lOMoARcPSD|18918331
MUHAMMAD MUSA BIN YUSOF
A17KA0105
SKAW - BACHELOR IN CIVIL ENGINEERING
SCHOOL OF CIVIL ENGINEERING
SSCE2193
ENGINEERING STATISTICS
9 SETS
***There might me some errors/miscalculation in this document. Kindly, contact me through any one of the medium below if you spot the
mistake(s). Thank you.
Follow my telegram channel to find out more about what stuffs I have
regarding civil course.
http://t.me/Musa_Civil_UTM
Downloaded by Mike Tan ()
, lOMoARcPSD|18918331
Universiti Teknologi Malaysia
Semester I Session 2010/2011
SSE 2193 ENGINEERING STATISTICS
TEST 1 (15%) Duration: 1 hour
INSTRUCTION: Answer ALL 4 questions and provide your final answers in 4 decimal
places.
1. Aliyaa is a basketball player from Sekolah Tun Fatimah, Johor Bahru. She is a 70% free
throw shooter which means her probability of making a free throw is 0.70. During the
National Boarding School tournament, calculate the following probability that
(a) Aliyaa makes her third free throw on his fifth shot? [3 marks]
(b) Aliyaa makes her first free throw on his fifth shot [3 marks]
2. (a) An exponential distribution is a continuous probability distribution which describes
a process in which events occur continuously and independently at a constant average
rate. If a random variable X has an exponential distribution with mean 60,
i. write down the probability density function for X. [1 mark]
ii. Hence, calculate that the probability that X is less than 56. [3 marks]
(b) Medical employees at a hospital in Johor Bahru have determined that the average
time between patient arrivals at the emergency room is exponentially distributed
with a mean time of 11 minutes. On a given day, it has been 11 minutes since a
patient has arrived. What is the probability that a patient will arrive within the
next 5 minutes? [4 marks]
3. Intravenous fluid bags are filled by an automated filling machine. Assume that the fill
volumes of the bags are normal random variables with a standard deviation of 0.10 fluid
ounces.
(a) What is the standard deviation of the sample mean fill volume of 25 bags? [1 mark]
(b) If the mean fill volume of the bags is 6.16 ounces, what is the probability that the
sample mean fill volume of 25 bags is more 6.10 ounces? [3 marks]
(c) Find the new mean fill volume if the probability that the sample mean fill volume
of 25 bags below 6.11 ounces is 0.04. [4 marks]
4. A manufacturing of semiconductor chips produces 2% defective chips. Assume that the
chips are independent and a random sample of 500 chips is selected. What is the proba-
bility that
(a) more than 1% chips are defective. [4 marks]
(b) at least 98.5% chips are non-defective. [4 marks]
Downloaded by Mike Tan ()
, lOMoARcPSD|18918331
2
List of Formula
Probability Distributions
Name Probability Mean Variance
Distribution
!
n
Binomial px (1 − p)n−x np np(1 − p)
x
x = 0, 1, ..., n, 0 6 p 6 1
e−λ λx
Poisson ; x = 0, 1, . . ., λ > 0 λ λ
x!
!
