Date | November 2020 | Marks available | 2 | Reference code | 20N.2.SL.TZ0.T_6 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | T_6 | Adapted from | N/A |
Question
Emlyn plays many games of basketball for his school team. The number of minutes he plays in each game follows a normal distribution with mean m minutes.
In any game there is a 30 % chance he will play less than 13.6 minutes.
In any game there is a 70 % chance he will play less than 17.8 minutes.
The standard deviation of the number of minutes Emlyn plays in any game is 4.
There is a 60 % chance Emlyn plays less than x minutes in a game.
Emlyn will play in two basketball games today.
Emlyn and his teammate Johan each practise shooting the basketball multiple times from a point X. A record of their performance over the weekend is shown in the table below.
On Monday, Emlyn and Johan will practise and each will shoot 200 times from point X.
Sketch a diagram to represent this information.
Show that m=15.7.
Find the probability that Emlyn plays between 13 minutes and 18 minutes in a game.
Find the probability that Emlyn plays more than 20 minutes in a game.
Find the value of x.
Find the probability he plays between 13 minutes and 18 minutes in one game and more than 20 minutes in the other game.
Find the expected number of successful shots Emlyn will make on Monday, based on the results from Saturday and Sunday.
Emlyn claims the results from Saturday and Sunday show that his expected number of successful shots will be more than Johan’s.
Determine if Emlyn’s claim is correct. Justify your reasoning.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)
Note: Award (A1) for bell shaped curve with mean m or 13.6 indicated. Award (A1) for approximately correct shaded region.
[2 marks]
P(T>17.8)=0.3 (M1)
OR
(M1)
Note: Award (M1) for correct probability equation using 0.3 OR correctly shaded diagram indicating 17.8. Strict or weak inequalities are accepted in parts (b), (c) and (d).
13.6+17.82 (17.8-17.8-13.62) OR (13.6+17.8-13.62) (M1)
Note: Award (M0)(M1) for unsupported 13.6+17.82 OR (17.8-17.8-13.62) OR (13.6+17.8-13.62) OR the midpoint of 13.6 and 17.8 is 15.7.
Award at most (M1)(M0) if the final answer is not seen. Award (M0)(M0) for using known values m=15.7 and σ=4 to validate P(T<17.8)=0.7 or P(T<13.6)=0.3.
15.7 (AG)
[2 marks]
P(13≤T≤18) (M1)
OR
(M1)
Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating 13 and 18.
0.468 (46.8%, 0.467516…) (A1)(G2)
[2 marks]
P(T≥20) (M1)
OR
(M1)
Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating 20.
0.141 (14.1%, 0.141187…) (A1)(G2)
[2 marks]
P(T<t)=0.6 (M1)
OR
(M1)
Note: Award (M1) for correct probability equation OR for a correctly shaded region with x indicated to the right-hand side of the mean.
16.7 (16.7133…) (A1)(G2)
[2 marks]
0.467516…×0.141187…×2 (M1)(M1)
OR
(0.467516…×0.141187…)+(0.141187…×0.467516…) (M1)(M1)
Note: Award (M1) for the multiplication of their parts (c)(i) and (c)(ii), (M1) for multiplying their product by 2 or for adding their products twice. Follow through from part (c).
0.132 (13.2%, 0.132014…) (A1)(ft)(G2)
Note: Award (G0) for an unsupported final answer of 0.066007…
[3 marks]
69102×200 (M1)
Note: Award (M1) for correct probability multiplied by 200.
135 (135.294…) (A1)(G2)
[2 marks]
(6798×200=) 136.734… (A1)
Note: Award (M1) for 137 or 136.734… seen.
Emlyn is incorrect, 135<137 (135.294…<136.734…) (R1)
Note: To award the final (R1), both the conclusion and the comparison must be seen. Award at most (A0)(R1)(ft) for consistent incorrect methods in parts (f) and (g).
OR
(6798=) 0.684 (0.683673…) (69102=) 0.676 (0.676470…) (A1)
Note: Award (A1) for both correct probabilities seen.
Emlyn is incorrect, 0.676<0.684 (R1)
Note: To award the final (R1), both the conclusion and the comparison must be seen. Award at most (A0)(R1)(ft) for consistent incorrect methods in parts (f) and (g).
[2 marks]