M248 Analyzing Data: Comprehensive Statistical Problem Set

This document presents solutions to a series of statistical problems related to the course M248 Analyzing Data. The problems cover topics such as descriptive statistics, including calculating the mean, median, quartiles, inter-quartile range, standard deviation, and variance. It also addresses probability distributions, confidence intervals, and hypothesis testing. Specific calculations and interpretations are provided for scenarios involving university rankings, shoplifting cases, garage space allocation, cell phone lifespans, and patient visits to an urgent care center. Additionally, the document includes analysis of educational television viewership and bank deposit differences. Desklib offers this assignment as a resource, providing students with access to a variety of solved assignments and past papers to aid in their studies.

Running head: M248
M248
Name of the student
Name of the university
Author’s note

1M248
Table of Contents
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................2
Part a............................................................................................................................................2
Part b............................................................................................................................................2
Part c............................................................................................................................................2
Part d............................................................................................................................................2
Answer 3..........................................................................................................................................2
Answer 4..........................................................................................................................................2
Part a............................................................................................................................................2
Part b............................................................................................................................................2
Answer 5..........................................................................................................................................3
Part a............................................................................................................................................3
Part b............................................................................................................................................3
Answer 6..........................................................................................................................................3
Part a............................................................................................................................................3
Part b............................................................................................................................................4
Answer 7..........................................................................................................................................4

2M248
Answer 1
Part a
University Rank # of Faculty
members
# of Enrolled
Students
Ratio of Faculty
Members to
Enrolled
Students
University S B+ 100 7850 1.27%
University T A+ 150 5500 2.73%
University M B+ 110 7400 1.49%
University N B+ 90 8350 1.08%
University V B 85 6610 1.29%
University X A 105 5560 1.89%
University Y C 80 7856 1.02%
University Z A 124 4563 2.72%
Part b
University Rank Graduates of 2017 # of Enrolled
Students
Percentage
of
Graduates
to Enrolled
Students
University S B+ 1450 7850 18.47%
University T A+ 920 5500 16.73%
University M B+ 1350 7400 18.24%
University N B+ 1320 8350 15.81%
University V B 910 6610 13.77%
University X A 1105 5560 19.87%
University Y C 850 7856 10.82%
University Z A 980 4563 21.48%

3M248
Part c
A+; 17%
A; 23%
B+; 40%
B; 13%
C+; 7%
Number of Universities according to Rank

4M248
Answer 2
Part a
The average number of the cases of shoplifting = 6
The median number of cases of shoplifting = 5.5
Since the mean number of cases of shoplifting is more than the median number of cases hence,
the data is skewed to the right.
Part b
The first Quartile value for the number of cases of shoplifting = 5
The third Quartile value for the number of cases of shoplifting = 7.5
Thus, the inter-quartile range = 2.5
Hence, 1.5 times IQR = 1.5*2.5 = 3.75
Since the minimum number of cases of shoplifting = 4
First Quartile – 1.5times IQR = 5 - 3.75 = 1.25
Third Quartile + 1.5times IQR = 7.5 + 3.75 = 11.25
Since none of the number of cases of shoplifting are outside the interval of 1.25 and 11.25,
hence, there are no outliers in the data.
Part c
The standard deviation of the number of cases of shoplifting = 1.76
The variance of the number of cases of shoplifting = 3.1
Part d
The median line in the box plot is nearer to the left end and further away from the right end.
Hence, the data is skewed to the right. The median of the number of shoplifts is between 5 and 6.

5M248

6M248
Answer 3
Part a
garage space for X cars: X Probability: P(X=x) X*P(X=x)
0 0.4 0
1 0.3 0.3
2 0.2 0.4
3 0.1 0.3
E ( X ) =μ=Σxf ( x )=1
Part b
garage space for X
cars: X x-μ (x-μ)2
Probability:
P(X=x) (x-μ)2f(x)
0 -1 1 0.4 0.4
1 0 0 0.3 0
2 1 1 0.2 0.2
3 2 4 0.1 0.4
Sum 1
σ 2=Σ ( x−μ )2 f ( x )=1
The standard deviation =
 = 1
Part c
garage space for X cars:
X Probability: P(X=x) F(x)
0 0.4 0.4
1 0.3 0.7
2 0.2 0.9
3 0.1 1
Part d
The probability that a randomly selected home has a garage space for at
least one car
= 0.3 + 0.2 + 0.1 = 0.6

7M248
Answer 4
Part a
The average lifetime of the cell phones =
 = 24.3 months
The standard deviation of the cell phones =
 = 2.6 months
Thus, the probability that the average lifetime would lie between 23 and 26 months
¿ P ( 23< X <26 )=P(23< Z <26)
¿ P( 23−24.3
2.6 < Z< 26−24.3
2.6 )
¿ P (−1.3
2.6 < Z < 1.7
2.6 )=P(−0.5< Z <0.65)
¿ 0.7422−0.3085=0.4336
Thus, the probability that the lifetime of the cell would lie between 23 and 26 months = 0.4336
Part b
The probability that the lifetime of the cell would be less than 23.5 hours = P ( 23.5< X )
¿ P
( 23.5−24.3
2.6
√ 33
< Z
) =P (( −0.8
0.4526 ) <Z )=P(−1.7676 <Z )
=0.0386
The probability that the average lifetime of the cell would be less than 23.5 months = 0.0386

8M248
Answer 5
Part a
The total number of days analysed = 20
The average patients daily at the urgent care center = μ = 18
Thus the total number of patients who visited the urgent care center = 20*18 = 360
The standard deviation of the number of patients daily at the urgent care center ¿ σ =3.2
Thus, the 95% Confidence interval ¿ μ ± z∗( σ
√n )
18+1.96∗
( 3.2
√ 360 )=18± 1.96∗3.2
18.97
18 ±1.96∗0.17=18 ± 0.33
Thus, the 95% confidence interval = 17.67, 18.33
The lower limit of the proportion = 17.67
The higher limit of the proportion = 18.33
Part b
The 95% confidence interval of the number of patients visiting the urgent care center, when
another sample of the number of patients visiting the urgent care center is taken then, the average
number of patients daily would lie between 17.67 and 18.33

9M248
Answer 6
Part a
The number of people sampled = 200
The number of people who watched educational television = 154
Thus the proportion of people who watched educational television p= 154
200 =0.77
Thus, the 95% confidence interval of the proportion ¿ p ± z∗
√ p∗ (1− p )
n
¿ 0.77 ± 1.96∗ √ 0.77∗( 1−0.77 )
200
¿ 0.77 ± 1.96∗0.029=0.77 ± 0.06
Thus, the 95% confidence interval = 0.71, 0.83
The lower limit of the proportion = 0.71
The higher limit of the proportion = 0.83
Part b
The 95% confidence interval can be interpreted as when another sample of 200 people are
sampled, it can be said with 95% confidence that the proportion of the people who watch
educational television would lie between 0.71 and 0.83.

10M248
Answer 7
The mean difference of the deposits of the banks = -1.1290
The 95% confidence interval of the difference is -2.4401, 0.1821
Thus, if another sample of 10 banks is taken then it can be said with 95% confidence that the
mean difference of the deposits from 3 years ago to today ranges between -2.4401 and 0.1821.