At a summer camp an arithmetic test is taken by 250 children. The times taken, to the nearest minute, to complete the test were recorded. The results are summarised in the table.
| Time taken, in minutes | 1-30 | 31-45 | 46-65 | 66-75 | 76-100 |
| Frequency | 21 | 30 | 68 | 86 | 45 |
(a) Question
Draw a histogram to represent this information.
Answer/Explanation
Ans:
| Class Width | 30 | 15 | 20 | 10 | 25 |
| Frequency Density | 0.7 | 2 | 3.4 | 8.6 | 1.8 |
(b) Question
State which class interval contains the median.
Answer/Explanation
Ans:
66 – 75
(c) Question
Given that an estimate of the mean time is 61.05 minutes, state what feature of the distribution accounts for the median and the mean being different.
Answer/Explanation
Ans:
Distribution is not symmetrical
Question
Box A contains 7 red balls and 1 blue ball. Box B contains 9 red balls and 5 blue balls. A ball is chosen at random from box A and placed in box B. A ball is then chosen at random from box B. The tree diagram below shows the possibilities for the colours of the balls chosen.
(a) Complete the tree diagram to show the probabilities.
(b) Find the probability that the two balls chosen are not the same colour.
(c) Find the probability that the ball chosen form box A is blue given that given that the ball chosen form box B is blue
Answer/Explanation
Ans:
(a)
(b) \(\frac{7}{8} \times \frac{5}{15}+\frac{1}{8} \times \frac{9}{15}\)
\(=\frac{44}{120}[\frac{11}{30}or0.367]\)
(c) P(A blue|B blue) \(=\frac{A blue \bigcap B blue)}{P(B blue)}\)
\(=\frac{\frac{1}{8}\times \frac{6}{15}}{\frac{7}{8}\times \frac{5}{15}+\frac{1}{8}\times \frac{6}{15}}=\frac{\frac{1}{20}}{\frac{41}{120}}\)
\(=\frac{6}{41}\) or 0.146
Question
A driver records the distance travelled i each of 150 journeys. These distances, correct to the nearest km, are summarised in the following table.
| Distance(km) | 0-4 | 5-10 | 11-20 | 21-30 | 31-40 | 41-60 |
| Frequency | 12 | 16 | 32 | 66 | 20 | 4 |
(a) Draw a cumulative frequency graph to illustrate the data.
(b) For 30% of these journeys the distance travelled is d km or more.
Use your graph to estimate the value of d.
(c) Calculate an estimate of the mean distance travelled for the 150 journeys.
Answer/Explanation
Ans:
(a)
| Distance | 0-4 | 5-10 | 11-20 | 21-30 | 31-40 | 41-60 |
| Upper boundary | 4-5 | 10-5 | 20-5 | 30-5 | 40.5 | 60-5 |
| Cumulative frequency | 12 | 28 | 60 | 126 | 146 | 150 |
(b) 70% of 150 = 105
Approx. 27
(c) Midpoints: 2.25, 7.5, 15.5, 25.5, 35.5, 50.5
Mean = \(\frac{2.25 \times 12 + 7.5 \times 16+ + 15.5 \times 32 + 25.5 \times 66 + 65.5 \times 20 + 50.5 \times 4}{150}\)
\(=\frac{27+120+496+1686+710+202}{150}\)
\([=\frac{3238}{150}]=21.6, 21\frac{44}{75}\)
Question
The numbers of chocolate bars sold per day in a cinema over a period of 100 days are summarised in
the following table.
(a) Draw a histogram to represent this information. [5]
(b) What is the greatest possible value of the interquartile range for the data? [2]
(c) Calculate estimates of the mean and standard deviation of the number of chocolate bars sold. [4]
Answer/Explanation
Ans
7 (a) Class widths: 10, 5, 15, 20, 10
Frequency density = frequency/their class width: 1.8, 4.8, 2, 1, 0.8
All heights correct on diagram (using a linear scale)
Correct bar ends
Bar ends: 10.5, 15.5, 30.5, 50.5, 60.5
7 (b) 11 – 15 and 31 – 50 B1
Greatest IQR = 50 – 11 = 39
7 (c) \(Mean=\frac{18\times 5.5+24\times 13+30\times 23+20\times 40.5+8\times 55.5}{100}=\frac{2355}{100}=23.6\)
\(Var=\frac{18\times 5.5^{2}+24\times 13^{2}+30\times 23^{2}+20\times 40.5^{2}+8\times 55.5^{2}}{100}-mean^{2}\)
\(\frac{77917.5}{100}-mean^{2}=224.57\)
Standard deviation = 15.0
(FT their variance)