CHAPTER 6 b)
No. Meas. 3
No. meas. 10, bin width 2 mm
2
1
0 48.7 48.9 49.1 49.3 49.5 49.7 49.9 50.1 50.3 50.5 50.7
Length range - cm
6.1 a) Data arranged in bins with width 2mm Bin (cm) 48.9-49.09 49.1-49.29 49.3-49.49 49.5-49.69 49.4-49.89 49.9-50.09 50.1-50.29 50.3-50.49 50.5-50.69
Number of Measurements 1 2 2 0 1 1 2 0 1
b) No. Meas. 8
No of meas. 10, bin width 2 in
7 6 5 4 3 2 1 0 48.1
Bin (in) 48.1-50.0 50.1-52.0 52.1-54.0 54.1-56.0 56.1-58.0 58.1-60.0 60.1-62.0 62.1-64.0 64.1-66.0 68.1-70.0
Number of Measurements 0 0 0 0 0 8 2 0 0 0
50.1
52.1
54.1
56.1
58.1
60.1
62.1
Length range - in
6.2 a) Data arranged in bins with width 2in
6.1
64.1
66.1
68.1
6.2
6.3 3 Total no. meas. = 12
No.
Bin width 5 psi
Meas.
2
1
0 80
85
90
95
100
105
110
115
120
125
Pressure range - psi
Bin 80-84.9 85-89.9 90-94.9 95-99.9 100-104.9 105-109.9 110-114.9 115-119.9 120-124.9
No. 0 1 2 0 2 2 3 1 0
No of meas. 12, bin width 0.5 bar
No. Meas. 4
3
2
1
0 8
8.5
9
9.5
10
Pressure range - bar
6.4 Bin 8-8.49 8.5-8.99 9-9.49 9.5-9.99 10-10.49 10.5-10.99
No. 0 3 3 4 2 0
6.3
10.5
11
6.5
Using the data from problem 6.1: Mean: n x1 x 2 x n x i n i 1 n 49.3 50.1 48.9 49.2 49.3 50.5 49.9 49.2 49.8 50.2 x 10
x
x 49.6cm
Median: Arranging data in ascending order: 48.9, 49.2, 49.2, 49.3, 49.3, 49.8, 49.9, 50.1, 50.2, 50.5 Median
49.3 49.8 49.6cm 2
Standard Deviation: S
( x i x )2 i 1 ( n 1) n
(49.3 49.64)2 (50.1 49.64 )2 (48.9 49.64)2 S (10 1) (10 1) (10 1) S 0.53 cm
Modes: 49.2cm, 49.3cm
6.4
1 2
6.6
Using the data from problem 6.2: Mean: n x x2 xn x x 1 i n i 1 n 59.3 60.0 58.8 59.1 59.2 60.4 59.8 59.3 59.8 60.3 x 10 x 59.6in
Median: Arranging data in ascending order: 58.8, 59.1, 59.2, 59.3, 59.3, 59.8, 59.8, 60.0, 60.3, 60.4 Median
59.3 59.8 59.6in 2
Standard Deviation: S
n
i 1
( xi x ) 2 ( n 1)
(58.8.1 59.6) 2 (59.1 59.6) 2 (59.2 59.6) 2 S (10 1) (10 1) (10 1)
S 0.54in
Modes: 59.3in,59.8in
6.5
1 2
6.7 x
110 104 106 94 92 89 100 114 120 108 110 115 12
x 105 psi
Median: Arranging data in ascending order 89, 92, 94, 100, 104, 106, 108, 110, 110, 114, 115, 120 Median
106 108 2
Median 107 Standard Deviation: S
( x i x )2 i 1 ( n 1) n
(110 105.2) 2 (104 105.2)2 (106 105.2) 2 S (12 1) (12 1) (12 1)
S 9.7 psi (10psi )
Mode: 110psi
6.6
1 2
6.8 x
9.5 9.3 9.4 8.9 8.8 8.7 9.0 9.8 10.2 10 9.5 9.9 12
x 9.4bar
Median: Arranging data in ascending order 8.7, 8.8, 8.9, 9.0, 9.3, 9.4, 9.5, 9.5, 9.8, 9.9, 10.0, 10.2 Median
9.4 9.5 2
Median 9.45
Standard Deviation: S
( x i x )2 i 1 ( n 1) n
(8.7 9.4) 2 (8.8 9.4) 2 (8.9 9.4) 2 S ( 12 1 ) ( 12 1 ) ( 12 1 )
1 2
S 0.50bar 1bar
Mode: 9.5bar
6.9 Probability of having a 6 and a 3 in tossing two fair dice: P (6 and 3) 2
1 1 2 0.056 6 6 36
5.6%
There are two ways of getting a 6 and a 3 - 6,3 and 3,6 6.10 Probability of having a 4 and a 2 in tossing two fair dice: P (64and 2) 2
1 1 2 0.056 5.6% 6 6 36
There are two ways of getting a 4 and a 2 - 4,2 and 2,4 6.11
Probability of an undergraduate electrical engineering student to be a woman: P 0.15 0.8 0.120
12%
6.7
6.12
Probability of an undergraduate biology student to be a woman: P 0.55 0.85 0.468 46.8%
6.13
Probability of all three components being defective: P 0.03 0.03 0.03 27 10 6 or 0.0027%
6.14
Probability of all three components being defective: P 0.02 0.02 4 10 4 or 0.04%
6.15 Binomial distribution all 5 > 12 oz.; p=0.99, n=5, r=5
n 5 5! 1 r 5 5!0
P (5) 1 0.99 5 (1 0.99 ) 0 0.951
all 5 < 12 oz.; p=0.99, n=5, r=0
n 5 5! 1 r 0 0!5
P(0) 1 0.99 0 (1 0.99) 5 10 5
6.16 Binomial distribution 6.8
all 5 > 8 oz.; p=0.98, n=5, r=5
n 5 5! 1 r 5 5!0
P (5) 1 0.98 5 (1 0.98) 0 0.904
all 5 < 12 oz.; p=0.98, n=5, r=0
n 5 5! 1 r 0 0!5
P (0) 1 0.98 0 (1 0.98) 5 3.2 10 9
6.17 Binomial distribution all 6 > 3000 hours; p=0.9, n=6, r = 6
6.9
n 6 6! 1 r 6 6!0
P (6 ) 1 0.9 6 (1 0.9) 0 0.531
6.10
6.18 Binomial distribution all 6 > 3600 hours; p=0.95, n=6, r = 6
n 6 6! 1 r 6 6!0
P (6) 1 0.95 6 (1 0.95) 0 0.735
6.19 Binomial distribution Success is failure before 1000 hours. We want probability of 1 or 2 failures. p=0.2, n=2, r=1 and r=2
2 2! 2 1 1!
2 2! 1 2 2!0
P (1) 2 0.21 (1 0.2)1 0.32
P (2) 1 0.2 2 (1 0.2)0 0.04
The probability of 1 or 2 is then P(1)+P(2) = 0.36
6.11
6.20
Probability distribution function: 3x 2 35 0
f ( x)
2 x3
(a) f(x) satisfies the requirement of a probability distribution function because: f (x) 0
x
P ( x )
2
3
2
3
f ( x )dx f ( x )dx f ( x )dx f ( x )dx
3
2
3
3x x dx 35 35 2
3
2
27 8 1 35 35
(b) Expected (mean) value of x:
xf ( x )dx
3
3
3x 2 dx 35
x
2
3
3x 3 x4 dx 35 4 35 2
3
2
3 81 16 140
1393 .
