Question
On a Monday at an amusement park, a sample of 40 visitors was randomly selected as they were leaving the park. They were asked how many times that day they had been on a ride called The Dragon. This information is summarized in the following frequency table.
It can be assumed that this sample is representative of all visitors to the park for the following day.
(a) For the following day, Tuesday, estimate
(i) the probability that a randomly selected visitor will ride The Dragon;
(ii) the expected number of times a visitor will ride The Dragon.
It is known that 1000 visitors will attend the amusement park on Tuesday. The Dragon can carry a maximum of 10 people each time it runs.
(b) Estimate the minimum number of times The Dragon must run to satisfy demand.
▶️Answer/Explanation
(a) (i) To estimate the probability that a randomly selected visitor will ride \textit{The Dragon} on Tuesday, we consider the frequency data provided from Monday’s sample. The total number of visitors sampled is 40. The number of visitors who rode \textit{The Dragon} at least once is the sum of the frequencies for one or more rides. This can be calculated as follows:
16 visitors rode \textit{The Dragon} once,
13 visitors rode it twice,
2 visitors rode it three times, and
3 visitors rode it four times.
Thus, the total number of visitors who rode \textit{The Dragon} is:
\(
16 + 13 + 2 + 3 = 34
\)
The probability that a randomly selected visitor will ride \textit{The Dragon} is therefore the number of visitors who rode the ride divided by the total number of visitors sampled. Hence, the probability is:
\(
\frac{34}{40}
\)
which simplifies to:
\(
\frac{17}{20}
\)
or \( 0.85 \) when expressed as a decimal.
Therefore, the estimated probability for a randomly selected visitor to ride \textit{The Dragon} on Tuesday is:
\(
0.85
\)
(ii) The expected number of times a visitor will ride \textit{The Dragon} is calculated using the expected value formula in probability, which is the sum of the products of the values and their corresponding probabilities:
\(
\textbf{Expected Value (}E(X)\textbf{)} = \sum [x \times P(X = x)]
\)
To find the expected value, we first determine the probability of each outcome by dividing the frequency of each outcome by the total number of responses (40 in this case). Then, each outcome is multiplied by its probability:
\(
P(0 \text{ rides}) = \frac{6}{40}
\)
\(
P(1 \text{ ride}) = \frac{16}{40}
\)
\(
P(2 \text{ rides}) = \frac{13}{40}
\)
\(
P(3 \text{ rides}) = \frac{2}{40}
\)
\(
P(4 \text{ rides}) = \frac{3}{40}
\)
These probabilities are then used to calculate the expected value:
\(
E(X) = (0 \times \frac{6}{40}) + (1 \times \frac{16}{40}) + (2 \times \frac{13}{40}) + (3 \times \frac{2}{40}) + (4 \times \frac{3}{40})
\)
\(
E(X) = \left(0\right) + \left(\frac{16}{40}\right) + \left(\frac{26}{40}\right) + \left(\frac{6}{40}\right) + \left(\frac{12}{40}\right)
\)
\(
E(X) = \frac{60}{40}
\)
\(
E(X) = 1.5
\)
Therefore, the expected number of times a visitor will ride \textit{The Dragon} is \textbf{1.5 times}.
(b) The first step in determining the minimum number of times The Dragon must run to satisfy demand is to calculate the expected number of rides per visitor. This is achieved by finding the weighted average of rides based on the provided frequency data. The process involves multiplying the number of times on The Dragon by the corresponding frequency and then summing these products. The sum is then divided by the total number of sampled visitors, which yields the average number of rides per visitor.
For the sample data provided, the calculation is as follows:
\(
(0 \text{ rides} \times 6 \text{ visitors}) + (1 \text{ ride} \times 16 \text{ visitors}) + (2 \text{ rides} \times 13 \text{ visitors}) + (3 \text{ rides} \times 2 \text{ visitors}) + (4 \text{ rides} \times 3 \text{ visitors})
\)
\(
= 0 + 16 + 26 + 6 + 12 = 60 \text{ rides}
\)
\(
\frac{60 \text{ rides}}{40 \text{ visitors}} = 1.5 \text{ rides per visitor}
\)
Given that 1000 visitors are expected on Tuesday, the total number of rides needed is:
\(
\frac{1.5 \text{ rides}}{\text{visitor}} \times 1000 \text{ visitors} = 1500 \text{ rides}
\)
The Dragon has a maximum capacity of 10 people per run, therefore, the minimum number of runs required to meet the total demand of rides is:
\(
\frac{1500 \text{ rides}}{10 \text{ people per run}} = 150 \text{ runs}
\)
Therefore, to satisfy the demand, The Dragon must run a minimum of 150 times.