x−1
Negative Binomial (1 − p)x−r pr r/p r(1 − p)/p2
r−1
x = r, r + 1, r + 2, . . ., 0 6 p 6 1
Geometric (1 − p)x−1 p ; x = 1, 2, . . ., 0 6 p 6 1 1/p (1 − p)/p2
Exponential λe−λx ; x > 0, λ > 0 1/λ 1/λ2
Sampling Distributions
σ2
2
N −n
X̄ ∼ N µ, σn X̄ ∼ N µ,
N −1 n
p(1 − p)
σ2 σ22
P ∼N p, X̄1 − X̄2 ∼ N µ1 − µ2 , n11 + n2
n
X̄ − µ X̄ − µ
Z= σ ∼ N (0, 1) T = ∼ t(n−1)
√ S
n √
n
X̄1 − X̄2 − (µ1 − µ2 ) P −p
Z= s ∼ N (0, 1) Z=r ∼ N (0, 1)
σ12 σ22 p(1 − p)
+ n
n1 n2
Downloaded by Mike Tan ()
, lOMoARcPSD|18918331
MUHAMMAD MUSA BIN YUSOF - SSCE2193 TEST 1 2010/2011 SEM 1
1. a) NEGATIVE BINOMIAL DISTRIBUTION
No. of success (free throw) :r=3
Probability of getting success (free throw) : p = 0.7
*** Number of shot is the trial in this problem X variable
X ~ b∗ (r , p) → X ~ b∗ (3 , 0.7)
P(X = 𝑥) = (𝑥−1 Cr−1 ) × (pr ) × (1 − p)𝑥−r
P(X = 5) = (5−1 C3−1 ) × (0.73 ) × (1 − 0.7)5−3 = 0.18522
1. b) GEOMETRIC DISTRIBUTION
No. of success (free throw) :r=1
Probability of getting success (free throw) : p = 0.7
*** Number of shot is the trial in this problem
X ~ b∗ (r , p) → X ~ b∗ (1 , 0.7)
P(X = 𝑥) = (𝑥−1 Cr−1 ) × (pr ) × (1 − p)𝑥−r
P(X = 5) = (5−1 C1−1 ) × (0.71 ) × (1 − 0.7)5−1 = 0.00567
2. a) EXPONENTIAL DISTRIBUTION
1 1
E(X) = 60 → E(X) = μ = = 60 → λ=
λ 60
1
X ~ Exp(λ) → X ~ Exp ( )
60
1
1
i) 𝑓(𝑥) = λe−λ𝑥 = e−60𝑥 ; 𝑥≥0 , λ≥0
60
1
ii) P(X < 56) = 1 − 𝑒 −λ𝑥 = 1 − 𝑒 −(60)(56) = 0.6068
2. b) EXPONENTIAL DISTRIBUTION
1 1
E(X) = 11 → E(X) = μ = = 11 → λ=
λ 11
1
X ~ Exp(λ) → X ~ Exp ( )
11
1
P(X < 5) = 1 − 𝑒 −λ𝑥 = 1 − 𝑒 −(11)(5) = 0.3653
Downloaded by Mike Tan ()
MUHAMMAD MUSA BIN YUSOF
A17KA0105
SKAW - BACHELOR IN CIVIL ENGINEERING
SCHOOL OF CIVIL ENGINEERING
SSCE2193
ENGINEERING STATISTICS
9 SETS
***There might me some errors/miscalculation in this document. Kindly, contact me through any one of the medium below if you spot the
mistake(s). Thank you.
Follow my telegram channel to find out more about what stuffs I have
regarding civil course.
http://t.me/Musa_Civil_UTM
Downloaded by Mike Tan ()
, lOMoARcPSD|18918331
Universiti Teknologi Malaysia
Semester I Session 2010/2011
SSE 2193 ENGINEERING STATISTICS
TEST 1 (15%) Duration: 1 hour
INSTRUCTION: Answer ALL 4 questions and provide your final answers in 4 decimal
places.
1. Aliyaa is a basketball player from Sekolah Tun Fatimah, Johor Bahru. She is a 70% free
throw shooter which means her probability of making a free throw is 0.70. During the
National Boarding School tournament, calculate the following probability that
(a) Aliyaa makes her third free throw on his fifth shot? [3 marks]
(b) Aliyaa makes her first free throw on his fifth shot [3 marks]
2. (a) An exponential distribution is a continuous probability distribution which describes
a process in which events occur continuously and independently at a constant average
rate. If a random variable X has an exponential distribution with mean 60,
i. write down the probability density function for X. [1 mark]
ii. Hence, calculate that the probability that X is less than 56. [3 marks]
(b) Medical employees at a hospital in Johor Bahru have determined that the average
time between patient arrivals at the emergency room is exponentially distributed
with a mean time of 11 minutes. On a given day, it has been 11 minutes since a
patient has arrived. What is the probability that a patient will arrive within the
next 5 minutes? [4 marks]
3. Intravenous fluid bags are filled by an automated filling machine. Assume that the fill
volumes of the bags are normal random variables with a standard deviation of 0.10 fluid
ounces.
