S1 January 2007 Q7

EdexcelOld spec10 marksNormal Distribution

7. The measure of intelligence, IQ, of a group of students is assumed to be Normally distributed with mean 100 and standard deviation 15.

(a) Find the probability that a student selected at random has an IQ less than 91. (4)

The probability that a randomly selected student has an IQ of at least \(100 + k\) is 0.2090.

(b) Find, to the nearest integer, the value of \(k\). (6)

Mark scheme (a)

Scheme	Marks
\(\mathrm{P}(X \lt 91) = \mathrm{P}\left(Z \lt \dfrac{91 - 100}{15}\right)\)	M1
\(= \mathrm{P}(Z \lt -0.6)\)	A1
\(= 1 - 0.7257\)	M1
\(= 0.2743\)	A1
	(4)

Notes

M1 Attempt standardisation

A1 awrt 0.274

1^st M1 for attempting standardisation. \(\pm\dfrac{(91 - \mu)}{\sigma \text{ or } \sigma^2}\). Can use of 109 instead of 91. Use of 90.5 etc is M0

1^st A1 for −0.6 (or +0.6 if using 109)

2^nd M1 for 1 – probability from tables. Probability should be > 0.5)

Mark scheme (b)

Scheme	Marks
\(1 - 0.2090 = 0.7910\)	B1
\(\mathrm{P}(X \gt 100 + k) = 0.2090\) or \(\mathrm{P}(X \lt 100 + k) = 0.7910\)	M1
Use of tables to get \(z = 0.81\)	B1
\(\dfrac{100 + k - 100}{15}, = 0.81\)	M1, A1ft
\(\underline{k = 12}\)	A1 cao
	(6)
	(10 marks)

Notes

B1 0.791

M1 (May be implied)

M1, A1ft (ft their \(z\) = 0.81, but must be \(z\) not prob.)

1^st B1 for 0.791 seen or implied.

1^st M1 for a correct probability statement, but must use \(X\) or \(Z\) correctly. Shown on diagram is OK

2^nd B1 for awrt 0.81 seen (or implied by correct answer - see below) (Calculator gives 0.80989…)

2^nd M1 for attempting to standardise e.g. \(\dfrac{100 + k - 100}{15}\) or \(\dfrac{k}{15}\)

\(\tfrac{X - 100}{15}\) scores 2^nd M0 until the \(100 + k\) is substituted to give \(k\), but may imply 1^st M1 if \(k = 112.15\) seen

1^st A1ft for correct equation for \(k\) (as written or better). Can be implied by \(k = 12.15\) (or better)

2^nd A1 for \(k = 12\) only.

Answers only
\(k = 112\) or 112.15 or better scores 3/6 (on EPEN give first 3 marks)
\(k = 12.15\) or better (calculator gives 12.148438…) scores 5/6 (i.e loses last A1 only)
\(k = 12\) (no incorrect working seen) scores 6/6

NB Using 0.7910 instead of 0.81 gives 11.865 which might be rounded to 12. This should score no more than B1M1B0M1A0A0.