Download StpT2Key

Test 2 Key:
Form A:
1. (10 points) Given P(notA) = 0.8 and P(B) = 0.6 and P(A&B) = 0.1 ,
compute P(A or B).
P(A)=1-.8=.2 P(AorB)=.2+.6-.1=.7
2. (10 points) Toss two balanced dice.
What is the probability that the sum of both "faces" is not 11?
P(sum is not 11)=1-P(sum=11) = 1-P{ (6,5), (5,6)} =1-2/36=34/36
3. (10 points) A frequency distribution for the number of siblings of 45 students in
one of the STP 226 classes is given in the table below.
Number of
siblings
Number of
students
(frequency)
0
1
2
3
4
5
6
7
7
6
12
8
5
4
2
1
For a student selected at random from that class, let
A= event that student has at most 3 siblings
B= event that student has at least 2 siblings
a.
Compute P(A&B)= P(student has 2 or 3 siblings)=20/45
b. Are events A , B mutually exclusive? Clearly explain why or why not?
No, P(A&B)>0 (there is a common part)
4. (10 points) The following table cross-classifies students participating in Varsity
sports in a certain high school.
Male
Female
Totals
Freshman
50
39
89
Sophomore
71
56
127
Junior
73
45
118
Senior
61
32
93
Totals
255
172
427
Suppose a student is selected at random from among all those participating in
varsity sports in that school.
a) Find the probability that this student is a Senior.
93/427
b) Find the probability that this student is a Female or a Junior.
172/427+118/427-45/427=245/427
5. (15 points) Use the standard normal curve to find the following:
a.
area between - 1.54 and 1.73
.9582-.0618=.8964
b.
z-value with the area equal to 0.97 to the right of it.
z = -1.88
(area=.9699)
c.
first quartile of the standard normal distribution
(i.e find z value with .25 area to the left of it)
z = -.67
6. (20 points) The lifetime of a particular brand of fuse is normally distributed with
mean  = 1500 hours and standard deviation  = 100 hours.
a.
Determine the percentage of that brand of fuses with lifetime
exceeding 1600 hours?
z=(1600-1500)/100=1
P(z>1)=.1587=15.87%
b.
Fill in the blanks:
95.44 % of that brand of fuses has lifetime between
1300 hrs and 1700hrs.
c)
Find 85th percentile of the distribution of the lifetime for that brand of
fuses
(i.e. find x with 85% of the area to the left of it)
(x-)/=z
x=+z
z=1.04 (85th percentile of st normal distribution)
x=1500+100(1.04)=1604
7. (25 points) According to a psychologist at the large university, IQ scores of the
students at that university are normally distributed with mean
=112 and standard deviation  =12. Let X denote the IQ score for
the students at that university.
a). Sketch the distribution curve of the variable X, point on the graph and
and describe in words its main features.
X has normal distribution: bell shape, symmetric, centered at 112, area under=1,
most area berween 96 and 148
b) Find the probability :
P(100 < X < 130)=P(-1<z<1.5)=.9332-.1587=.7745
z=(x-112)/12, (100-112)/12=-1, (130-112)/12=1.5
c)
Let X denote the sample mean of a random sample of IQ-s of 4
students from that university.
Name the distribution of X and give mean and standard deviation of X .
Normal distribution with mean =112 and st deviation=6
d)
Let X be defined as in part c)
True or false: The distribution of X has wider
spread than the distribution of X.
False, distribution is narrower (st deviation of X is smaller )
e)
Let X be defined as in part c)
Find probability:
P ( X < 121 ) =P(z<1.5)=.9332
z=( X -112)/6
(121-112)/6=1.5
TEST 2 KEY, Form B
1. (10 points) Given P(notA) = 0.2 and P(B) = 0.6 and P(A or B) = 0.10 ,
compute P(A&B).
P(A)=1-.2=.8
.10=.8-.6-P(A&B)
P(A&B)=1.3, but it can't be a probability since it is >1, so the situation is not
possible, there is no solution here.
2. (10 points) Toss two balanced dice.
What is the probability that the sum of both "faces" is not 10?
P(sum is not 10)=1-P(sum=10) = 1-P{ (4,6), (5,5), (6,5)} =1-3/36=33/36
3. (10 points) A frequency distribution for the number of siblings of 45 students in
one of the STP 226 classes is given in the table below.
Number of
siblings
Number of
students
(frequency)
0
1
2
3
4
5
6
7
7
6
12
8
5
4
2
1
For a student selected at random from that class, let
A= event that student has at most 2 siblings B= event that student has at least 4
siblings
a.
Compute P(A or B)=1-8/45=37/45
b.
Are events A , B mutually exclusive? Clearly explain why or why not?
Yes, there is no common part, P(A&B)=0.
4. (10 points) The following table cross-classifies students participating in Varsity
sports in a certain high school.
Male
Female
Totals
Freshman
50
39
89
Sophomore
71
56
127
Junior
73
45
118
Senior
61
32
93
Totals
255
172
427
Suppose a student is selected at random from among all those participating in
varsity sports in that school.
a) Find the probability that this student is a Junior.
118/427
b)Find the probability that this student is a Female or a Senior.
172/427+93/427-32/427=233/427
5. (15 points) Use the standard normal curve to find the following:
a.
area between - 1.55 and 1.77
.9616-.0606=.901
b.
z-value with the area equal to 0.96 to the right of it.
z= -1.75
c.
third quartile of the standard normal distribution
(i.e. z value with are .75 to its left)
z = .67
6. (20 points) The lifetime of a particular brand of fuse is normally distributed with
mean  = 1500 hours and standard deviation  = 100 hours.
a.
Determine the percentage of that brand of fuses with lifetime
less than 1450 hours?
z=(1450-1500)/100= -.5
P(z<-.5)=.3085=30.85%
b.
Fill in the blanks:
99.74 % of that brand of fuses has lifetime between
1200 hrs and 1800hrs.
Find 65th percentile of the distribution of the lifetime for that brand of
fuses
(i.e. find x with 65% of the area to the left of it)
c)
(x-)/=z
x=+z
z=.39 (65th percentile of st normal distribution)
x=1500+100(.39)=1539
7. (25 points) According to a psychologist at the large university, IQ scores of the
students at that university are normally distributed with mean
=112 and standard deviation  =12. Let X denote the IQ score for
the students at that university.
a). Sketch the distribution curve of the variable X, point on the graph and
and describe in words its main features.
X has normal distribution: bell shape, symmetric, centered at 112, area under=1,
most area berween 96 and 148.
b) Find the probability :
P(95 < X < 124)= P(-1.42<z<1)=.8413-.0778=.7635
z=(x-112)/12, (95-112)/12=-1.42, (124-112)/12=1
c)
Let X denote the sample mean of a random sample of IQ-s of 9
students from that university.
Name the distribution of X and give mean and standard deviation of X .
Normal distribution with mean =112 and st deviation=4
d)
Let X be defined as in part c)
True or false: The distribution of X has wider
spread than the distribution of X.
False, distribution of X is narrower, since st. dev. is smaller
e)
Let X be defined as in part c)
Find probability:
P ( X < 118 ) =P(z<1.5)=.9332
z=( X -112)/4
(118-112)/4=1.5