Download Homework#4 solution

Homework #4 Solution
• Daily amount of raining in winter at
Gaza is by ml cube, A random
variable X have a continuous uniform
distribution with A = 100 ml cube , B
= 170 ml cube .
• Find the probability that a given day
the amount of raining is:
Homework #4 Solution
• At most 110 ml cube.
A= 100 , B = 170
At most 110
110  100
P( x  110) 
 0.143
170  100
Homework #4 Solution
1. More than 120 ml cube but less than
160 ml cube.
160  120
P(120  X  160) 
 0.571
170  100
Homework #4 Solution
• At least 145
170  140
P( X  140) 
 0.429
170  100
Homework #4 Solution
•
A given normal distribution finds the
value of k such:
1. P( z < k) = 0.0078
From table k = - 2.42
2. P( z > k) = 0.0099.
P(z > k ) = 1-.0099=0.9901
Then k = 2.33
Homework #4 Solution
•
A given normal distribution finds the
value of k such:
3. P( -0.88 < z <k ) = 0.8106
Area left of z = - 0.88= 0.1894
Then
0.1894 + 0.8106 = 1 = k
K = 3.49
4. t15,0.02 = k
k = 2.249
Homework #4 Solution
• 9 basketball players through balls into
a ring if the probability to score is 0.4
find:
• At least 4 players are scored
n= 9 , p = 0.4 , q = 0.6
9
P(0)   (0.4) 0 (0.6) 9  0.01
 0
Homework #4 Solution
9
1
8

P(1)  
(
0
.
4
)
(
0
.
6
)
 0.06
1 
 
9
2
7

P(2)  
(
0
.
4
)
(
0
.
6
)
 0.161
 2
 
9
P(3)   (0.4) 3 (0.6) 6  0.251
3
At least 4 players are scored = 1-P(0)+P(1)+P(2)+P(3)
=1- (0.01 + .06 + 0.161 + 0.251) = 0.518
Homework #4 Solution
• At least 5 players failed to score
p=0.6, q=0.4, At least 5 players failed to score = 1 – (P(0)+P(1)+P(2)+P(3)+P(4)
9
P(0)   (0.6) 0 (0.4) 9  0.0.000262
 0
9
P(1)   (0.6)1 (0.4) 8  0.00354
1 
9
2
7


P(2)   (0.6) (0.4)  0.0122
 2
Homework #4 Solution
9
3
6

P(3)  
(
0
.
6
)
(
0
.
4
)
 0.0743
3
 
9
4
5
P(4)  
 4
(0.6) (0.4)  0.167
 
At least 5 players failed to score = 1 – (P(0)+P(1)+P(2)+P(3)+P(4)
= [ 1 – (0.000262 + 0.00354 + 0.0122 + 0.0743 + 0.167)]=0.7427
• Exactly 6 players are scored
9
P(6)   (0.4) 6 (0.6) 3  0.0743
 6
Homework #4 Solution
• An experiment was run to test many of
thermometers to test the freezing point
of water. Assume that the mean reading
is 0 c and the standard deviation is 1.1
c and assume that the readings are
normal distribution.
• If one thermometer selected randomly
find:
Homework #4 Solution
1. The reading of freezing point is less
than 1.58 c.
1.58  0
z
 1.436
1.1
Then the area is 0.9251, which is the probability of
reading less than 1.58.
Homework #4 Solution
•
•
The reading of freezing point is above 1.19c.
Above -1.19
z
 1.19  0
 1.08
1.1
Homework #4 Solution
• The reading of freezing point is
between 1.6c and -1.2c.
 1.2  0
 1.09
1.1
1.6  0
z (1.6) 
 1.45
1.1
z (1.2) 
Area = 0.9265 - 0.1379 = 0.7886
Homework #4 Solution
•
A fabric manufacturer believes that the
proportion of orders for row material
arriving late is P=0.62.
• If random sample of 50 orders show that 4
or fewer arrived late, the hypothesis that
P=0.62 should be rejected in favor of the
alternative p < 0.62, use binomial
distribution:
1. Find the probability of committing a type I
error.
Homework #4 Solution
  P( x  4 | p  0.62)   p(0)  p(1)  p(2)  p(3)  p(4) | p  0.62
 50 
p (0)   (0.62 0 )(0.38 50 )  9.8  10  22
0 
 50 
p (1)   (0.621 )(0.38 49 )  7.96  10  20
1 
 50 
p (2)   (0.62 2 )(0.38 48 )  3.18  10  18
2 
 50 
p (3)   (0.62 3 )(0.38 47 )  8.3  10 17
3 
 50 
p (4)   (0.62 4 )(0.38 46 )  1.6  10 15
4 
  p(1)  p(2)  p(3)  p(4)
Homework #4 Solution
• Find the probability of committing type
II error for alternatives p=0.29, p=0.38,
p=0.48
 50 

