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IB Mathematics SL 4.1 Concepts of population, sample AI SL Paper 2 - Exam Style Questions - New Syllabus

IB DP Maths AI SL Paper 2 - All Topics

Question

Grace, a librarian, wants to study the amount of time, \(T\) minutes, that people spent in her library on a given day.
(a) State whether the variable \(T\) is discrete or continuous. [1]
Grace’s data for \(160\) people who visited the library on that day is presented in the following table.
\(T\) (minutes)\(0 \le T < 20\)\(20 \le T < 40\)\(40 \le T < 60\)\(60 \le T < 80\)\(80 \le T < 100\)
Frequency5062\(k\)148
(b) Find the value of \(k\). [2]
(c) (i) Write down the modal class.
    (ii) Write down the mid-interval value for this class. [2]
(d) Use Grace’s data to calculate an estimate of the mean time that people spent in the library. [2]
(e) Using the table, write down the maximum possible number of people who spent 35 minutes or less in the library on that day. [1]
Grace assumes her data to be typical of future visitors to the library.
(f) Find the probability a visitor spends at least 60 minutes in the library. [2]
The following box and whisker diagram shows the times, in minutes, that the 160 visitors spent in the library. 
(g) Write down the median time spent in the library. [1]
(h) Find the interquartile range. [2]
(i) Hence show that the longest time that a person spent in the library is not an outlier. [3]
Grace believes the box-and-whisker diagram suggests that the times spent in the library are not normally distributed.
(j) Identify one feature of the box-and-whisker diagram which might support Grace’s belief. [1]
▶️ Answer / Explanation
Solution

(a) \(T\) is a continuous variable (time). [1]

(b) Total frequency is \(160\): \(50+62+k+14+8=160 \Rightarrow k=160-50-62-14-8=\boxed{26}\). [2]

(c) The largest frequency is \(62\) in the class \(20\le T<40\); so
(i) modal class: \(\boxed{20\le T<40}\).   (ii) mid-interval: \(\dfrac{20+40}{2}=\boxed{30}\) minutes. [2]

(d) Use class midpoints \(10,30,50,70,90\) with frequencies \(50,62,26,14,8\):

\[ \bar T=\frac{50(10)+62(30)+26(50)+14(70)+8(90)}{160} =\frac{5360}{160}=\boxed{33.5\text{ minutes}}. \] [2]

(e) Maximum possible \(\le 35\) minutes: take all \(50\) from \(0\text{–}20\) and all \(62\) from \(20\text{–}40\) (if they happened to be \(\le 35\)). Total \(\boxed{112}\). [1]

(f) “At least 60 minutes” means classes \(60\text{–}80\) and \(80\text{–}100\): \(14+8=22\) people.

\[ P(T\ge 60)=\frac{22}{160}=0.1375=\boxed{13.75\%}\ (\text{to }3\text{ s.f., }0.138). \] [2]

(g) From the boxplot, the median is \(\boxed{26\ \text{minutes}}\). [1]

(h) From the boxplot, \(Q_1=16\), \(Q_3=50\). So \(\text{IQR}=Q_3-Q_1=50-16=\boxed{34\ \text{minutes}}\). [2]

(i) Upper outlier fence \(=Q_3+1.5\,\text{IQR}=50+1.5(34)=\boxed{101}\).

The longest observed time on the diagram is \(92\) minutes, and \(92<101\). Therefore, the longest time is not an outlier. [3]

(j) The diagram is not symmetric (e.g., the median is not central in the box and the whiskers differ markedly in length), indicating right skew, so the data are unlikely to be normally distributed. [1]

Total: 17 marks
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