solution to homework 7

solution to homework 7 - Statistics – V3100018.001-006...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Statistics – V3100018.001-006 Solutions to homework 7 Giuseppe Arbia, Catholic University of the Sacred Hearth, Roma, Italy 1 Ques%on
1
A
brewery
has
a
beer
dispensing
machine
that
pours
beer
into
 the
company’s
12
ounce
bo8les.
The
distribu<on
of
the
amount
of
beer
 actually
dispensed
by
the
machine
follows
a
Normal
distribu<on
with
a
 standard
devia<on
of
0.2
ounces.
The
company
can
control
the
mean
 amount
of
beer
dispensed
by
the
machine.
 (a)
What
value
of
the
mean
should
the
company
set
if
it
wants
to
guarantee
 that
98.5%
of
the
bo8les
contain
at
least
12
ounces
(the
amount
on
the
 label)?
 (b)
Suppose
now
that
the
distribu<on
of
pours
is
Normal
with
mean
12.1
 and
standard
devia<on
0.5.
Customers
get
very
angry
when
their
beers
 have
less
than
the
adver<sed
12
ounces
of
beer
in
them.
Calculate
the
 propor<on
of
customers
that
could
get
angry
because
of
this
reason
.
 X ! N (µ;! = 0.2 ) P( X > x ) = 0.985 P( X < x ) = 1 ! 0.985 = 0.015 " % x!µ P$Z < = z0.015 ' = 0.015 # & ! ‐2.17
 z0.015 = !2.17 x!µ = !2.17 ! 12 ! µ = !2.17 0.2 µ = 12 + 0.2 " 2.17 = 12 + 0.434 = 12.434 The
company
should
aim
at
having
12.434
ounces
so
that
in
most
cases
it
has
more
than
12.

 Q1b:
(b)
Suppose
now
that
the
distribu<on
of
pours
is
Normal
with
mean
12.1
and
 standard
devia<on
0.5.
Customers
get
very
angry
when
their
beers
have
less
than
the
 adver<sed
12
ounces
of
beer
in
them.
Calculate
the
propor<on
of
customers
that
 could
get
angry
because
of
this
reason

 
 X ! N (12.1;! = 0.5) 12 ! 12.1 P ( X < 12 ) = P ( Z < )= 0.5 " 0.1 % P$Z < ! ' = P ( Z < !0.2 ) = 0.4207 # 0.5 & 42.07
%
could
get
angry
 Ques%on
2
Given
the
random
variable
X
following
a
Normal
distribu<on
 with
mean
=
150
and
standard
devia<on
=
12,
it
is
more
likely
to
observe
at
 random
a
value
such
that:
 a)  145

≤

x1

≤

155

or

 b)  165

≤

x2

≤

175

?
 Show
interac<ve
excel
file
Norm_CLT_Confint
 Worksheet
Normal
X
 Ques%on
3
We
observe
the
height
of
Italians
(variable
X)
and
of
Americans
 (variable
Y).
We
know
that
in
both
cases
the
distribu<on
of
height
is
Normal,
 with
the
same
means,
but
different
standard
devia<ons.
For
the
Italians
we
 have
a
mean
of
5.7
and
a
standard
devia<on
of
0.33.
For
the
Americans
we
 have
a
mean
of
5.7
and
a
standard
devia<on
of
0.7
 
 a)  What
is
the
probability
that
an
Italian
randomly
chosen
has
a
height
 between
5.5
and
5.8?

 X ! N (5.7;! = 0.33) P(5.5 < X < 5.8) = P( X < 5.8) ! P( X < 5.5) = " " 5.8 ! 5.7 % 5.5 ! 5.7 % P$Z < ' ! P$Z < ' # # 0.33 & 0.33 & ! ! 0.1 $ 0.2 $ P#Z < & ' P#Z < ' &= " " 0.33 % 0.33 % = P ( Z < 0.30 ) ' P ( Z < '0.61) = 0.6179 ' 0.2709 = 0.347 What
is
the
probability
that
an
American
randomly
chosen
has
a
height
 between
5.5
and
5.8?
 