(c) Variance of population: 2
(x )
2
f ( x )dx
3
3x 2 (x ) dx 35 2 2
3
2
3
3
( x 2 2x 2 )
2 3
3x 2 dx 35
3x 4 3x 3 3x 2 3 dx 2 dx 2 dx x5 35 35 35 5 35 2 2
3 3 5 25 194 . 2.774 5 35
1666 .
6.12
3 2
2 2 2
6.21
Probability distribution function: f ( x)
x3 1 x 3 20
0
(a) f(x) satisfies the requirement of a probability distribution function because: f ( x) 0 x
1
3
1
3
f ( x)dx f ( x)dx f ( x)dx f ( x)dx
P( x )
3
x
3
1 x
4
20 dx 4 20
1
3
1
1 (81 1) 1 80
(b) Expected (mean) value of x:
3
xf ( x)dx x
3
1
x
4
1x
x3 dx 20
5 3
20 dx 5 20
1
1
1 243 1 100
2.42
(c) Variance of population: 2
(x )
2
f ( x ) dx
3
(x
)2
1
3
1
x3 dx 20
3
(x 2 2 x 2)
1
3 x5 x4 dx 2 dx 20 1 20 6
1 x [ 20 6
3
1
2 x 5
5
3
1
3
2
1
2
x3 dx 20
x3 dx 20
x4 4
3
] 1
1 1 2 2.42 ( 2.42) 2 [ (729 1) ( 243 1) (81 1)] 20 6 5 4 1 [121.333 234.256 117 .128] 20 0.210
0.458
6.13
6.22
Probability of the following cases of problem 6.20: 0
(a) for x 0
P ( x 0)
f ( x )dx
0
2
f ( x )dx
0
0
1
3x 2 x3 1 dx 0.0286 or 2.86% 35 35 0 35
Cumulative distribution of random variable x: x
x
f ( x )dx
0dx 0
x 2
x
3x 2 x3 8 dx 2 35 35 35
2 x 3
3
3x 2 35 dx 1 2
f ( x )dx
2
3x 2 x3 8 dx 0.2286 or 22.86% 35 35 2 35
(b) for 0 x 1 P (0 x 1)
F( x )
0
1
6.23
2
3 x
F(-2) = 0 , F(0) = 0.229 and F(3) = 1
6.14
6.24
Binomial distribution can be used because of the satisfactory/ unsatisfactory outcome of the process. n r n r p (1 p ) r
P (r )
In this case:
n n! r !( n r )! r
,
p=0.95 n=4
(a) All four parts be satisfactory: 4 P (4) (0.95)4 (1 0.95)0 4 1 0.8145 1 0.8145 or 8145% .
(b) For at least two parts to be satisfactory, we should calculate the probability that 2,3 and 4 parts be satisfactory: 4 2 2 ( 0.95 ) (1 0.95) 2
P( 2)
4 4! 6 ( 3 !)(1!) 2
6 0.9025 0.0025 0.0135 4 3 1 ( 0.95) (1 0.95) 3
P( 3)
4 4! 4 (3 !)(1!) 3
4 0.8574 0.05 0.1715
Probability of having at least two satisfactory parts: P (2) P (3) P (4)
0.0135 0.1715 0.8145 0.9995
6.25
or
99.95%
We want the probability that at most 2 computers will fail. The is the probability that 0, 1 or 2 failures. Define “success” as a computer failure. Then p=0.1. Then we want the probability that 0, 1 or 2 will fail. 20 0 20 0.1 (1 0.1) 0.121577 0
P (0)
20 1 19 0.1 (1 0.1) 0.27017 1
P (1)
20 2 18 0.1 (1 0.1) 0.28518 2
P (2)
P(2 or less) = .121577 + 0.27017 + 0.28518 = 0.677
6.15
6.26
Define success = failure. P = 0.1 We want the probability that 2, 3, 4, or 5 will fail. P(2-5) = P(2)+P(3)+P(4)+P(5) for example: 20 2 18 ( 0.1) (1 0.1) 0.2852; 2
20 20 ! 190 2! 18 ! 2
P ( 2)
20 3 17 (0.1) (1 0.1) .1901; 2
P ( 3)
20 20 ! 1140 3 ! 17 ! 3
20 4 16 (0.1) (1 0.1) 0.098; 4
P ( 4)
20 20 ! 4845 4 ! 16 ! 4
20 5 158 0.032; (0.1) (1 0.1) 5
P ( 5)
20 20 ! 15504 5 ! 15! 5
P(2-5) = 0.2852+.1901+.098+.032 = 0.597 6.27
(a) For all 6 parts to be satisfactory, using binomial distribution: 6 6 0 ( 0.95) (1 0.95) 6
P( 6)
0.7351
6 6! 1 6! 0! 6
73.51%
(b) For at least two parts to be satisfactory, we should find the sum of probabilities for 2,3,4,5 and 6 parts to be satisfactory: P (number of successes 2) 1 P (0) P (1) 6 0 6 ( 0.95) (1 0.95) 0
P (0)
6 6! 1 0! 6! 0
156 . 10 8 6 1 5 (0.95) (1 0.95) 1
P (1)
6 6! 6 1! 5 ! 1
6 0.95 (3.15 10 7 ) 178 . 10 6 so P (# of success 2) 100%
6.16
6.28
Probability of one or more power failure = 0.05 Probability of no power failure (success) = 0.95 Binomial distribution will be used here: n r n r p (1 p ) r
n n! r !( n r )! r
P (r )
,
a- No power failure in three months: n=3, r=3 3 3 0 (0.95 ) (1 0.95) 0.857 3
P(3)
85.7%
b- Exactly one month with power failure in four months: n=4, r=3 4 3 1 ( 0.95) (1 0.95) 3
P( 3)
4 0.8574 0.05 0.1715
4 4! 4 3 !(4 3)! 3 17.15%
c- At least one power failure in the nest five months: n=5 5 0 5 7 (0.95 ) ( 0.05) 3.125 10 0
P ( 0)
5 1 4 5 ( 0.95 ) ( 0.05 ) 2.969 10 1
5 5! 5 1! 4 ! 1
5 2 3 . 10 3 ( 0.95 ) (0.05) 1128 2
5 5! 10 2! 3 ! 2
5 3 2 (0.95 ) ( 0.05) 0.0214 3
5 5! 10 3 ! 2! 3
5 4 1 (0.95 ) ( 0.05) 0.2036 4
5 5! 5 4 ! 1! 4
P (1)
P ( 2)
P ( 3) P( 4) 4
P(i ) 0.226
Chance of at least on power failure in 5 months.
i 0
6.29
Binomial distribution can be used: success = a failure n=16 r=0 p=0.01 n r n r p (1 p ) r
P (r )
n n! r !( n r )! r
,
16 0 16 (0.01) (0.99) 0
P (r 0)
P (r 0) 0.85
16 16 ! 1 0 ! 16! 0
85% chance of no failures
6.17
6.30
Binomial distribution
(a) p=0.05, n=100, r = 2, 5, 10
n 10 10! 4950 r 2 !98 n 10 0! 7528 0 r 5 !95 n 10 10! 13 1.73x0 r 10 !90
P ( 2) 4950 0.05 2 (1 .05) 98 0.081
P (2) 75287520 0.05 5 (1 .05)95 0.180
P (2) 1.731x1013 0.05 10 (1 .05 )90 0.0167
Note: One cannot normally compute 100! with a hand calculator – it is a very large number. However 100!/(2!98!) can be rewritten 100x99x98!/(2!98!) =100x99/2!, which can be readily computed.