 50 
 50 
0
50
1
49
2
48
 (0.29 )(0.71 )   (0.29 )(0.71 )   (0.29 )(0.71 )  
0 
1 
2 


1  1 
 50 

 50 
3
47
4
46
 (0.29 )(0.71 )   (0.29 )(0.71 )

 3 

4 
Homework #4 Solution
 50 

 50 
 50 
0
50
1
49
2
48
 (0.38 )(0.62 )   (0.38 )(0.62 )   (0.38 )(0.62 )  
0 
1 
2 


2  1
 50 

 50 
3
47
4
46
 (0.38 )(0.62 )   (0.38 )(0.62)

 3 

4 
 50 

 50 
 50 
0
50
1
49
2
48
 (0.48 )(0.52 )   (0.48 )(0.52 )   (0.48 )(0.52 )  
0 
1 
2 


3  1 
 50 

 50 
3
47
4
46
 (0.48 )(0.52 )   (0.48 )(0.52 )

 3 

4 
Homework #4 Solution
• The average height of girls in a class at
past year was 162.5 cm with standard
deviation of 6.9cm. If random sample of
20 girls were selected test the claim
that the average height are 165.5 cm at
alpha = 0.01.
H0: µ=165.5
Ha: µ≠165.5
Homework #4 Solution
• Calculated t
t
n( X  )


20 (165.5  162.5
 1.9544
6.9
Tabulated t
t
0.01

,19
2
 2.861
Calculated t < tabulated t then
We cannot reject the H0
Homework #4 Solution
• State the regression equation with
suitable hypothesis and calculate errors
for :
y 10 37 40 44 38 47 60
x 12 23 24 22 29 35 38
Homework #4 Solution
Sum
y
10
37
40
44
38
47
60
276
x
12
23
24
22
29
35
38
183
X^2
144
529
576
484
841
1225
1444
5243
xy
120
851
960
968
1102
1645
2280
7926
Estimate (y)
17.528
34.567
36.116
33.018
43.861
53.155
57.802
errors
-7.528
2.433
3.884
10.982
-5.861
-6.155
2.198
Homework #4 Solution
• N = 7;
;
(∑X)2 = 5243; (∑X) (∑Y) = 7926
y  39.43, x  26.14
S xy  
 X  Y 
183  276
XY 
 7926 
 710.57
n
7
Homework #4 Solution
S xx   X 2
 X 

2
n
(183 2 )
 5243 
 458.857
7
S XY
710.57
b1 

 1.549
S XX 458.857
b0  y  b1 x  39.43  (1.549  26.14  1.06
Homework #4 Solution
• For b0
H 0 : b0  0
H a : b0  0
For b1
H 0 : b1  0
H a : b1  0
 b0  1.06  0
Then it’s affected on model
Homework #4 Solution
 b1  1.549  0
Then it’s affected on model
Then the model is:
yˆ  1.06  1.549 x
ˆ
Errors  Y  Y