 Y ! N (5.7;! = 0.7) P(5.5 < Y < 5.8) = P(Y < 5.8) ! P(Y < 5.5) = " " 5.8 ! 5.7 % 5.5 ! 5.7 % P$Z < ' ! P$Z < ' # # 0.7 & 0.7 & ! ! 0.1 $ 0.2 $ P#Z < & ' P#Z < ' &= " " 0.7 % 0.7 % = P ( Z < 0.14) ' P ( Z < '0.29 ) = 0.5557 ' 0.3859 = 0.1698 What
is
the
probability
that
an
Italian
 randomly
chosen
is
taller
than
6
?
 X ! N (5.7;! = 0.33) P( X > 6) = " " 6 ! 5.7 % 0.3 % P$Z > ' = P$Z > ' = P ( Z > 0.91) = # # 0.33 & 0.33 & = 1 ! P ( Z < 0.91) = 1 ! 0.8186 = 0.1814 What
is
the
probability
that
an
american
 randomly
chosen
is
taller
than
6
?
 Y ! N (5.7;! = 0.7) P(Y > 6 ) = " " 6 ! 5.7 % 0.3 % P$Z > ' = P$Z > ' = P ( Z > 0.428) = P ( Z > 0.43) = # # 0.7 & 0.7 & = 1 ! P ( Z < 0.43) = 1 ! 0.6664 = 0.3336 Ques%on
4
We
randomly
draw
an
observa<on
from
a
Normal
random
variable
X.
 The
observed
value
is
x1
=
152.

If
the
mean
of
X
is
=
145,
and
z1
=
1.3
(that
is
the
 observed
 value
 of
 x1
 is
 1.3
 standard
 devia<ons
 above
 the
 mean)
 what
 is
 the
 standard
devia<on
of
X
?

 X ! N (145;! ) z1 = 1.3 x1 ! µ 152 ! 145 z1 = = = 1.3 ! ! 152 ! 145 != = 5.38 1.3 Ques%on
5
Suppose
that
a
variable
X
is
Normally
distributed
with
the
median
=
 120,
and
the
third
quar<le
=
140.

 a.  What
is
the
probability
that
X
is
greater
than
120
?
 b.  What
is
the
probability
that
X
is
less
than
110
?
 Due
to
normality,
hence
symmetry,
Mean
=
Median
=
120
 P( X > 120 )= P( Z > 0 ) = 0.5 The
informa<on
on
the
quar<le
implies
 
 
 
 
 
 
 From
the
tables
 
 
 P( X < 140 ) = 0.75 P( Z < z0.75 ) = 0.75 z0.75 = 0.68 z0.75 = 0.68 Q3 ! µ = 0.68 ! 140 ! 120 = 0.68 ! 140 ! 120 != = 29.41 0.68 Standardized
value
of
the
3rd
quar<le
 X ! N (120;! = 29.41) P( X < 110 ) = " " 110 ! 120 % 10 % P$Z < ' = P$Z < ! ' = P ( Z < !0.34) = # # 29.41 & 29.41 & = 0.3669 Ques%on
6
Given
the
random
variable
X,
Normally
distributed
with
mean
=
 100
and
standard
devia<on
=
10,
calculate
two
values
x1
and
x2,
symmetrical
 with
respect
to
the
mean,
such
that
P(
x1
≤

X

≤

x2
)
=
0.95.
 X ! N (100;! = 10 ) P( x1 < X < x2 ) = 0.95 P( X < x2 ) = 0.975 P( Z < z2 ) = 0.975 z2 = 1.96 x2 ! µ x2 ! 100 z2 = = = 1.96 ! 10 x2 = 19.6 + 100 = 119.6 For
symmetry
we
have
that

 µ ! x1 = x2 ! µ 100 ! x1 = x2 ! 100 = 119.6 ! 100 = 19.6 x1 = 100 ! 19.6 = 80.4 Alterna<ve
solu<on.
By
symmetry
we
have
 z2 = 1.96 z1 = !1.96 x1 ! µ x1 ! 100 z1 = = = !1.96 ! 10 x1 = !19.6 + 100 = 80.4 ...
View Full Document

This note was uploaded on 04/05/2012 for the course STATS V3100018.0 taught by Professor Giuseppearbia during the Spring '12 term at NYU.

Ask a homework question - tutors are online