(b) We want the probability of 0 or 1 failing to be 0.99. i.e. P(1)+P(2)=0.99 6.18
n 10 10! 1 r 0 0!1
n 10 10! 10 r 1 1!9
P (0) P (1) 1 p 0 (1 p )10 10 p 1 (1 p ) 9 0.99
This must be solved by trial and error for p. Using a spreadsheet, the answer is p=0.0155 6.31
Binomial distribution
(a) p=0.03, n=100, r = 1, 4, 15
n 10 10 ! 10 r 1 !9
P (1) 100 0.031 (1 .03) 99 0.147
6.19
n 10 10! 3921 5 r 4 !96 n 10 10! 17 2.5310 r 15 !85
P (4) 3921225 0.03 4 (1 .03) 96 0.171
P (15) 2.533 x1017 0.0315 (1 .03) 85 2.729 10 7
Note: One cannot normally compute 100! with a hand calculator – it is a very large number. However 100!/(1!99!) can be rewritten 100x99x98!/(1!99!) =100/1!, which can be readily computed.
(b) We want the probability of 0 or 1 failing to be 0.99. i.e. P(1)+P(2)=0.99
n 10 10! 1 r 0 0!1
n 10 10! 10 r 1 1!9
P (0) P (1) 1 p 0 (1 p )10 10 p1 (1 p ) 9 0.99
6.20
This must be solved by trial and error for p. Using a spreadsheet, the answer is p=0.0155 6.32 We are looking at the probability that more than 175 passengers will show up. This can be solved as a binomial distribution problem. Consider success that a passenger shows up, so p=0.95. We then want more than 175 successes out of 180 trials.
n180 !18079 16! 6 42.9710 r176 !4 176!xx 4 n180 !18079 1! 95,860 17 6!3 17!xxr 3 6.21
n 180 ! 180 79 ! 160 178 !2 178!xxr 2 n180 ! 18079! 180 1 79 !1 79!xxr 1 n180 180! 1 r 180 180!x P (176 ) 42.297 10 6 x 0.95176 (1 0.95) 4 0.03174 P (177 ) 955860 x 0.95177 (1 0.95) 3 0.01363 P (178 ) 16110 x 0.95 178 (1 0.95 ) 2 0.004363
6.22
P (179 ) 180 x 0.95179 (1 0.95 )1 0.0009263 P (180) 1x 0.95180 (1 0.95 )0 0.00009778
P(r>175) = P(176)+ P(177)+ P(178)+ P(179)+ P(180) = 0.03174+0.01363+0.004363+0.0009263+0.00009778 = 0.0508
6.23
6.33 We are looking for the probability that there will 5 or less defective components. Consider success to be a defective component then p = 0.05, n=55 and r = 0,1,2,3,4,5
r=0:
r=1:
r=2:
r=3:
r=4:
5 5!5 1 0 0! 5!5 5 5! 5 1 1!54! 5 5! 1485 2 2!53! 5 5! 26 35 3 3!52 5 5! 34105 4 4!51
P (0) 1x 0.05 0 (1 .05 ) 55 0.0595
P (1) 55 x 0.051 (1 .05) 54 0.1725
P ( 2) 1485 x 0.05 2 (1 .05 ) 53 0.2449
P (3) 26235 x 0.05 3 (1 .05) 52 0.2277
P (0) 341055 x 0.05 4 (1 .05 ) 51 0.1558
6.24
r=5:
5 5! 3478 61 5 5! 0
P (5) 3478761x 0.05 5 (1 .05 )50 0.0836
So the probability of 5 or less defective components is 0.0595+0.1725+0.2449+0.2277+0.1558+0.0836 = 0.944. This is the probability that there are 50 or more good components. 6.34 This is a Poisson distribution problem. = 40/8 = 5 visits/hour. The probability of more than 5 visits in an hour is 1 – [P(5)+P(4)+ P(3)+P(2)+P(1)+P(0)]. P (0) e 5 5 0 / 0! 6.74 x10 3
P (1) e 5 5 1 / 1! .03369 P ( 2) e 5 5 2 / 2! 0.0842
P (3) e 5 5 3 / 3! 0.1404 P ( 4) e 5 5 4 / 4! 0.1755 P (5) e 5 5 5 / 5! 0.1755
So P(x>5) = 1 ( 6.74 x10 3 0.03369 0.0842 0.1404 0.1755 0.1755 ) =0.384. The probability that there will be more than 5 visits in an hour is 0.384 6.35 This can be solved as either a binomial distribution (p=0.001, n=4, r=1) or as a Poisson distribution. For the poisson distibution, the expected occurrence () will be 4/1000 = 0.004. We are then looking for the probability of x = 1. P (1) e 0.004 .004 1 / 1! 3.984 x10 3 . 6.36 This is a Poisson distribution. The expected number of crashes, , is (1/3)5=1.666. Then, P (0) e 1.666 1.666 0 / 0! 0.189 6.37 This is a Poisson distribution problem although it can be done as a binomial distribution. The expected frequency () of defects will be (1/50)x10 = 0.2 for the typical 10 m2 kitchen. The probability of 1 or more defects is 1-P(0). P (0 ) e 0.2 0.2 0 / 0! 0.818 P(x1) = 1-0.818 = 0.182. If solved as a binomial distribution , p= 0.02, n=10. For r = 0,
1 0 ! 0 10 1 (0)1xP .2(1.02) 0.81 0 !10
. So P(x1) = 1-0.818 = 0.182
6.38 Poisson distribution. The fact that a bulb has failed does not affect the probability of other failures. We need to know what the probability of two or more failures during the day. The probability of 2 or more is: P(x2) = 1 – P(0) – P(1). = 2. P (0) e x / x! e 2 2 0 / 0! 0.13533
6.25
P (1) e x / x! e 2 21 / 1! 0.2707
P(x2) = 1 – 0.1355 – 0.2707 = 0.5938 6.39 Poisson distribution. The expected number of failures in 50 calls will be (5/100)x50 = 2.5. (a) The probability of exactly 5 failures is: P (5) e x / x! e 2.5 2.5 5 / 5! 0.0668
(b) P(5 or less) = P(0)+P(1)+P(2)+P(3)+P(4)+p(5) Similar to part (a) P(5 or less) = 0.08208+0.2052+0.2565+0.2137+0.1336+.0668 = 0.958 (c) P(more than 5) = 1 – P(5 or less) = 1 – 0.958 =0.042 6.40 Poisson distribution. The average value of customers in that 1 hour is 20 = . (a) P (25) e 20 20 25 / 25! 0.0446 (b)P(20 to 25) = P(20)+ P(21)+ P(22)+ P(23)+ P(24)+ P(25)=0.0888+0.0846+0.0769+0.0669+.0557 = 0.417558 10 (c)P(10 or less) = i 0 P ( i ) = 0.010812 (d) P(x>10) = 1 – P(x10) = 1 – 0.010812 = 0.989 6.41 Poisson distribution. =3 for 1 sheet (a) P (25) e 3 310 / 10! 0.00081 (b) P(0) = 0.049787 (c) =3/6 = 0.5 for a single board. P (25 ) e 0.5 0.51 / 1! 0.3033 (d) P(more than 1) = 1 – P(1) – P(0) = 1 – 0.3033 - 0.6065 = 0.0902 6.42 For P(1or more) = 0.01, P(0) = 0.99 = e 0 0! Solving for we get = 0.01 defects per board or 0.06 defects per sheet.
