20N.2.SL.TZ0.S_9

pestleMathematicsAISLPaper 220N· sl-4-3-mean-median-mode-mean-of-grouped-data-standard-deviation-quartiles-iqrsource ↗

Fiona walks from her house to a bus stop where she gets a bus to school. Her time, $W$ minutes, to walk to the bus stop is normally distributed with $W ~ N (12, 3^{2})$ .

Fiona always leaves her house at 07:15. The first bus that she can get departs at 07:30.

The length of time, $B$ minutes, of the bus journey to Fiona’s school is normally distributed with $B ~ N (50, σ^{2})$ . The probability that the bus journey takes less than $60$ minutes is $0.941$ .

If Fiona misses the first bus, there is a second bus which departs at 07:45. She must arrive at school by 08:30 to be on time. Fiona will not arrive on time if she misses both buses. The variables $W$ and $B$ are independent.

Find the probability that it will take Fiona between $15$ minutes and $30$ minutes to walk to the bus stop.

[2]

Find $σ$ .

[3]

Find the probability that the bus journey takes less than $45$ minutes.

[2]

Find the probability that Fiona will arrive on time.

[5]

This year, Fiona will go to school on $183$ days.

Calculate the number of days Fiona is expected to arrive on time.

[2]

Markscheme / solution

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

$0.158655$

$P (15 < W < 30) = 0.159$ A2 N2

[2 marks]

finding standardized value for $60$ (A1)

eg $z = 1.56322$

correct substitution using their $z$ -value (A1)

eg $\frac{60 - 50}{σ} = 1.56322, \frac{60 - 50}{1.56322} = σ$

$6.39703$

$σ = 6.40$ A1 N3

[3 marks]

$0.217221$

$P (B < 45) = 0 .217$ A2 N2

[2 marks]

valid attempt to find one possible way of being on time (do not penalize incorrect use of strict inequality signs) (M1)

eg $W \leq 15$ and $B < 60$ , $15 < W \leq 30$ and $B < 45$

correct calculation for $P (W \leq 15 and B < 60)$ (seen anywhere) (A1)

eg $0.841 \times 0.941, 0.7917$

correct calculation for $P (15 < W \leq 30 and B < 45)$ (seen anywhere) (A1)

eg $0.159 \times 0.217, 0.03446$

correct working (A1)

eg $0.841 \times 0.941 + 0.159 \times 0.217, 0.7917 + 0.03446$

$0.826168$

$P$ (on time) $= 0.826$ A1 N2

[5 marks]

recognizing binomial with $n = 183, p = 0.826168$ (M1)

eg $X ~ B (183, 0.826)$

$151.188$ ( $151.158$ from $3 sf$ )

$151$ A1 N2

[2 marks]

Examiners’ report

[N/A]