Exam in “Data processing” (2015-02-20)
Problem 1: Computing Percentages
In 1912 in the North Atlantic there was a disaster of „Titanic”.
A summary of people who survived or did not survive is provided in the table below:
Survived
Died
First class
passengers
203
122
Second class
passengers
118
167
Third class
passengers
178
528
Crew members
212
673
1.1. Royal mail ship “Carpathia” saved the people who survived the “Titanic” disaster.
Which barchart shows the distribution of the survivors according to their class? None of
the diagrams has numbers on the vertical axis; you should identify the diagram, which most
accurately displays the proportions given in the above table.
1.2. Which statement is correct?
(A) Second class passengers had better chance to survive than Third class passengers.
(B) Third class passengers had better chance to survive than Second class passengers.
(C) Second and Third class passengers had the same chance to survive.
1.3. Great grandmother of Maija traveled by „Titanic”, she was Second class passenger. If
we do not know anything else about this great grandmother, what were her chances to
survive the disaster? Write the result as percentages, round to the nearest tenth.
Problem 2: Histogram and Median Age
Ādaži is one of those municipalities in Latvia, which has relatively many children and young people.
According to the Latvian population database (PMLP), on January 1, 2015 there were the following
age groups in Ādaži



2295 people “below the working age” (i.e. younger than 15)
6943 people “at the working age” (at least 15 years old, but younger than the retirement
age)
1476 people “above the working age” (i.e. at the retirement age or older)
(The retirement age in 2015 is 62 years, 6 months.)
2.1. Which histogram shows correct distribution of the Ādaži population into the age groups?
2.2. Assuming that every rectangle shown in the histogram has people therein distributed
uniformly, find the median age of the Ādaži inhabitants. (Median age is such that one half is
younger than it, and another half – older.) Express the result in years, round to the nearest tenth.
3. Uzdevums: S80/S20 indekss
The current government of Latvia has expressed the desire to reduce income inequality. One of the
goals stated in the government declaration (quote): “the income inequality will reduce, the S80/S20
index will reach 5.4 in 2017.”
The S80/S20 ratio measures the relative disparity in the distribution of some quantity (wage, income,
standard of living, etc.). For an income distribution, S80/S20 compares the mass of income held by 20%
of the richest persons to that held by 20% of the poorest persons.) Accordingly to Eurostat, in Latvia
in 2013 the index was 6.3.
When he learned about the plans of Latvian government, the Wizard of Oz decided to group all the
100 inhabitants of the Emerald City into 10 groups – 10 people per each group (first – the 10 richest,
after than – 10 next richest, etc. and finally 10 – poorest ones). The average incomes of these groups
he denoted by letters A, B, C, D, E, F, G, H, I, J.
3.1. Write a formula, which expresses S80/S20 index in this city, using the quantities that were
denoted by A, B, C, D, E, F, G, H, I and J.
3.2. Annual income of these 10 population groups of the Emerald city are shown in the figure.
(Each number denotes average annual income in thousands of USD.) What is the value of the
S80/S20 index in this case?
3.3. Has the S80/S20 index in the Emerald City reached the goal declared by the government of
Latvia? Write your answer (Yes or No) with a short explanation.
Problem 4: Finding averages
The picture shows the average results of the 9th grade history exams in 2014. Average score was
computed for each school.
4.1. Find the history result average in the Ādaži municipality (Ādažu novads) – i.e. the grand total
of all children in the “Ādažu vidusskola” and “Ādažu Waldorf school”, if it is known that in the
Waldorf school exam was taken by 9 people, and in the “Ādažu vidusskola” – by 42 people.
4.2. Find the history result average for Carnikava+Garkalne municipalities combined. (“Carnikavas
pamatskola” and “Berģu Mūzikas un mākslas pamatskola”). The exact number of exam takers is
not known, but it is known that in Garkalne grade 9 had twice as many children than in Carnikava.
Problem 5: Normal Distribution
In the Spring of 2014 there was the centralized exam on mathematics. The results are shown in the
table below.
Grade Level
“A”
“B”
“C”
“D”
“E”
“F”
Level not
assigned
The number of exam
submissions
930
2834
4692
4184
2635 1042
113
Level “A” at this exam was given to those exam submission, which had grades 85% or more.
The average result of this exam is
μ = 40.5, the standard deviation σ = 20.7.
2.1. Please find the z-score for the exam result 85% - i.e. by how many standard deviations this result
exceeds the average value.
2.2. Assume that the results of 2014 are normally distributed with μ = 40.5 and σ = 20.7. Using
RStudio function pnorm(…), find the probability that the exam result will reach or exceed value
85%.
2.3. What is the actual number of all exam submissions, that in the Spring of 2014 exceeded 85%?