Transcript Review of distributions

Review of distributions
• Question 1
The daily amount of coffee in litters, dispensed
by a machine located in airport lobby is a
random variable X having a continuous
uniform distribution with A = 7 and B = 10. Find
the brobability that on a given day the amount
of coffee dispensed by this machine will be:
a) At most 8.8 litters.
b) More than 7.4 litters but less than 9.5 litters.
c) At least 8.5 litters.
Review of distributions
Solution:
a)
8.8  7
P( X  8.8) 
 0.60
10  7
Review of distributions
• Solution:
b)
9.5  7.4
P(7.4  X  9.5) 
 0.70
10  7
Review of distributions
• Solution:
c)
10  8.5
P( X  8.5) 
 0.50
10  7
Review of distributions
Question 2:
• Given a standard normal distribution,
find the value of k such that
a) P( Z > k ) = 0.2946
b) P( Z < k ) = 0.0427
c) P(-0.93 < Z < k ) = 0.7235
Review of distributions
Solution:
a) P ( Z < k ) = 1 – 0.2946 = 0.7054, from table
this area is equal to Z = 0.54
Then k = 0.54.
b) P (Z < k) = 0.0427. Directly from table this
area is equal to Z = -1.72 = k.
c) P (- 0.93 < Z < k) = 0.7235.
The total area equal to 0.7235 + area left of
Z= - 0.93
= 0.7235 + 0.1762=0.8997
Now from table this area is equal to Z = 1.28=k
Review of distributions
Q3: The finished inside diameter of a
piston ring is normally distribution with
a mean of 10 cm and standard
deviation of 0.03 cm.
a) What proportion of rings will have
inside diameters exceeding 10.075 cm?
b) What is the probability that a piston
rings will have an inside diameter
between 9.97 and 10.03 cm?
c) Below what value of inside diameter
will 15% of the piston rings fall?
Review of distributions
Solution:
a)
z
X 

10.075  10

 2.5
0.03
P (Z > 2.5) = 1-P (Z < 2.5) =1 - .0.9938 = 0.0062
So that 0.62% of the rings has inside diameters
exceeding 10.075 cm
Review of distributions
Solution:
b)
z1 
X 


9.97  10
 1.0
0.03
, z2 
10.03  10.0
 1.0
0.03
Then P (9.97 < X< 10.03) = P (-1.0 < Z <1.0)
= P(Z < 1.0) – P (Z < -1)
= 0.8413 - 0.1587 = 0.6826
Review of distributions
Solution:
c) Area = 0.15 and from table the value of
Z = -1.04 then
X  10
 1.04 
0.03
 X  10  0.03(1.04)  9.9688cm
Value of inside diameter will 15% of the piston
rings fall below 9.9688 cm
Review of distributions
Question 4:
•
A lawyer commutes daily from his
suburban home to his midtown office.
The average time for a one-way trip is
24 minutes, with a standard deviation
of 3.8 minutes. Assume the distribution
of trip times to be normal distribution.
a) What is the probability that a trip will
take at least 0.5 hour?
Review of distributions
• Question 4:
b) If the office opens at 9:00 A.M and the
lawyer leaves his house at 8:45 A.M
daily, what percentage of the time is he
late for work?
c) If he leaves the house at 8:35 A.M and coffee
is served at the office from 8:50 A.M until
9:00 A.M , what is probability that he misses
coffee?
d) Find the length of time above which we find
the slowest 15% of the trip?
Review of distributions
Solution:
a)
z
X 

30  24

 1.58
3.8
P(X > 30) = P(Z > 1.58) = 1 – P (Z < 1.58) = 1 - 0.9428 = 0.0571
b)
z
X 

15  24

 2.37
3.8
P( X > 15) = P( Z > -2.37 ) = 1-P( Z < -2.37 )
= (1-0.0089) = 0.9911
Therefore he is late at 99.11% of time.
Review of distributions
Solution:
c)
z
X 

25  24

 0.26
3.8
P( X > 25) = P( Z > 0.26 ) = 1-P( Z < 0.26 ) = (1- 0.6026) = 0.3974
The probability that he misses coffee is 39.74%
d) The length of time above which we find the slowest 15% of
the trip is 1-.15=0.85 ,Then from the table Z is equal 1.04
d)
X    z    (3.8)(1.04)  24  27.952 minutes
Review of distributions
• Question 5:
•
Twelve people are given two
identical speakers, which they are
asked to listen to for differences, if any.
Suppose that these people answer
simply by guessing, find the probability
that 3 people claim to have heard a
difference between the speakers.
Review of distributions
• Solution:
• We can solve this problem by binomial
distribution with n=12 and p = 0.5 as:
n k
12 
n k
  p (1  p)   (0.5) 3 (0.5) 9  0.0537
k 
3 
The probability that 3 people claim to have heard a
difference between the speakers is 0.0537.
Review of distributions
• Question 6:
• Public opinion reported that 5% of Americans
are afraid of being alone in a house at night.
If a random sample of 20 Americans is
selected, find these probabilities.
a) There are exactly 5 people in the sample
who are afraid of being alone at night.
b) There are most 3 people in sample who are
afraid of being alone at night.
c) There are at least 3 people in the sample
who are afraid of being alone at night.
Review of distributions
Solution
a)
n k
 20 
n k
  p (1  p)   (0.05) 5 (0.95)15  0.0022
k 
5 
Review of distributions
Solution
b) At most 3 peoples are afraid means
that P(0) +P(1)+P(2)+P(3)
n k
 20 
nk
0
20



P(0)  
p
(
1

p
)

(
0
.
05
)
(
0
.
95
)
 0.358
k 
0 
 


 20 
1
19
P (1)  
1 
(0.05) (0.95)  0.377


 20 
2
18

P ( 2)  
(
0
.
05
)
(
0
.
95
)
 0.189
2 


 20 
3
17

P (3)  
(
0
.
05
)
(
0
.
95
)
 0.06
3 


P (0)  P (1)  P ( 2)  P (3)  0.358  0.377  0.189  0.06  0.984
Review of distributions
Solution
• There are at least 3 people in the
sample who are afraid of being alone at
night.
P (at least 3 are afraid) = 1 – [P (0) +P (1)
+P (2)]
= 1- [0.358+.0377+0.189) = 0.076