(a) What is the standard deviation of the sample mean fill volume of 25 bags? [1 mark]
(b) If the mean fill volume of the bags is 6.16 ounces, what is the probability that the
sample mean fill volume of 25 bags is more 6.10 ounces? [3 marks]
(c) Find the new mean fill volume if the probability that the sample mean fill volume
of 25 bags below 6.11 ounces is 0.04. [4 marks]
4. A manufacturing of semiconductor chips produces 2% defective chips. Assume that the
chips are independent and a random sample of 500 chips is selected. What is the proba-
bility that
(a) more than 1% chips are defective. [4 marks]
(b) at least 98.5% chips are non-defective. [4 marks]
Downloaded by Mike Tan ()
, lOMoARcPSD|18918331
2
List of Formula
Probability Distributions
Name Probability Mean Variance
Distribution
!
n
Binomial px (1 − p)n−x np np(1 − p)
x
x = 0, 1, ..., n, 0 6 p 6 1
e−λ λx
Poisson ; x = 0, 1, . . ., λ > 0 λ λ
x!
!
x−1
Negative Binomial (1 − p)x−r pr r/p r(1 − p)/p2
r−1
x = r, r + 1, r + 2, . . ., 0 6 p 6 1
Geometric (1 − p)x−1 p ; x = 1, 2, . . ., 0 6 p 6 1 1/p (1 − p)/p2
Exponential λe−λx ; x > 0, λ > 0 1/λ 1/λ2
Sampling Distributions
σ2
2
N −n
X̄ ∼ N µ, σn X̄ ∼ N µ,
N −1 n
p(1 − p)
σ2 σ22
P ∼N p, X̄1 − X̄2 ∼ N µ1 − µ2 , n11 + n2
n
X̄ − µ X̄ − µ
Z= σ ∼ N (0, 1) T = ∼ t(n−1)
√ S
n √
n
X̄1 − X̄2 − (µ1 − µ2 ) P −p
Z= s ∼ N (0, 1) Z=r ∼ N (0, 1)
σ12 σ22 p(1 − p)
+ n
n1 n2
Downloaded by Mike Tan ()
, lOMoARcPSD|18918331
MUHAMMAD MUSA BIN YUSOF - SSCE2193 TEST 1 2010/2011 SEM 1
1. a) NEGATIVE BINOMIAL DISTRIBUTION
No. of success (free throw) :r=3
Probability of getting success (free throw) : p = 0.7
*** Number of shot is the trial in this problem X variable
X ~ b∗ (r , p) → X ~ b∗ (3 , 0.7)
P(X = 𝑥) = (𝑥−1 Cr−1 ) × (pr ) × (1 − p)𝑥−r
P(X = 5) = (5−1 C3−1 ) × (0.73 ) × (1 − 0.7)5−3 = 0.18522
1. b) GEOMETRIC DISTRIBUTION
No. of success (free throw) :r=1
Probability of getting success (free throw) : p = 0.7
*** Number of shot is the trial in this problem
X ~ b∗ (r , p) → X ~ b∗ (1 , 0.7)
P(X = 𝑥) = (𝑥−1 Cr−1 ) × (pr ) × (1 − p)𝑥−r
P(X = 5) = (5−1 C1−1 ) × (0.71 ) × (1 − 0.7)5−1 = 0.00567
2. a) EXPONENTIAL DISTRIBUTION
1 1
E(X) = 60 → E(X) = μ = = 60 → λ=
λ 60
1
X ~ Exp(λ) → X ~ Exp ( )
60
1
1
i) 𝑓(𝑥) = λe−λ𝑥 = e−60𝑥 ; 𝑥≥0 , λ≥0
60
1
ii) P(X < 56) = 1 − 𝑒 −λ𝑥 = 1 − 𝑒 −(60)(56) = 0.6068
2. b) EXPONENTIAL DISTRIBUTION
1 1
E(X) = 11 → E(X) = μ = = 11 → λ=
λ 11
1
X ~ Exp(λ) → X ~ Exp ( )
11
1
P(X < 5) = 1 − 𝑒 −λ𝑥 = 1 − 𝑒 −(11)(5) = 0.3653
Downloaded by Mike Tan ()
