Date | November 2020 | Marks available | 1 | Reference code | 20N.1.SL.TZ0.S_8 |
Level | Standard Level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Write down | Question number | S_8 | Adapted from | N/A |
Question
Each athlete on a running team recorded the distance (MM miles) they ran in 3030 minutes.
The median distance is 44 miles and the interquartile range is 1.11.1 miles.
This information is shown in the following box-and-whisker plot.
The distance in miles, MM, can be converted to the distance in kilometres, KK, using the formula K=85MK=85M.
The variance of the distances run by the athletes is 169 km2169km2.
The standard deviation of the distances is bb miles.
A total of 600600 athletes from different teams compete in a 5 km5km race. The times the 600600 athletes took to run the 5 km5km race are shown in the following cumulative frequency graph.
There were 400400 athletes who took between 2222 and mm minutes to complete the 5 km5km race.
Find the value of aa.
Write down the value of the median distance in kilometres (km).
Find the value of bb.
Find mm.
The first 150150 athletes that completed the race won a prize.
Given that an athlete took between 2222 and mm minutes to complete the 5 km5km race, calculate the probability that they won a prize.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg Q3-Q1 , Q3-1.1 , 4.5-a=1.1Q3−Q1 , Q3−1.1 , 4.5−a=1.1
a=3.4a=3.4 A1 N2
[2 marks]
325 (=6.4)325 (=6.4) (km) A1 N1
[1 mark]
METHOD 1 (standard deviation first)
valid approach (M1)
eg standard deviation=√variance , √169standard deviation=√variance , √169
standard deviation=43=43 (km) (A1)
valid approach to convert their standard deviation (M1)
eg 43×58 , √169=85M43×58 , √169=85M
20242024 (miles) (=56) A1 N3
Note: If no working shown, award M1A1M0A0 for the value 43.
If working shown, and candidate’s final answer is 43, award M1A1M0A0.
METHOD 2 (variance first)
valid approach to convert variance (M1)
eg (58)2 , 6425 , 169×(58)2
variance =2536 (A1)
valid approach (M1)
eg standard deviation=√variance , √2536 , √169×(58)2
2024 (miles) (=56) A1 N3
[4 marks]
correct frequency for 22 minutes (A1)
eg 20
adding their frequency (do not accept 22+400) (M1)
eg 20+400 , 420 athletes
m=30 (minutes) A1 N3
[3 marks]
27 (minutes) (A1)
correct working (A1)
eg 130 athletes between 22 and 27 minutes, P(22<t<27)=150-20600 , 1360
evidence of conditional probability or reduced sample space (M1)
eg P(A B) , P(t<27 22<t<30) , P(22<t<27)P(22<t<m) , 150400
correct working (A1)
eg 130600400600 , 150-20400
130400 (1340=78000240000=3901200=0.325) A1 N5
Note: If no other working is shown, award A0A0M1A0A0 for answer of 150400.
Award N0 for answer of 38 with no other working shown.
[5 marks]