Statistics Assignment-2: Probability Distribution, Standard Deviation, Confidence Interval

This Statistics Assignment-2 covers topics like probability distribution, standard deviation, and confidence interval. It includes questions related to a class of 50 students, employees in a company, normal distribution of weights of men, and more.

CSTS-SEU-KSA
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Statistics (STAT-101)
Assignment-2 (Weeks: 5-7)
2nd Semester, 1439-1440 (2018-2019)
Due date :23/02/2019 (Time: 11:00 AM)
Student’s Name
Student’s ID
Section/CRN
Location
Marking Scheme
Question Score Obtained Score
Q-1 3
Q-2 3
Q-3 3
Q-4 3
Q-5 3
Q-6 3
Total 18
Note: You are required to fill your full name, ID and CRN.
Solve the following questions 6 ×3 mark =18 mark

CSTS-SEU-KSA
1. In a class on 50 students, 35 students passed in all subjects, 5 failed in one subject, 4
failed in two subjects and 6 failed in three subjects.
a. Construct a probability distribution table for number of subjects a student from the
given class has failed in.
b. Calculate the Standard Deviation.
Solution:
a) Probability distribution table;
Number of subjects failed Probability
0 35
50 =0.7
1 5
50 =0.1
2 4
50 =0.08
3 6
50 =0.12
b) Standard deviation
S . D= √∑ x2 p ( x )−x2
x= ( 0∗0.7 )+ ( 1∗0.1 ) + ( 2∗0.08 ) + ( 3∗0.12 )=0.62
S . D= √ ( 02∗0.7 ) + ( 12∗0.1 ) + ( 22∗0.08 ) + ( 32∗0.12 )−0.622
¿ √ 1.5−0.622
¿ √ 1.1156=1.05622
2. 45 % of the employees in a company take public transportation daily to go to work.
For a random sample of 7 employees, what is the probability that at most 2 employees
take public transportation to work daily?
Solution:
P ( At most 2 )=P ( X ≤ 2 )=P ( X =0 ) + P ( X =1 )+( X =2)
P ( X=0 ) = 7 !
0 ! ( 7−0 ) !∗0.450 ¿ ( 1−0.45 ) 7=0.015224
P ( X=1 )= 7 !
1 ! ( 7−1 ) ! ∗0.451 ¿ (1−0.45 )6=0.087194

CSTS-SEU-KSA
P ( X=2 ) = 7 !
2! ( 7−2 ) ! ∗0.452 ¿ ( 1−0.45 ) 5 =0.214022
P ( At most 2 ) =0.015224+0.087194+0.214022=0.31644
3. Find
a) P( z <1.87)
b) P(z >−1.01)
c) P(−1.01<z <1.87)
Solution:
a) P ( z<1.87 ) =0.969258
b) P ( z>−1.01 )=0.8437524
c) P ( −1.01< z< 1.87 ) =0.969258−0.8437524=0.1255056
4. Assume the population of weights of men is normally distributed with a mean of
175 lb. and a standard deviation 30 lb. Find the probability that 20 randomly
selected men will have a mean weight that is greater than 178 lb.
Solution:
We seek to find P(x >178)
Z= x−μ
σ / √n =178−175
30/ √20 = 3
6.70820 =0.4472
P ( Z >0.4472 )=0.3273604
Thus the probability that 20 randomly selected men will have a mean weight that is
greater than 178 lb is 0.3274.
5. We have a random sample of 100 students and 75 of these people have a weight less than 80
kg. Construct a 95% confidence interval for the population proportion of people who have a
weight less than 80 kg.
Solution:
^P= x
n = 75
100 =0.75
C . I → ^P± Zα/2 √ ^P ( 1− ^P )
n
→ 0.75 ±1.96∗
√ 0.75 ( 1−0.75 )
100
→ 0.75 ±1.96∗0.0433

CSTS-SEU-KSA
→ 0.75 ±0.08487
Lower limit :0.75−0.08487=0.66513
Upp er limit : 0.75+0.08487=0.83487
Thus the 95% confidence interval for the population proportion of people who have a weight
less than 80 kg is between 66.513% and 83.487%.
6. We have a sample of size n = 20 with mean x=12 and the standard deviations=2. What is a
95% confidence interval based on this sample?
Solution:
C . I → x ± Zα /2
s
√ n
→ 12± 1.96∗2
√ 20
→ 12± 1.96∗0.4472
→ 12± 0.8765
Lower limit :12−0.8765=11.1235
Upper limit :12+0.8765=12.8765
Thus the 95% confidence interval based on this sample is between 11.1235 and 12.8765