6.26
6.43 The area from 100 to 100.5 is 0.2. From Table 6.3, z = 0.52 0.52
20%
20%
110 .5 100 , 0.96
Probability of error greater than 0.75 Volts: z
100.75 100 0.78125 0.96
99.5
100
100.5
28.26%
100
From the normal distribution curve(Table 6.3) for z = 0.7812 P(z) = 0.2826
for error greater than 100.75 or less than 99.25 (1000.75) we will have: P ( z ) 2 (0.5 0.2826) 0.4348
43.48%
6.44 The area from 110 to 110.5 is 0.25. From Table 6.3, z = 0.67 0.67
110.5 110 , 0.75
Probability of error greater than 1 Volts: z
111 110 1.33 0.75
From the normal distribution curve(Table 6.3) for z = 1.33 P(z) = 0.4082 for error greater than 111 or less than 109 (1101) we will have: P ( z ) 2 (0.5 0.4082) 0.1836 18.36%
6.27
100.75
6.45 6.827 6.832 0.5 0.01
a) z1
From Table 6.3 area = 0.1915 100(0.1915)(2) = 38 readings within 0.5 cm 6.812 6.832 2 0.01
b) z1
From Table 6.3 area = 0.4772 100(0.4772)(2) = 95 readings within 2 cm 6.782 6.832 5 0.01
c) z1
From Table 6.3 area = 0.5 100(0.5)(2) = 100 readings within 5 cm d) z1
6.831 6.832 0.1 0.01
From Table 6.3 area = 0.0398 100(0.0398)(2) = 8 readings within 10 cm
6.28
6.46 a) z1
7.76 7.75 1 .0 0.01
From Table 6.3 area = 0.3413 20(0.3413)(2) = 14 readings within 1 cm 7.77 7.75 2 0.01
b) z1
From Table 6.3 area = 0.4772 20(0.4772)(2) = 19 readings within 2 cm 7.80 7.75 5 0.01
c) z1
From Table 6.3 area = 0.5 20(0.5)(2) = 20 readings within 5 cm d) z1
7.85 7.75 10 0.01
From Table 6.3 area = 0.5 20(0.5)(2) = 20 readings within 10 cm
6.29
6.47 (a) The average is 71.3, the median is 70 and S = 12.62 (b) The grades according to the criterion will be: score grade 95 A 86 B+ 83 B+ 79 B 79 B 78 B 75 B70 C+ 70 C+ 68 C+ 63 C 63 C 55 C 55 C50 D
(c) There are 15 students and the division will be: grade A AB+ B BC+ C CD F
no. of students 0.342 0.66 1.3785 2.247 2.8725 2.8725 2.247 1.3785 0.66 0.342
Of course, the number of students with each grade is an integer. If we round off, we will only get 14 students total so some judgement is required for the additional student.
6.30
40%
10%
6.48 a)
From Table 6.3 for area0.4; z = 1.28 1.28
x 10000 500
x = 9,360 hrs b)
z1
7000 10000 6 area 0.5( fromTable 6.3) 500
z2
10000 10000 0 area 0 500
0.5 0 0.5 50% or1000 bulbs will fail
6.49 Normal distribution problem. 9 8 7 6 5 4 3 2 1
(a) The probability of the weight being less than 7.9 is the area under the curve to the left of x = 7.9. For this value of x, z ( x ) / (7.9 8.05) / .05 3.0 . From the nomral distribution table, for z = 3.0, the area under the curve (from the mean to the value of interest) is 0.4987. The area to the left of x = 7.9 is thus 0.5 – 0.4987 = 0.0013. So 0.13% of the cans will be rejected. (b) For this case, z ( x ) / (7.9 8.00) / .05 2.0 . For this value of z, the area under the normal distribution curve from the mean to x=7.9 is 0.4772. So the area to the left of 7.9 is 0.5 – 0.4772 = 0.0228. 2.28% of the cans will be rejected. (c) for this case, z ( x ) / (7.9 8.05) / .03 5 The table does not give a value for z this high but for 4.9, it is given as 0.5. So the area to the left of 7.9 is 0.5 – 0.5 = 0. For this z value the number of rejected cans is negligible. (actually 3x10 -5%. )x f(
0
7 .85
7.9
7.95
8
8.0 5
8.1
8.15
8.2
8.25
x
6.50 (a) A 2 tolerance means that values greater than z = 2 or less than z =-2. For z = 2.0, the value from the normal distribution table is 0.4772 . The right hand tail area (for z>2) is 0.5 – 0.4772 = 0.0228 The left hand tail has the same area. The probability of rejection is then 2x0.0228 = 0.0456 or 4.56%. (b) This part is best solved using the binomial distribution. For 2 rejections, we define success as a rejection. Then for 2 rejections, r = 2, n=20 and p = 0.056. The probability 6.31
of 2 rejected parts is then 0.1705. Similarly, for r = 4 and r = 10 we get 0.0099 and 4.5x10-9 respectively.
6.51 0.2 inches x 7.25 ft 87 inches d x x 0.3 acceptable d 0.3 z 15 . 0.2 From Table 6.3, A 0.4332
P d 0.3 2 0.4332 0.8664
From Table 6.3
% rejection = 100(1-0.8664) = 13.36% To reduce the rejection rate to 3% 1-P = 0.03 P = 0.97 z(for P/2 = 0.485) = 2.17 z
d
0.3 0.14 inch 2.17
This means that the manufacturer has to cut the columns close to the standard size. 6.52
x 10.500 inches 0.005 inches
10.500
10.520 10.500 0.020 4 0.005 0.005 P ( x 10.520) 0.5 0.5 1 100% U sin gTable 6.3
a) z
10.485 10.500 3 0.005 10.515 10.500 z2 3 0.005
b) z1
From Table 6.3, A = 0.4987
P (10.485 x 10.515) 2 0.4987 0.9974 99.87% From Table 6.3
c) From Table 6.3,
z 2.5,P ( x 2.5 x x 2.5 ) 2 0.4938 0.9876 98.76%
6.32
10.520
P (rejection ) 1 0.9876 0.012
6.53
x 25.00cm
0.02cm
25.02 25.00 0.02 1 0.02 0.02 P ( x 25.02) 0.5000 0.3413 0.8413 84.13%
a) z
Using Table 6.3
24.95 25.00 2 .5 0.02 25.05 25.00 z2 2.5 0.02
b) z1
From Table 6.3, A = 0.4938
P(24.95 x 25.05) 2 0.4938 0.9876 98.76% From Table 6.3
c) From Table 6.3,
z 2 P ( x 2 x x 2 ) 2 0.4772 0.9544 95.44%
P (rejection ) 1 0.9544 0.046
6.54 40%
10%
50,000
a) = 5000 From table 6.3 for area = 0.40
z 1.28
x 50,000 128 . 5000
x = 43,600 miles
6.33
b) z1
60,000 50,000 2 5000
area = 0.4772
50,000
z2
70,000 50,000 4 5000
60,000
70,000
area = 0.5
0.5-0.4772 = 0.0228 (100,000 tires)(0.0228) = 2280 tires fail between 60,000 to 70,000 miles
c)
20,000 50,000 6 5000
No tires expected to have life less than 20,000 d) Major assumption: Life span of tire follow normal distribution.
40%
10%
6.55 = 160 hrs For A = 0.4, z 1.28 x 3600 128 . 160 x 3395.2
Recommend replacing bulbs after 3395 hours. 6.56 x 6.686 lb, S 0.0486 lb, estimate of x S / n 0.0486 / 10 0.01536 lb 6.57a Confidence level: 95% 1- = 0.95 = 0.05 0.5 - /2 = 0.475
z = 1.96 (Table 6.3)
6.34
n 2 (196 . )( 2) 30 40 30 0.620 mph with confidence of 95% x z
6.57b For a large sample, n>30, x z / 2 / n For 90, 95, and 99% confidence levels, x/2 = 1.64, 1.96, and 2.58 respectively. For 90% confidence level, the confidence interval is 1.64x0.2/(40)1/2 = 0.052 oz. For 95% and 99% confidence levels, the intervals are 0.061 oz and 0.082 oz. 6.58 In this case the sample size is small, less than 30, so we must use the tdistribution. The interval on the mean is given by: x t / 2S / n . The sample size is 20 so the degrees of freedom is 19. From the t-distribution table, the values of t for = 19 and the confidence levels of 90%, 95% and 99% (/2 = 0.05, 0.025 and 0.005) are 1.729, 2.093, and 2.861. For 90% confidence level, the confidence interval is 1.729x0.2/(20)1/2 = 0.0773 oz. For 95% and 99% confidence levels, the intervals are 0.0936 oz and 0.128 oz. 6.59 (a) The mean is 16.042 oz. and the sample standard deviation is 0.079 oz. The standard deviation of the mean is 0.07941/(12) 1/2 = 0.0229 oz. (b) This is a t-distribution problem with = 12-1 = 11 and /2 = 0.025. From the tdistribution table, t has a value of 2.201. The confidence interval on the mean is then t/2S/(n)1/2 = 2.201x0.0229 = 0.0504 oz.
