Solved: The time, ${T}$ minutes, taken to complete a crossword puzzle can be modelled by... [IB Mathematics Analysis and Approaches (AA)]

Question

SLPaper 2

The time, ${T}$ minutes, taken to complete a crossword puzzle can be modelled by a normal distribution with mean ${\mu}$ and standard deviation ${8.6}$ .

It is found that ${30\%}$ of times taken to complete the crossword puzzle are longer than ${36.8}$ minutes.

Use ${\mu}=32.29$ in the remainder of the question.

Six randomly chosen individuals complete the crossword puzzle.

Find the probability that a randomly chosen individual will take more than $30$ minutes to complete the crossword puzzle.

[2]

Verified

Solution

evidence of identifying the correct area under the normal curve**(M1)** Note: Award M1 for a clearly labelled sketch. $P(T>30)=0.605$ A1

[2****marks]

By stating and solving an appropriate equation, show, correct to two decimal places, that $\mu=32.29$ .

[4]

Verified

Solution

$T \sim N(\mu, 8.6^2)$

$P(T \leq 36.8) = 0.7$ (A1)

states a correct equation, for example, $\frac{36.8-\mu}{8.6} = 0.5244...$ A1

attempts to solve their equation (M1)

$\mu = 36.8 - 0.5244... \times 8.6 = 32.2902...$ A1

the solution to the equation is $\mu = 32.29$ , correct to two decimal places AG

[4 marks]

Having spent $25$ minutes attempting the crossword puzzle, a randomly chosen individual had not yet completed the puzzle. Find the probability that this individual will take more than $30$ minutes to complete the crossword puzzle.

[4]

Verified

Solution

recognizes that $P(T>30 \cap T \geq 25)$ is required**(M1)**

Note: Award M1 for recognizing conditional probability.

$=\frac{P(T>30 \cap T \geq 25)}{P(T \geq 25)}$ (A1)

$=\frac{P(T>30)}{P(T \geq 25)}=\frac{0.6049...}{0.8016...}$ M1

$=0.755$ A1

[4 marks]

Find the probability that at least five of them will take more than $30$ minutes to complete the crossword puzzle.

[3]

Verified

Solution

let $X$ represent the number of individuals out of the six who take more than $30$ minutes to complete the crossword puzzle $X \sim B(6, 0.6049...)$ (M1) for example, $P(X=5) + P(X=6)$ or $1 - P(X\leq 4)$ (A1) $P(X\geq 5) = 0.241$ A1

[3 marks]

Find the 86th percentile time to complete the crossword puzzle.

[2]

Verified

Solution

let $t_{0.86}$ be the 86th percentile attempts to use the inverse normal feature of a GDC to find $t_{0.86}$ (M1) $t_{0.86}=41.6$ (mins)A1

[2 marks]

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