17M.1.SL.TZ2.S_3

pestleMathematicsAASLPaper 117M· sl-4-9-normal-distribution-and-calculationssource ↗

The random variable $X$ is normally distributed with a mean of 100. The following diagram shows the normal curve for $X$ .

M17/5/MATME/SP1/ENG/TZ2/03

Let $R$ be the shaded region under the curve, to the right of 107. The area of $R$ is 0.24.

Write down ${\text{P}}(X > 107)$ .

[1]

Find ${\text{P}}(100 < X < 107)$ .

[3]

Find ${\text{P}}(93 < X < 107)$ .

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\text{P}}(X > 107) = 0.24\,\,\,\left( { = \frac{6}{{25}},{\text{ }}24\% } \right)$ A1 N1

[1 mark]

valid approach (M1)

eg $\,\,\,\,\,$ ${\text{P}}(X > 100) = 0.5,{\text{ P}}(X > 100) - {\text{P}}(X > 107)$

correct working (A1)

eg $\,\,\,\,\,$ $0.5 - 0.24,{\text{ }}0.76 - 0.5$

${\text{P}}(100 < X < 107) = 0.26\,\,\,\left( { = \frac{{13}}{{50}},{\text{ }}26\% } \right)$ A1 N2

[3 marks]

valid approach (M1)

eg $\,\,\,\,\,$ $2 \times 0.26,{\text{ }}1 - 2(0.24),{\text{ P}}(93 < X < 100) = {\text{P}}(100 < X < 107)$

${\text{P}}(93 < X < 107) = 0.52\,\,\,\left( { = \frac{{13}}{{25}},{\text{ }}52\% } \right)$ A1 N2

[2 marks]

Examiners’ report

[N/A]