6.35
6.60
Find 95% confidence interval on the mean x 50,000 miles S 5000 miles n 100
1- = 0.95 = 0.05 0.5-/2 = 0.475 z = 1.96 (From Table 6.3) x z 2
n
50,000 196 . ( 5000 / 50,000 980 miles
100)
6.36
6.61
n 10 VCRs x 1500 hours S 150 hours 0.05
Find the 95% confidence interval on the mean of the life of the VCRs.
n 1 9 a) . (FromTable 66. ) t 2262 0025 2 . 2 x t (S 2
n)
1500 2.262(150 1500 107 hours
10 )
Confidence interval on the mean is 107 hours b) For 95% confidence interval on the mean of 50 hrs:
t S 50 hrs n 2
( 0.025) 2
Assuming S =150 hrs will remain the same (in reality should be recalculated and it may change), using Table 6.6 we find n by trial and error. n 15 20 25 36
t
2
2.145 2.093 2.064 2
So 36 systems should be tested.
t S 2 n
(hrs )
83 70 62 50
t 2 is an approximate value for n > 30, in which case the t-distribution 2
approaches normal distribution.
6.37
6.62
For this problem, we need to find a one-sided confidence interval of the form x t
S n
t-distribution (a) x = 41.25x106, S = 0.30x106, n=10. For 99% confidence level, = 0.01 and = 10 – 1 = 9 from Table t = 2.821. The tolerance interval is then: x t
S n
41.25x10 6 2.821
0.3 x10 6 10
40.982 10 6
41.0x106 falls into this interval so the manufacturer cannot be confident that the average strength exceeds this limit. (b) If we change n to 20, then = 20 – 1 =19 and t = 2.539. 41.25x10 6 2.539
0.3 x10 6 20
41.080 10 6
Now the manufacturer can be confident the mean exceeds 41.0x10 6 psi.
6.38
6.63
n 10 x 910% . S 0.8% 0.05
n 1 9 0.025 t 2 2.262 ( From Table 6.6)
2
x t S 2
91 2.262 0.8 n 10
910 . 0.57 Cut confidence interval by one half > 0.57/2 = 0.285 Assuming that n > 30, and S to remain the same: z 2 z 0.025
0.285 n 1.96
( S )
S 0.8 z 2
n
2
0.285
n 30
20 more motors should be tested.
6.39
6.64
n8 x 88.5% S 0.5%
0.05 n 1 7 0.025 t 2 2.365 ( FromTable 6.6) 2 x t S 2
88.5 2.365 0.5 n
88.5 0.42 Cut confidence interval by one half > 0.42/2 = 0.21 Assuming that n > 30, and S to remain the same: z 2 z 0.025
0.21 n 1.960
( S )
S 0. 5 z 2
n
2
0.21
n 30
22 more motors should be tested.
6.40
8
6.65
Given: Confidence level = 99% Sheight = 2.44 in. Shand = 0.34 in.
Find: N
Soln.: Assuming N > 30 0.01 2 0.005
0.5 2 0.495
From Table 6.3
z 2.575
For height: the confidence interval: z 2
1 in, n
2.575
2.44 n
1 n 40
For sleeve length: z 2
0.4 in, n
2.575
0.34 0.2 n 18 n
In order to satisfy both conditions, 40 members should be chosen. 6.66
n 15 x 25 ppm
Find the 95% confidence interval.
S 3 ppm
0.05 n 1 14 2 0.025 2 14 , 0.025 26.119 ( FromTable 6.7 ) 2 14 ,1.025 5.6287
( n 1) S 2 2 14 ,.025
14 32 ( n 1) S 2 14 32 4.82; 22.17 2 26.119 14 5.6287 ,1.025
4.82 2 22.17 or 2.20 4.71 for 95% confidencelevel
6.41
6.67
The distribution governing the time interval between the arrival successive cars is give to be f(t,λ) = λe-λt Where
1 1 8
(a) P (t 6) 1 e ( 6 ) 1 e
6 8
0.5276
(b) P(t 15) 1 P(t 15) 1 (1 e
(c)
15 8
) 0.1534
In order to have only two arrivals in 10 minutes, if the interval between the first 1
two arrivals is t, mins, the interval between the next arrival should be greater than (10-t ) Thus, 1
2
1
P[no more than 2 arrivals] = P[0≤t ≤10 and t ≥(10-t )] 10 10 1 f (t1 ) P (t 2 (10 t1 ) dt1 e t1 (e (10t1 ) ) dt1 0 0
1 10 1 810 e dt1 e 10 1.25e 1.25 0.3581 8 0 10
6.68 We are given that TBF has an exponential distribution with μ = 450 hours. Therefore
1 1 450
(a) P ( failure 300hours ) 1 e
( 300 )
1 e
300 450
0.4856
(b) P ( failure 500hours ) e ( 500 ) 1 e
(c)
500 450
0.3292
P[300 failuretim e 600] P[ failure 600] P[ failure 300]
[1 e
600 450
] [1 e
300 450
] 0.7364 0.4866 0.2498
6.42
6.43
6.69 Let the pollution level, X, be in ppm. If we define y = log(X), then y is N(µ,σ), where μ = 1.9031 and σ = 1.3010 Note: Here we have used y = log(X) rather than y = ln(X) (a)
P[ X 90] P[ y log(90)] P[Y 1.9542]
We now convert this problem to a standard normal distribution, P[ y 1.9542] P[ Z
1.9542 1.9031 ] P[ Z 0.393] 1.3010
Using the standard normal distribution in Table 6.3, P[ X 90] P[ Z 0.393] 0.5 0.1528 0.3472
(b)
P[ X 20] P[ y log(20)] P[Y 1.3010] 1.3010 1.9031 P[ Z ] P[ Z 0.4628] 1.3010 0.5 0.1782 0.3218
(c) If the mean pollutant level is reduced to 40 ppm, then the new value of μ = log(40 = 1.6021 P[ X 20] P[ y log(20)] P[Y 1.3010] 1.3010 1.6021 P[ Z ] P[ Z 0.2304] 1.3010 0.5 0.0915 0.4085
6.44
6.70 Let X be the time taken to fix the software bug. X has a mean of 200 mins, a standard deviation of 30 mins. If y = log(X) is N(μ,σ) with μ = log(200) = 2.3010, σ = log(30) = 1.4771 (a)
P[ X 500] P[ y log(500)] P[Y 2.6990]
We now convert this problem to a standard normal distribution, P[ y 2.6990] P[ Z
2.6990 2.3010 ] P[ Z 0.2694] 1.4771
P[ X 500] P[ Z 0.2694] 0.5 0.1062 0.3938
(b)
P[100 X 200] P[2.000 y 2.3010]
2 2.3010 2.3010 2.3010 Z ] P[0.2038 Z 0] 1.4771 1.4771 0.0808
P[
(c)
P[ X 50] P[ y log(50)] P[ y 1.6990] 1.6990 2.3010 P[ Z ] P[ Z 0.4076] 1.4771 0.5 0.1582 0.3418
6.71 Chi-squared distribution. S = 0,002 in., n = 10, = n – 1 = 9. 95% confidence level: = 1-0.95 = 0.05. /2 = 0.025, 1-/2 = 0.975. From Table 6.7, 2/2 = 19.023, 21-/2 = 2.704. (n 1)S 2 (n 1)S 2 (10 1)0.002 2 (10 1)0.002 2 2 2 2 / 2 12 / 2 19.023 2.704 6 2 5 or 0.00137 0.00364 1.89 10 1.33 10 6.72 Chi squared distribution. S = 5500, n = 8, = 7. = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995. 2/2 = 20.278, 21-/2 = 0.9893. (n 1)S 2 (n 1)S 2 (8 1)5500 2 (8 1)5500 2 2 2 2 / 2 12 / 2 20.278 0.9893 2 6 10442351 214.0 10 or 3231 14360
6.45
6.73 Chi squared distribution. S = 10, n = 12, = 11. = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995. 2/2 = 26.757, 21-/2 = 2.6032. (n 1)S 2 (n 1)S 2 (12 1)10 2 (12 1)10 2 2 2 2 / 2 12 / 2 26.757 2.6032 2 41.11 422.5 or 6.41 20.55 We cannot be 99% confident that the standard deviation is less 15 mA.
