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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	Determine and Justify	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]

Show that $m = 15.7$ .

[2]

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]

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

[3]

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

[2]

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]

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)

(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).

$\frac{13.6 + 17.8}{2}$ $(17.8 - \frac{17.8 - 13.6}{2})$ OR $(13.6 + \frac{17.8 - 13.6}{2})$ (M1)

Note: Award (M0)(M1) for unsupported $\frac{13.6 + 17.8}{2}$ OR $(17.8 - \frac{17.8 - 13.6}{2})$ OR $(13.6 + \frac{17.8 - 13.6}{2})$ 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]

$P (13 \leq T \leq 18)$ (M1)

(M1)

Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating $13$ and $18$ .

$0.468 (46.8 %, 0.467516 \dots)$ (A1)(G2)

[2 marks]

c.i.

$P (T \geq 20)$ (M1)

(M1)

Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating $20$ .

$0.141 (14.1 %, 0.141187 \dots)$ (A1)(G2)

[2 marks]

c.ii.

$P (T < t) = 0.6$ (M1)

(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.

$16.7 (16.7133 \dots)$ (A1)(G2)

[2 marks]

$0.467516 \dots \times 0.141187 \dots \times 2$ (M1)(M1)

$(0.467516 \dots \times 0.141187 \dots) + (0.141187 \dots \times 0.467516 \dots)$ (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).

$0.132 (13.2 %, 0.132014 \dots)$ (A1)(ft)(G2)

Note: Award (G0) for an unsupported final answer of $0.066007 \dots$

[3 marks]

$\frac{69}{102} \times 200$ (M1)

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

$135 (135.294 \dots)$ (A1)(G2)

[2 marks]

$(\frac{67}{98} \times 200 =) 136.734 \dots$ (A1)

Note: Award (M1) for $137$ or $136.734 \dots$ seen.

Emlyn is incorrect, $135 < 137 (135.294 \dots < 136.734 \dots)$ (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).

$(\frac{67}{98} =) 0.684 (0.683673 \dots) (\frac{69}{102} =) 0.676 (0.676470 \dots)$ (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]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.8—Binomial distribution

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Hide related questions

Topic 4—Statistics and probability » SL 4.9—Normal distribution and calculations

Topic 4—Statistics and probability

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Markscheme

Examiners report

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