Question

1. maximum mark: 17

elsie, a librarian, wants to investigate the length of time, t minutes, that people spent in her library on a particular day.
(a) state whether the variable t is discrete or continuous. 1
elsie’s data for 160 people who visited the library on that particular day is shown in the following table.

t (minutes)	( 0 leq t < 20 )	( 20 leq t < 40 )	( 40 leq t < 60 )	( 60 leq t < 80 )	( 80 leq t < 100 )
frequency	50	62	( k )	14	8

(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 elsie’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
elsie assumes her data to be representative of future visitors to the library
(f) find the probability a visitor spends at least 60 minutes in the library 2

Explanation:

Step1: Classify variable type

Time can take any value in an interval, so it is continuous.

Step2: Calculate $k$ via total frequency

Total frequency = 160, so $k = 160 - (50+62+14+8)$
$k = 160 - 134 = 26$

Step3: Identify modal class

Modal class has highest frequency: $20 \leq T < 40$

Step4: Compute modal class midpoint

Midpoint = $\frac{20+40}{2} = 30$

Step5: Estimate the mean

First, find mid-interval values for all classes: 10, 30, 50, 70, 90.
Mean = $\frac{(10 \times 50) + (30 \times 62) + (50 \times 26) + (70 \times 14) + (90 \times 8)}{160}$
Calculate numerator: $500 + 1860 + 1300 + 980 + 720 = 5360$
Mean = $\frac{5360}{160} = 33.5$

Step6: Max people ≤35 minutes

All from $0 \leq T <20$ and all from $20 \leq T <40$: $50 + 62 = 112$

Step7: Calculate probability of ≥60 minutes

Frequency for $60 \leq T <100$: $14 + 8 = 22$
Probability = $\frac{22}{160} = \frac{11}{80}$

Answer:

(a) Continuous
(b) $k=26$
(c) (i) $20 \leq T < 40$; (ii) 30
(d) 33.5
(e) 112
(f) $\frac{11}{80}$