6.74 There is an error in the problem. The required variance should be 0.0004 mm 2. (not m2). (a) Chi squared distribution. S = 0.01, n = 10, = 9. = 1-0.95 = 0.05, /2 = 0.025, 1-/2 = 0.975. 2/2 = 19.023, 21-/2 = 2.7004. (n 1)S 2 (n 1)S 2 2 2 / 2 12 / 2
(10 1)0.012 (10 1)0.012 2 19.023 2.7004
4.73 10 5 2 3.333 10 4 or 6.88 10 3 0.018
(b) The maximum of the confidence interval is less than the desired variance so the part is acceptable. (c) S = 0.01, n = 5, = 4. = 1-0.95 = 0.05, /2 = 0.025, 1-/2 = 0.975. 2/2 = 11.143, 21-/2 = 0.4844. (n 1)S 2 (n 1)S 2 (5 1)0.012 (5 1)0.012 2 2 2 / 2 12 / 2 11 .143 0.4844 5 2 4 3.590 10 8.258 10 The upper end of the confidence interval exceeds the allowable so the part is not acceptable.
6.46
6.75 There is an error in the problem. The required variance should be 0.0004 mm 2. (not m2). (a) Chi squared distribution. S = 0.01, n = 10, = 9. = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995. 2/2 = 23.589, 21-/2 = 1.7349. (n 1)S 2 (n 1)S 2 (10 1)0.012 (10 1)0.012 2 2 2 2 / 2 1 / 2 23.589 1.7349 5 2 4 3.81 10 5.188 10
(b) The maximum of the confidence interval is greater than the desired variance so the part is not acceptable. 6.76 There is an error in the problem. The required variance should be 0.0004 mm 2. (not m2). (a) Chi squared distribution. S = 0.01, = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995. (n 1)S 2 (n 1)S 2 2 2 / 2 12 / 2 We need to determine n so that the maximum of the confidence interval is 0.0004. We are told that n = 10 was too small so n must be larger. The following calculations were performed on a spreadsheet program. n 11 12 13 14
nu 10 11 12 13
chisq 2.155845 2.603202 3.073785 3.565042
upper lim 0.000463855 0.000422557 0.000390398 0.000364652
It can be concluded that 13 measurements will make the part acceptable.
6.47
6.77
Data from problem one arranged in ascending order: 48.9, 49.2, 49.2, 49.3 ,49.3, 49.8, 49.9, 50.1, 50.2, 50.5 x 49.64 S
xi
x
n 1
2
0.530
xi x Low po int
48.9 49.64 0.74
High po int
50.5 49.64 0.86
From the Table 6.8: n = 10 = 1.798 S = (0.530)(1.798) = 0.95294 Since S > deviations, then: No data should be required. 6.78
From problem 6.4: x 105.2 S 9.71
n = 12 = 1.829 (Table 6.8) 1 Pl arg est P 120 105.2 14.8 2 Psmallest P 89 105.2 16.2 S (9.71)(1829 . ) 17.76
Neither 1 or 2 exceeds S so No points rejected
6.48
6.79 x
y
xx
( x x) 2
y y
( y y) 2
20 30 40 50
1.02 1.53 2.05 2.55
-33.5714 -23.5714 -13.5714 -3.5714
1127.04 555.611 184.183 12.7549
-1.5271 -1.0171 -0.4971 0.0029
51.2669 23.9745 6.7463 -0.01036
60 75 100
3.07 3.56 4.05
6.4286 21.4286 46.4286
41.3269 459.185 2155.61
0.5229 1.0129 1.5029
2.3320 1.0345 0.25711 -6 8.4110 0.27342 1.0260 2.2587
7.1717
176.8214
4535.71
( x x)( y y)
3.3615 21.7050 69.7775
x 53.5714 y 2.5471 rxy rxy
(x
(x
i
i
x )( y i y )
x )2 ( y i y )2
12
176.8214
435.71 7.1717
12
rxy 0.9804
This value of rxy which is close to 1 shows a strong linear relationship.
6.49
6.80 (a)
x y n x x x y x x y b n x x
a
n xi y i
i
i
2
2
i
i
2
i
i
i
i
i
2
2
i
i
2
xi
yi
(xi)
20 30 40 50 60 75 100
1.02 1.53 2.05 2.55 3.07 3.56 4.05
400 900 1600 2500 3600 5625 10000
375
17.83
24,625
(yi)
2
xiyi
1.0404 2.3409 4.2025 6.5025 9.4249 12.6736 16.4025
20.4 45.9 82.0 127.5 184.2 267.0 405
52.5873
1132
7(1132) ( 375)(17.83) 0.0398 7( 24625) ( 375) 2 ( 24625)(17.83) ( 375)(1132) b 0.4587 7( 24625) ( 375 ) 2
a
y 3.898 10 2 x 0.4587 y V( mV) x T(C)
(b)
4.5 4 3.5 3 2.5
Best Fit Line
2 Output - mV 1.5 1 20
40
60 80 Temperature - C
6.50
100
6.51
6.81 (a) xi(t)
yi(t)
( x i x)
( x i x) 2
( y i y)
( y i y) 2
0.0000 0.1000 0.2000 0.3000 0.4000 0.5000
4.98 1.84 0.68 0.25 0.09 0.03
-0.2500 -0.1500 -0.0500 0.0500 0.1500 0.2500
0.0625 0.0225 0.0025 0.0025 0.0225 0.0625
3.6683 0.5283 -0.6317 -1.0617 -1.2217 -1.2817
13.4567 0.2791 0.3990 1.1271 1.4925 1.6427
-0.9171 -0.0793 0.0316 -0.0531 -0.1833 -0.3204
=18.3971
=-1.5215
=0.1750
( y i y)( x i x )
x 0.25 sec. y
y
rxy
rxy
N
i
13117 . Volts
(x
(x
i
i
x )( y i y )
x )2 ( y i y )2
15215 .
