Statistics for Academics Purposes 700045 Exam 1 Term 1 2020 Analysis

This document presents the solutions to the Statistics for Academics Purposes Exam 1, administered in Term 1 of 2020. The solutions cover a range of statistical concepts, including data types (ordinal, ratio, nominal), sampling methods (cluster sampling), population vs. sample, parameter vs. statistic, measures of central tendency (median, mode, mean, range), data distribution analysis, calculations of sample mean, sample variance, and sample standard deviation. The document also includes frequency distributions, histograms, and calculations of weighted averages and the coefficient of variation. The solutions are presented in a clear, step-by-step manner, providing detailed explanations and calculations for each question. This resource is designed to aid students in understanding and mastering the concepts covered in the exam.

EXAM 1
TERM 1 2020
[DATE]

Question 1
Part a
(i) It would be ordinal since data is categorical and can be arranged in an order.
(ii) Ratio level data since numerical data with defined zero.
(iii) Nominal level data since categorical values without any automatic arrangement.
Part b
Cluster sampling
Part c
The population is all primary school children while the sample is 2350 school children
who have been selected
Part d
i) This is a parameter since it describes the characteristic of a population
ii) This is a statistic since it describes the characteristic of a sample.
Question 2
Part a
(i) Median age would be (17+1)/2 or 9th value Based on the given stem and leaf plot, this
value is 37 years. The modal age is 35 years since this is the only value which is
repeated twice. All the other values have frequency 1.
(ii) Maximum value = 60 years
Minimum value = 23 years
Range = 60 -23 = 37 years
1

(iii) Mean will be impacted since 60 is the largest value in the given data set and thus
the new mean would be lower than the current mean. Mode and median would
remain unaffected.
Part b
(i) 15 minutes
(ii) Since 20 minutes corresponds to 75 percentile of the female data, hence 75% of
the females completed their test within 20 minutes
(iii) The distribution is asymmetric as the difference between median and Q1 is
significantly greater than the difference between median and Q3. This implies that a
right skew or positive skew is present.
(iv) Option D
Question 3
(i) Sample mean x bar= (115+94+61+58+57+53)/2=73
(ii) Table
x (X- xbar) (X- x bar)^2
115 42 1764
94 21 441
61 -12 144
58 -15 225
57 -16 256
53 -20 400
(iii) Sample variance and sample standard deviation
Sample variance s2= (25.4 )2=646
S ample standard deviations= √∑ ¿ ¿ ¿ ¿ ¿
Question 4
2

(a) Classes = 6
Min value = 37
Max value = 95
Class width = (95-37)/6 = 9.66 which means 10
Class width = 10
(b) Table
Class
limits
Frequenc
y
Class
boundaries
Class
midpoint
Cumulative
Frequency
Relative
Frequency
37-46 4 41.5 4 0.11
47-56 6 47 50.5 10 0.17
57-66 9 57 61.5 19 0.25
67-76 5 47 71.5 24 0.14
77-86 7 57 81.5 31 0.19
87-96 5 47 91.5 36 0.14
Total 36 255 398 124 1
(c) Frequency histogram
37-46 47-56 57-66 67-76 77-86 87-96
0
1
2
3
4
5
6
7
8
9
10
4
6
9
5
7
5
Frequency histogram
(Final exam scores%)
FInal exam scores %
Frequency
3

Question 5
Part a
Weighted average marks = 82*(0.15) + 75*(0.15) +81*(0.15) + 90*(0.15) + 84*(0.10) + 88*(0.30) =
84%
Since, average is less than 85%, hence I did not get A+.
Part b
(i) Coefficient of variation for sales = Standard deviation of sales/Mean of sales = 6/85 = 0.071
Coefficient of variation for commission = Standard deviation of commission/Mean of commission =
(720/5350) = 0.135
(ii) Commissions are more variable since the coefficient of variable for it is higher than the
corresponding value for sales.
4