S1 June 2014 (R) Q6

EdexcelOld spec13 marksNormal Distribution

6. The time taken, in minutes, by children to complete a mathematical puzzle is assumed to be normally distributed with mean \(\mu\) and standard deviation \(\sigma\). The puzzle can be completed in less than 24 minutes by 80% of the children. For 5% of the children it takes more than 28 minutes to complete the puzzle.

(a) Show this information on the Normal curve below. (2)

(b) Write down the percentage of children who take between 24 minutes and 28 minutes to complete the puzzle. (1)

(c)

(i) Find two equations in \(\mu\) and \(\sigma\).

(ii) Hence find, to 3 significant figures, the value of \(\mu\) and the value of \(\sigma\). (7)

A child is selected at random.

(d) Find the probability that the child takes less than 12 minutes to complete the puzzle. (3)

A blank Normal curve drawn above a horizontal axis

Normal curve with 24 and 28 marked above the mean on the horizontal axis; area 0.80 to the left of 24 and 0.05 to the right of 28 — Mark scheme (a)

Scheme	Marks
24 and 28 (above the mean)	B1
For 0.80 and 0.05 (clearly indicated)	B1
	(2)

Scheme	Marks
15%	B1
	(1)

Mark scheme (c)

Scheme	Marks
(i) \(\dfrac{(28 - \mu)}{\sigma} = 1.64(49)\) or \(\dfrac{(24 - \mu)}{\sigma} = 0.84(16)\)	M1
0.8416 and 1.6449 seen	B1
\(\mu = 28 - 1.64(49)\sigma\) , \(\mu = 24 - 0.84(16)\sigma\)	A1,A1
(ii) \(24 - 0.8416\sigma = 28 - 1.6449\sigma\) eliminating \(\mu\) or \(\sigma\)	M1
\(\sigma = 4.9794597\ldots\) awrt 4.98	A1
\(\mu = 19.809286\ldots\) awrt 19.8	A1
	(7)

Notes

1st M1 for \(\dfrac{\pm(28 - \mu)}{\sigma} = z_1\) or \(\dfrac{\pm(24 - \mu)}{\sigma} = z_2\) where \(|z_1| \gt 1.5\) and \(|z_2| \lt 1\)
Condone \(z_2 = 0.8\)

B1 for both values 0.8416 and 1.6449 or better seen. Calc: 0.8416212.., 1.644853..

1st A1 for \(\mu = 28 - 1.64(49)\sigma\) or any correct arrangement (allow 1.64 ~1.65 inclusive)

2nd A1 for \(\mu = 24 - 0.84(16)\sigma\) or any correct arrangement (allow 0.84 or better)

2nd M1 for an attempt to solve simultaneous equations by eliminating \(\mu\) or \(\sigma\)

3rd A1 for awrt 4.98 (Condone \(\sigma = 5\) or awrt 5.0 if B0 scored)

4th A1 for awrt 19.8

SC For use of 0.84 and 1.64 giving \(\sigma = 5\) and \(\mu\) = awrt 19.8 score M1B0A1A1M1A1A1
or 0.84 and 1.65 giving \(\sigma\) = awrt 4.94 and \(\mu\) = awrt 19.9 score M1B0A1A1M1A1A1

Mark scheme (d)

Scheme	Marks
\(z = \dfrac{(12 - \text{'}19.8\ldots\text{'})}{\text{'}4.97\ldots\text{'}}\)	M1
\(\mathrm{P}(Z \lt -1.57) = 1 - \mathrm{P}(Z \lt 1.57)\)	dM1
\(1 - 0.9418 = 0.0582\) awrt 0.06	A1
	(3)
	(13 marks)

Notes

1st M1 for standardising with 12, their \(\mu\) and \(\sigma\) provided \(\sigma \gt 0\)
If \(\sigma \lt 0\) from their equations in (c) allow M1 if they use \(|\sigma|\)

2nd dM1 for \(1 - \mathrm{P}(Z \lt \text{'}1.57\text{'})\) dependent on the 1st M1 being scored i.e. leads to prob < 0.5

A1 for awrt 0.06 from correct working