0.1750 18.397
12
12
0.848
Note: The relationship between Voltage and time is not linear! (b) xi(t)
yi(lnV)
( x i x)
0.0000 0.1000 0.2000 0.3000 0.4000 0.5000
1.6054 0.6098 -0.3857 -1.3863 -2.4079 -3.5066
-0.2500 -0.1500 -0.0500 0.0500 0.1500 0.2500
( x i x) 2 0.0625 0.0225 0.0025 0.0025 0.0225 0.0625
( y i y) 2.5173 1.5216 0.5262 -0.4744 -1.4961 -2.5947
=0.1750
( y i y) 2 6.3368 2.3154 0.2769 0.2251 2.2382 6.7324
=18.124
( y i y)( x i x ) -0.6293 -0.2282 -0.0263 -0.0237 -0.2244 -0.6487
=-1.780
x 0.25 y 0.912
rxy rxy
(x
(x
i
i
x )( y i y )
x )2 ( y i y )2
1780 .
0.175 18.124
12
12
0.99984 , showing a closer linear relationship between
ln(V) and than between V and t.
6.52
6.82 For the best fit line, the equ. is reading=1.0525weight-1.1661 For the forced zero, reading=0.9987weight.
40 reading
30 20
y = 1.0525x - 1.1661 y = 0.9987x
10 0 -10 0
10
20 weight - lb
6.53
30
40
6.83 DelP 1.96 4.2 4.9 5.48 5.91 7.3 7.73 9 9.9
Sy.x= rsquared=
DelP - in H2O
Voltage 0.31 0.65 0.75 0.85 0.91 1.12 1.19 1.38 1.52
0.022862 [Excel function:(steyx(yrange,xrange)] 0.99988 [Excel function: rsq(yrange,xrange)]
12 10 8 6 4 2 0
y = 6.5606x - 0.0629
0
0.5
1
1.5
2
Voltage
6.84 For the data given in Problem 6.79, we first compute the following quantities, using x = T and y = V
x
i
375, xi 2 24265, yi 17.86, xi yi 1133 .9, yi 52.7712 2
We can now compute the slope, a, and the intercept, b, as n xi yi xi y i 7 1133 .9 375 17.86 a 0.0390 7 24625 (375) 2 n xi 2 ( xi )2
b
n xi 2 yi ( xi )( xi y i ) n xi ( xi ) 2
2
14590 0.4590 31750
Therefore the best fit is V = 0.0390T + 0.4590
α/2,n-2
(a)
The two-sided 95% confidence interval implies α/2 = 0.025 so that t
2.571, the standard error, S yx
( y
Yi ) 2 0.2394 n2 i
The 95% two-sided confidence interval for a is S yx t ,n 2 0.2394 2.571 2 (a ) 0.0390 0.0390 0.0089 2 24625 7 (53.17) 2 xi 2 n x or (0.0302 ≤ a ≤ 0.0479) Similarly the 95% confidence interval for the intercept, b, is
6.54
0.025,5
=t
=
b S yx t 2
( xi ) 2 1 2 n n ( xi 2 n x 2 )
, n 2
1 375 2 2 7 7 ( 24625 7 53.17 2 ) 0.4595 0.5284 (0.0688,0.9879)
0.4595 2.571 0.2394
o
(b)
When T = 70 C, V = aT + b = 3.1929V
The 95% two-sided prediction interval for the above V is y * S yx t 0.025,5
n 1 ( x * x) 2 n xi 2 n x 2 where
x* = 70 and y* = 3.1929V o
Therefore the 95% prediction interval for V at T = 70 C is 7 1 (70 53.57) 2 3.1929 0.2394 2.571 7 24625 7 53.212 3.1929 0.6739 (2.5191,3.8668)
o
(c)
The upper limit of V at T = 70 C at 95% confidence level involves a one-sided
prediction interval with α = 0.05.
α,n-2
At α = 0.05, t
0.05,5
=t
= 2.015
Thus, the upper limit will be 7 1 (70 53.57) 2 3.1929 0.2394 2.051 7 24625 7 53.212 3.1929 0.5281 3.7211volts
6.85 Based on the given data, we compute the following quantities, where x = T oF and y = σ in ksi.
x
9000, xi 2 8180000, y i 314.8, xi y i 182970, y i 10348 2
i
6.55
o
(a)
The independent variable is the temperature, T F, and the dependent variable is
the tensile strength, σ (ksi). The coefficients of the linear regression model are a
12 182970 9000 314.8 0.0372 12 818 10 4 (9000) 2
818 10 4 314.8 9000 182.97 10 3 b 54.099 12 818 10 4 (9000) 2
Therefore the linear regression model for the data is σ = -0.0372T + 54.099
o
(b)
When the temperature is T = 670 F, the expected σ is
exp
σ
= -0.0372x670+54.099 = 29.2056 ksi.
For a 90% two-sided confidence interval, α/2 = 0.05, n = 12. α/2,n-2
From the tables, t
0.05,10
=t
= 1.812 o
The confidence interval for strength, σ, at T = 670 F is
exp S yx t 2
,n 2
n 1 ( x * x) 2 n xi 2 n x 2 where
x* = 670 and S yx
( y
Yi ) 2 115 .6 3.34 n2 10 i
o
Therefore, 90% confidence interval for σ at T = 670 F is 13 (670 750) 2 ( 29.2056 3.34 1.812 ) 12 818 10 4 12 750 2 ( 29.2056 6.3153) ( 22.89,35.52) ksi
o
(c)
The expected loss of strength for each 100 F increase is
Δσ = a x 100 = -0.0372 x 100 = -3.72 ksi The 95% confidence interval for this change in strength is 100 x (95% confidence interval for a) 6.56
The 95% confidence interval for the slope parameter, a, is S yx t ,n 2 3.34 2.228 s a 0.0372 818 10 4 12 750 2 xi 2 n x 0.0372 0.0062 (0.0434,0.0309)
o
Therefore, 95% confidence interval for change in strength for a 100 F change is = (-4.34, -3.09) ksi o
(d)
At T = 550 F, the expected tensile strength is
σ = a x 550 + b = -0.0372 + 550 + 54.09 = 33.641 ksi The one-sided 95% confidence level for the strength is (ax * b) S yx t 2
,n 2
n 1 ( x * x) 2 n xi 2 n x 2 where
x* = 550, α = 0.05, n = 12 α,n-2
From the tables, t
0.05,10
=t
= 1.812
Then, 0.0372 550 54.099 3.34 1.812
12 1 (550 750) 12 818 10 4 12 750 2
= 33.641-0.0051 = 33.6359 ksi o
Therefore, theh minimum strength at T = 550 F at a 95% confidence level is min
σ
= 33.6359 ksi
6.57
6.86 a) Best fit line for t and V 2
xi (t)
yi (V)
xi 0 0.01 0.04 0.09 0.16 0.25
0.0 0.1 0.2 0.3 0.4 0.5
4.98 1.84 0.68 0.25 0.09 0.03
2
yi 24.800 3.3856 0.4624 0.0625 0.0081 0.0009
xiyi 0 0.184 0.136 0.075 0.036 0.015
. 0.55 7.87 28.720 0.446 15
a
x y n x x
n xi y i
i
i
a
i
2
2
i
6( 0.446) (15 . )(7.87) 8.694 6(0.55) (15 . )2
x y x x y b n x x 2
i
i
i
i
i
b
i
2
2
i
(0.55)(7.87) (15 . )(0.446 ) 3.485 6( 0.55) (15 . )2
The best fit line is therefore: y = -8.694x + 3.485 Best fit line for t and lnV xi
2
0.0 0.1 0.2 0.3 0.4 0.5
xi 0 0.01 0.04 0.09 0.16 0.25
. 15
0.55
2
yi
xiyi
1.6054 0.60977 -0.38566 -1.3863 -2.4079 3.5066
yi 2.5774 0.37181 0.14874 1.9218 5.7982 12.2959
0 0.060977 -0.077132 -0.41589 -0.96316 -1.7533
5.47129
. 231139
. 31485
6.58
a
x y n x x
n xi y i
i
i
a
i
2
2
i
6( 3.1485) (15 . )( 5.47129) 10.175 6(0.55) (15 . )2
x y x x y b n x x 2
i
i
i
i
i
b
i
2
2
i
(0.55)( 5.47129) (15 . )( 3.1485) 16319 . 6(0.55) (15 . )2
Best fit line is therefore: y = -10.175x + 1.6319 b)
Standard error of the estimate for T vs V S y .x
y
2 i
b y i a x i y i n2
( 28.720) ( 3.485)(7.87) ( 8.694 )( 0.446) 62 11369 .
