User interface language: English | Español

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.

[2]
a.

Show that m=15.7.

[2]
b.

Find the probability that Emlyn plays between 13 minutes and 18 minutes in a game.

[2]
c.i.

Find the probability that Emlyn plays more than 20 minutes in a game.

[2]
c.ii.

Find the value of x.

[2]
d.

Find the probability he plays between 13 minutes and 18 minutes in one game and more than 20 minutes in the other game.

[3]
e.

Find the expected number of successful shots Emlyn will make on Monday, based on the results from Saturday and Sunday.

[2]
f.

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.

[2]
g.

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]

a.

PT>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 PT<17.8=0.7 or PT<13.6=0.3.


15.7         (AG)


[2 marks]

b.

P13T18         (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]

c.i.

PT20         (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]

c.ii.

PT<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]

d.

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]

e.

69102×200         (M1)


Note: Award (M1) for correct probability multiplied by 200.


135  135.294         (A1)(G2)


[2 marks]

f.

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]

g.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.
[N/A]
e.
[N/A]
f.
[N/A]
g.

Syllabus sections

Topic 4—Statistics and probability » SL 4.8—Binomial distribution
Show 94 related questions
Topic 4—Statistics and probability » SL 4.9—Normal distribution and calculations
Topic 4—Statistics and probability

View options