Download Solutions to the Practice Midterm Exam – Fall 2009

Solutions to the Practice Midterm Exam – Fall 2009
Question 1: Multiple choice
1.1. The average salary of all male workers at a company is $50,000. The average salary of all
female workers at the same company is $54,000. What must be true of the average salary of all
workers?
(a)
(b)
(c)
(d)
It must be larger than the median salary.
It must be $52,000.
It must be larger than $52,000.
It could be any number between $50,000 and $54,000.
1.2. Suppose the numbers x1 , x2 ,  xn are measured in units of inches. What unit of measurement
will the variance of these numbers have?
(a)
(b)
(c)
(d)
Inches
Square inches
Cubic inches
None of the above
1.3. A medical researcher thinks that a new drug will help reduce blood pressure. The researcher
finds 25 men and 15 women that are willing to participate in the study. The researcher chooses 20
of the subjects at random to take the drug every day, while the remaining 20 take a placebo. This
is a
(a)
(b)
(c)
(d)
Simple random sample design.
Stratified random sample design.
Randomized block experimental design.
Completely randomized experimental design.
1.4. Suppose X is a random variable with mean 10 and standard deviation 4. Let Z=10-2X. What
is the variance of Z?
(a)
(b)
(c)
(d)
16
8
64
None of the above
Question 2: The fat content, x, and number of calories, y, was measured for 11 brands of fast
food chicken sandwiches. We are given the following summary information about the data:
y = 472.7 , s y = 144.2 , x = 20.6 , s x = 9.8 and r = 0.947
(a) Calculate the least square regression line for predicting number of calories from fat
content.
Solution:
b=r
sy
sx
= 0.947 ×
144.2
= 13.934
9.8
a = y − bx = 472.7 − b × 20.6 = 185.651
yˆ = a + bx = 185.651 + 13.934 x
(b) Calculate the proportion of variability in the calories that is explained by the regression
line.
Solution:
r 2 = 0.947 2 = 0.897
Question 3: You recently bought a new set of four tires from a manufacturer who just announced
a recall because 2% of that particular brand of tires are defective. What is the probability that at
least one of your tires is defective? You may assume that the tires are defective independently of
one another.
Solution:
P (at least one defective tire) = 1-P (No defective tires)
Let Ai = tire i is not defective
P ( Ai ) = 1 − 0.02 = 0.98
P (No defective tires) = P ( A1 ∩ A2 ∩ A3 ∩ A4 ) = P ( A1 ) P ( A2 ) P( A3 ) P( A4 ) = (0.98) 4
P (at least one defective tire) = 1-(0.98) 4 = 0.0776
Note: Students often try to solve the problem the following incorrect way:
Let Ai = tire i is defective
The mistake many students made was assuming that the events Ai were disjoint and that they
could solve the problem using the following formula:
P (at least one defective tire) = P ( A1 ∪ A2 ∪ A3 ∪ A4 ) = P ( A1 ) + P ( A2 ) + P ( A3 ) + P ( A4 )
= 0.02 + 0.02 + 0.02 + 0.02 = 0.08
However, the events Ai are not disjoint, they are independent. Hence, simply adding the
probabilities is incorrect. The easiest way to see how incorrect your results could get using this
formula is to think of the scenario that instead of buying four tires, you bought 51 tires. Using the
method outlined above you would get
P (at least one defective tire) = 51 × 0.02 = 1.02
which is obviously false.
Question 4: A researcher examined the effects of alcohol on reaction time. After the subjects
obtained a specified blood alcohol level, their reaction time (in milliseconds) to a certain stimulus
was measured. The mean and standard deviation of the reaction time were 392 and 87,
respectively. Assume that the reaction time for people with the specified blood alcohol level
follows a normal model.
(a) Find the percent of people whose reaction time is between 400 and 460 milliseconds?
(b) Fifteen percent of the subjects have reaction times above what value?
Solution:
(a)
Calculate the z-scores:
z=
460 − 392
= 0.78
87
According to the Normal Table 78.23% have reaction times below 460 ms
z=
400 − 392
= 0.09
87
According to the Normal Table 53.59% have reaction times below 400 ms
Proportion between 400 and 460 ms = 0.7823 − 0.5359 = 0.2464
(b)
To be in the top 15% you have to be z = 1.04 standard deviations above the
mean, this corresponds to x = zσ + µ = 1.04 × 87 + 392 = 482.48 .
Question 5: Events A and B in a sample space are such that P(A∪B) = 0.4 and P(A) = 0.3.
(a) What is P(B) if A and B are disjoint?
(b) What is P(B) if A and B are independent?
Solution:
(a)
If A and B are disjoint, then P ( A ∪ B ) = P ( A) + P ( B ) .
P( B) = P( A ∪ B) − P( A) = 0.4 − 0.3 = 0.1
(b)
We know that P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) .
Since A and B are independent, we also known that P ( A ∩ B ) = P ( A) P ( B ) .
So, P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) = P ( A) + P ( B ) − P ( A) P ( B )
P( A ∪ B) = P( A) + P( B) − P( A) P( B)
0.4 = 0.3 + P( B) − 0.3 * P( B)
0.1 = P( B)(1 − 0.3)
P( B) =
0.1 1
=
0.7 7
Question 6: A student who has never played basketball is introduced to the game. He takes two
shots. The probability he makes the first is 0.2 and the probability he makes the second is 0.3.
a. Let X be the number of shots the player makes. Find the probability model for X.
b. Find the mean and standard deviation of X.
Solution:
A = player makes shot 1
B = player makes shot 2
The branch (A and B) entails making two shots.
P ( A ∩ B ) = 0.2 × 0.3 = 0.06
P(player makes exactly two shot) = 0.06
The branches (A and BC) and (AC and B) entail making one shot.
P ( A ∩ B C ) = 0.2 × 0.7 = 0.14
P ( A C ∩ B ) = 0.8 × 0.3 = 0.24
P(player makes exactly one shot) = 0.14 + 0.24 = 0.38
The branch (AC and BC) entails making zero shots.
P ( AC ∩ B C ) = 0.8 × 0.7 = 0.56
P(player makes exactly zero shots) = 0.56
Probability Model:
Value of X
Probability
0
0.56
1
0.38
2
0.06
Calculate the mean and standard deviation of X.
µ = 0.56 × 0 + 0.38 × 1 + 0.06 × 2 = 0.5
σ 2 = 0.56 × (0 − 0.5) 2 + 0.38 × (1 − 0.5) 2 + 0.06 × (2 − 0.5) 2 = 0.37
σ = σ 2 = 0.37 = 0.61