12
S y .x S y .x
Standard error of the estimate for T vs ln(V) Sy .x
. )( 5.47129 ) ( 10.175)( 3.1485) ( 23.1139) (16319 62
12
Sy .x 0.0378
The standard error for (T vs ln(V)) is much less than for (t vs V) therefore there is less data scatter around this best fit curve.
6.59
6.87 DelP 0.05 0.07 0.09 0.12 0.15 0.17 0.19 0.21 0.23 0.25 Sy.x=
Q 2 2.35 2.7 3.12 3.5 3.72 3.85 4.1 4.35 4.45
ln(DelP) -2.99573 -2.65926 -2.40795 -2.12026 -1.89712 -1.77196 -1.66073 -1.56065 -1.46968 -1.38629
0.105469 Sy.x=
ln(Q) 0.693147 0.854415 0.993252 1.137833 1.252763 1.313724 1.348073 1.410987 1.470176 1.492904
pred(lnQ) 0.693541 0.862349 0.988434 1.132764 1.244715 1.307509 1.363311 1.413523 1.459164 1.500996
e 0.000394 0.007934 -0.00482 -0.00507 -0.00805 -0.00621 0.015238 0.002536 -0.01101 0.008092
e/Sy.x 0.04402 0.886565 -0.53839 -0.56645 -0.89932 -0.69442 1.702752 0.283388 -1.23055 0.904233
log10(DelP) -1.30103 -1.15490196 -1.04575749 -0.92081875 -0.82390874 -0.76955108 -0.7212464 -0.67778071 -0.63827216 -0.60205999
log10(Q) 0.30103 0.371068 0.431364 0.494155 0.544068 0.570543 0.585461 0.612784 0.638489 0.64836
0.008949
y = 0.5017x + 2.1965
5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0
1.6 1.4
y = 12.179x + 1.5506
1.2 1
ln(Q)
Q-cfm
Excel Sx.y function is STEYX(yrange,xrange)
0.8 0.6 0.4 0.2 0
0
0.05
0.1
0.15
0.2
0.25
0.3
-4
-3
DelP- psi
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
0
-0.5
2 1 e/Sx.y
log10(Q)
-1
-1
ln(DelP)
y = 0.5017x + 0.9539
-1.5
-2
0 -1
0
0
0.05
0.1
0.15
0.2
0.25
0.3
-2
log10(DelP)
DelP
The ln(Q) vs ln(DelP) is clearly the better fit than the straight day (Q vs. DelP). The fit with log10 is different but is equivalent to the ln fit and is just as good. The standardized residuals were computed for the ln regression analysis. The value of the standardized residual at DelP = 0.19 is 1.702 and this point is suspect although it is not clearly an outlier.
6.60
6.88 Data T 90 90 90 90 90 120 120 120 120 120
rpm 2000 3000 4000 5000 6000 2000 3000 4000 5000 6000
Use the regression function in Excel under Tools/Data Analysis/regression Regression Parameters Meas Q rpm T rpmsq rpmxT pred Q pred-meas 0.5 2000 90 4000000 180000 0.499563 -0.00044 0.85 3000 90 9000000 270000 0.855213 0.005213 1.13 4000 90 16000000 360000 1.124443 -0.00556 1.31 5000 90 25000000 450000 1.307253 -0.00275 1.4 6000 90 36000000 540000 1.403643 0.003643 0.5 2000 120 4000000 240000 0.495564 -0.00444 0.86 3000 120 9000000 360000 0.866214 0.006214 1.15 4000 120 16000000 480000 1.150444 0.000444 1.35 5000 120 25000000 600000 1.348254 -0.00175 1.46 6000 120 36000000 720000 1.459644 -0.00036
SUMMARY OUTPUT Regression Statistics Multiple R 0.99993859 R Square 0.99987718 Adjusted R Square 0.99977892 Standard Error 0.00530498 Observations 10 ANOVA df Regression Residual Total
SS MS 4 1.145549 0.286387 5 0.000141 2.81E-05 9 1.14569
F Significance F 10176.2 5.85E-10
CoefficientsStandard Error t Stat -0.369 0.038255 -9.64585 Intercept X Variable 1 0.000526714 1.16E-05 45.38555 X Variable 2 -0.001133333 0.000336 -3.37787 X Variable 3 -4.32143E-08 1E-09 -43.1045 5E-07 7.91E-08 6.322548 X Variable 4 The regression equation is:
P-value Lower 95%Upper 95%Lower 95.0% Upper 95.0% 0.000203 -0.46734 -0.27066 -0.46734 -0.27066 9.81E-08 0.000497 0.000557 0.000497 0.000557 0.01972 -0.002 -0.00027 -0.002 -0.00027 1.27E-07 -4.6E-08 -4.1E-08 -4.6E-08 -4.1E-08 0.001459 2.97E-07 7.03E-07 2.97E-07 7.03E-07
Q = -0.369+0.0005267T-4.3214E-8rpmsq+5E-7rpmxT
Q - l/s
2nd order fit 4 3 2 1 0
y = -0.0236x2 + 0.9779x - 2.5974
4
5
6
7
8
V - m/s
6.61
6.89 V 4 5 6 7 8
Use the plotting and trendline functions in Excel Q 0.94 1.69 2.44 3.08 3.72
This data is non-linear and the 2nd order fit is superior. Note the points at the ends –the linear fit is above them and the 2nd order passes through them.
Q - l/s
linear fit 4 3 2 1 0
y = 0.695x - 1.796 4
5
6
7
V - m/s
6.62
8
6.90 Velocity 0.07 0.08 0.11 0.135 0.145 0.185 0.19 0.22 0.24 0.285 0.295
Use the plotting and trendline functions of Excel
0.4
2
y = 0.0373x + 0.11x + 0.0698
0.3 Velocity
Voltage 0.01 0.115 0.29 0.48 0.59 0.81 0.88 1.02 1.12 1.325 1.4
0.2 0.1 0 0
0.5
1
1.5
Voltage
T 200 300 400 500 600 700 800 900 1000 1100 1200 1300
sigma 41.5 40.3 39.2 37.8 35 32.5 28.5 23 16 9.5 6.5 5
sigma
6.91 (a) 50 40 30 20 10 0
y = 1E-10x4 - 4E-07x3 + 0.0003x2 - 0.1061x + 53.842
0
500
1000
1500
Temp
(b) Due to the S shape of the curve, at least a 3 rd order curve will be required. (c) The 4th order curve fit is shown.
6.63
6.92 Let L1 be the length of the shelves and L2 be the length between sides of the cabinet. The clearance between the two is = L2 – L1. We want to be 99% confident that is greater than equal to zero. S SL22 SL21 0.06 2 0.03 2 0.067
Assume large sample size for the standard deviations so we can use the normal distribution. We want to be sure that 99% of the shelves will not be too long. Using Table 6.3, For 99% confidence level, A = 0.49 and z =2.33 .This value of z corresponds to
zS 2.33 0.067 0.156 in
The longest the nominal length the shelves can be is 24 – 0.156 = 23.844 in long. Note: We did not concern ourselves with how short the shelves can be.
6.64
