C)= 

Example 2. From the set of all families with two children a family is selected at random.

Assume that in a two child family all sex distributions are equally likely.  

a) What is the sample space?

b) What is the probability that the other child of the family is a girl if it is found to have a girl?

Solution.

a)   Sample space :

S=

b)   Let

A(1)="Family has a girl"

A(2)="Family has 2 girls"

 

%What we need to find?

 

P(A(2)|A(1))=?

P(A(1))=3/4   Why?

 

%Which formula we use?

Example 3.

For events A, and B: P(A)=0.52, P(B)=0.36, P(AB)=0.20.

Find

a)   P(A complement)=?

b)   P(B complement)=?

c)   P(A'B)=?

d)   P(AB')=?

e)   P(A'B')=?

 

B

B'

 

A

0.20   

    AB'

0.52   

A'

     A'B

       A'B'

     A'

 

0.36   

    B'

 

a) P(A')=1-P(A)=1-0.52= 0.48

b) P(B')=1-P(B)=1-0.36= 0.64

c) P(B)=0.36 ,

 P(A'B)+P(AB)=0.36      (How?)

So

 P(A'B)+0.20=0.36

Therefore

P(A'B)=0.36-0.20=0.16.

AND SO ON….

 

B

B'

 

A

0.20    AB

0.32      AB'

0.52    A

A'

0.16     A'B

0.32       A'B'

0.48     A'

 

0.36    B

0.64    B'

 

 

 

Tutorial 3

 

Example 1. Random variable X has  density function

f(x)= Cx , if 0<x<2 and

f(x)= 0 otherwise.

 

a)   What is value of C?

b)   What is the proportion of values of X less than 1?

c)   Find 25th percentile.

d)   Find median value of X

Example 2.

Suppose that a grocery store purchases 5 cartons of skim milk at the wholesale price of $1.20 per carton and retails

the milk at $ 1.65 per carton. After the expiration date, the unsold milk is removed from the shelf and the three fourths of the wholesale price will be paid to grocer. If the probability distribution of random variable X, the number of cartons that are sold from this lot is

 

x

0

1

2

3

4

5

F(x)

1/15

2/15

2/15

3/15

4/15

3/15

find the expected profit.

 

Solution:

Let  Y=profit.

Then it is clear that

Y=1.65X+(3/4)*1.20(5-X)-5*1.20.

%Why?

We can write it as

Y=1.65X+0.9*5-0.9X-6=0.75X-1.5;

%How we find?

E(Y)=E(0.75X-1.5)=0.75E(X)-1.5.

Now we calculate E(X).

E(X)=0*(1/15)+1*(2/15)+2*(2/15)+3*(3/15)+4*(4/15)+5*(3/15)=46/15=3.06.

E(X)=3.06.

%What does the last mean?

Now

E(Y)=0.75*(3.06)-1.5= 2.295-1.5=0.795.

E(Y)=80 cents.

 

 

 

Tutorial 4 (Ch 8)

 

If a certain machine makes electrical resistors having a mean resistance of 40ohms and a standard deviation of 2 ohms, what is the probability that a random sample of 36 of these resistors will have a combined resistance of more than 1458?

 

Steps to solving question.

 

1)    underline key words

If a certain machine makes electrical resistors having a mean resistance of 40 ohms and a standard deviation of 2 ohms, what is the probability that a random sample of 36 of these resistors will have a combined resistance of more than 1458?

 

2)    Organize your information

a.     Population parameter:

                                                             i.      mean (or m) = 40 ohms and

                                                           ii.      standard deviation = s =2 ohms

b.     Sample statistics

                                                             i.      Random sample

                                                           ii.      Sample size n = 36

c.     Problem information (what you need to find out)

                                                             i.      Combined resistance of > 1458

                                                           ii.      What is combined resistance? Total resistance

3)    Translate this to our usual problem statement

P(Total resistance > 1458) = ?

Or

P(Total resistance/n > 1458/36) = ?

Or more familiarly,

P(sample mean > 40.5) = ?

We can now use the central limit theorem to evaluate this probability with an approximate normal distribution

4)    Solve for this problem with the regular steps (these steps are omitted for your practice)

5)    Answer is 0.0668