7/28/2019 S-lab-2-2507-F11

1/3

STAT 2507 Solution-Assignment # 2 Total mark=20 Fall 2011

Question 1. Use the following minitab commands to generate 8 observations:

MTB > set c11

DATA> 5 6 2 8 8 1 2 7

DATA> end

MTB > random 8 c12;

SUBC> normal 0 0.7.MTB > let c1=c11+c12

(a) [1] Use Minitab to compute the standard deviation of the numbers in c1. The answer is

3.36. (any number between 2.4 and 3.6 is acceptable)

(b) [1] Add 2.5 to each number (let c2=c1+2.5 will do this, assuming the raw data are in c1).

Now compute the standard deviation: 3.36. (same as part a) How does your answer compare

with that in part (a)? Same.

(c) [1] Multiply the data in (a) by 4 and compute the standard deviation: 13.44. (part a

4) How does your answer compare with that in part (a)? 4 times. (If you are not sure, divide

the standard deviation of part (c) by the one of part (a).)

Question 2. The Minitab data set sat has the mean verbaland mathscores of the Scholastic

Aptitude Test (SAT) for the academic years 1967 to 1981. Retrieve the sat data set (by either

typing: retrieve sat , or by choosing file Open Worksheet (you may need to click on

look in Minitab Sample Data folder icon) and then choosing sat).

(a) [1] Fit a L-S regression line using yearto predict the verbalscores (verbal score is called the

response variable). What is the equation of the L-S regression line? verbal= 7373 3.51 year.

(b) [1] Fit a L-S regression line using year to predict the math scores (response variable).

What is the equation of the L-S regression line? Math= 4955 2.27 year.(c) [1] Fit a L-S regression line using verbal scores to predict the math scores (response vari-

able). What is the equation of the L-S regression line?Math= 193 + 0 .643 verbal.

(d) [1] What is the predicted score for math in 1970? 483.1. In 1983? 453.59. In 2000? 415.

Which of these seem to make sense? For the years in the range of data, 1970 and 1983.

(e) [1] Use your answer to part (c) to predict the mathscore if the verbal score is 450. Answer:

482.35. Can you predict the verbalscore if the mathscore is 480? Answer: Not based on equation in part (c

Why? Equation in part (c) is a model to predict y (Math), based on x (verbal).

Question 3. Suppose that X has a binomial distribution with n=30 and p=0.6. Use Minitabto simulate 50 values of X.

MTB > random 50 c1;

SUBC > binomial 30 0.6.

(a) [1] What proportion of your values are less than 20? 36/50. (any number between 25/50

and 40/50 is acceptable)

(b) [1] What is the exact probability that X will be less than 20? 0.70850.

Hint: To find P(X k), for any k 0, use the cdf command; this works by typing

MTB > cdf;

SUBC > binomial 30 0.6.

1

7/28/2019 S-lab-2-2507-F11

2/3

(c) [1] Find P(X < 24)= 0.9828, and P(12 X 15)= 0.1754 0.0083 = 0.1671 .

(d) [1] If you simulate 10000 values of X, what would be the expected number of values

(among the 10000) that are less than 24? (1000)(0.9828)=9828.

ALSO do the following questions:

1. [2] Suppose that a random variable X follows the distribution given by:

a -2 -1 0 1 2

P(X = a) 1/8 1/4 1/4 ? 1/8

Let Y = X2. The values of P(Y 1) and E(Y) are respectively

(a) 3/4 and 1 (b) 1/4 and 1/8. (c) 3/4 and 3/2 (**) (d) 1/4 and 0.

2. [2] A survey has revealed that 75% of all college students study. It is also known that

85% of all students who study will graduate, while only 35% of those students who donot study will graduate. A randomly selected student is observed to graduate. What is

the probability that the student studied?

(a) 0.825 (b) 0.8375 (c) 0.8793 (**) (d) 0.875

3. For events A and B we have

P(A) = 0.3 P(B) = 0.8 P(A B) = 0.9

(a) [1.5] Find P(A|B) P(A B) P(B A).

Solution:

P(A B) = P(A) + P(B) P(A B) P(A B) = 0.3 + 0.8 0.9 = 0.2

P(A|B) = P(AB)P(B)

= 0.20.8

= 0.25

P(A B) = P(B) P(A B) = 0.8 0.2 = 0.6

(B A) = 1 P(A B) = 1 0.2 = 0.8

(b) [0.5] Are A and B independent? Why?

Solution:

No there are not independent, since

P(A B) = P(A)P(B), 0.2 = (0.3)(0.8)

4. A box of candy contains 30 pieces. Twenty-six are made of chocolate and four are made

of vanilla.

(a) [1] Five pieces are selected at random, without replacement. What is the probability

that four of them are chocolates?

2

7/28/2019 S-lab-2-2507-F11

3/3

(a) 0.376 (b) 0.764 (c) 0.420 (**) (d) 0.125

It is Hypergeometric distribution

P(X = 4) =C264.C4

1

C305

= 0.420

(b) [1] What would be the answer to part (a) if the five pieces are selected at randomwith replacement (i.e., a selected piece is put back in the box before the next selection

is made)?

(a) 0.376 (**) (b) 0.764 (c) 0.420 (d) 0.125

It is Binomial distributin

P(X = 4) = C54(

2630)

4

(430) = 0.376

3