Question
A discrete random variable X has the probability distribution given by the following table.
x | 0 | 1 | 2 | 3 |
P( X = x ) | p | \(\frac{1}{4}\) | \(\frac{1}{6}\) | q |
Given that E (X) = \(\frac{19}{12}\) , determine the value of p and the value of q .
Answer/Explanation
Ans:
Question
A biased four-sided die, A, is rolled. Let X be the score obtained when die A is rolled. The
probability distribution for X is given in the following table.
(a) Find the value of p . [2]
(b) Hence, find the value of E (X ) . [2]
A second biased four-sided die, B, is rolled. Let Y be the score obtained when die B is rolled.
The probability distribution for Y is given in the following table.
(c) (i) State the range of possible values of r .
(ii) Hence, find the range of possible values of q . [3]
(d) Hence, find the range of possible values for E (Y ) . [3]
Agnes and Barbara play a game using these dice. Agnes rolls die A once and Barbara rolls die B once. The probability that Agnes’ score is less than Barbara’s score is \(\frac{1}{2}\)
(e) Find the value of E (Y ) . [6]
Answer/Explanation
Ans
Question
The random variable T has the probability density function
\[f(t) = \frac{\pi }{4}\cos \left( {\frac{{\pi t}}{2}} \right),{\text{ }} – 1 \leqslant t \leqslant 1.\]
Find
(a) P(T = 0) ;
(b) the interquartile range.
Answer/Explanation
Markscheme
(a) Any consideration of \(\int_0^0 {f(x){\text{d}}x} \) (M1)
0 A1 N2
(b) METHOD 1
Let the upper and lower quartiles be a and −a
\(\frac{\pi }{4}\int_a^1 {\cos \frac{{\pi t}}{2}{\text{d}}t = 0.25} \) M1
\( \Rightarrow \left[ {\frac{\pi }{4} \times \frac{2}{\pi }\sin \frac{{\pi t}}{2}} \right]_a^1 = 0.25\) A1
\( \Rightarrow \left[ {\frac{1}{2}\sin \frac{{\pi t}}{2}} \right]_a^1 = 0.25\)
\( \Rightarrow \left[ {\frac{1}{2} – \frac{1}{2}\sin \frac{{\pi a}}{2}} \right] = 0.25\) A1
\( \Rightarrow \frac{1}{2}\sin \frac{{\pi a}}{2} = \frac{1}{4}\)
\( \Rightarrow \sin \frac{{\pi a}}{2} = \frac{1}{2}\)
\(\frac{{\pi a}}{2} = \frac{\pi }{6}\)
\(a = \frac{1}{3}\) A1
Since the function is symmetrical about t = 0 ,
interquartile range is \(\frac{1}{3} – \left( { – \frac{1}{3}} \right) = \frac{2}{3}\) R1
METHOD 2
\(\frac{\pi }{4}\int_{ – a}^a {\cos \frac{{\pi t}}{2}{\text{d}}t = 0.5 = \frac{\pi }{2}\int_0^a {\cos \frac{{\pi t}}{2}{\text{d}}t} } \) M1A1
\( \Rightarrow \left[ {\sin \frac{{a\pi }}{2}} \right] = 0.5\) A1
\( \Rightarrow \frac{{a\pi }}{2} = \frac{\pi }{6}\)
\( \Rightarrow a = \frac{1}{3}\) A1
The interquartile range is \(\frac{2}{3}\) R1
[7 marks]
Examiners report
All but the best candidates struggled with part (a). The vast majority either did not attempt it or let t = 1 . There was no indication from any of the scripts that candidates wasted an undue amount of time in trying to solve part (a). Many candidates attempted part (b), but few had a full understanding of the situation and hence were unable to give wholly correct answers.