Solved: John rings a church bell 120 times. The time interval, T i ... [IB Mathematics Applications & Interpretation (Math AI)]

Question

HLPaper 3

John rings a church bell 120 times. The time interval, ${T_i}$ , between two successive rings is a random variable with mean of 2 seconds and variance of $\frac{1}{9}{\text{ second}}{{\text{s}}^2}$ .

Each time interval, ${T_i}$ , is independent of the other time intervals. Let $X = \sum\limits_{i = 1}^{119} {{T_i}}$ be the total time between the first ring and the last ring.

The church vicar subsequently becomes suspicious that John has stopped coming to ring the bell and that he is letting his friend Ray do it. When Ray rings the bell the time interval, ${T_i}$ has a mean of 2 seconds and variance of $\frac{1}{{25}}{\text{ second}}{{\text{s}}^2}$ .

The church vicar makes the following hypotheses:

${H_0}$ : Ray is ringing the bell; ${H_1}$ : John is ringing the bell.

He records four values of $X$ . He decides on the following decision rule:

If $236 \leqslant X \leqslant 240$ for all four values of $X$ he accepts ${H_0}$ , otherwise he accepts ${H_1}$ .

Find

(i) ${\text{E}}(X)$ ;

(ii) ${\text{Var}}(X)$ .

[3]

Verified

Solution

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

(i) ${\text{mean}} = 119 \times 2 = 238$ A1

(ii) ${\text{variance}} = 119 \times \frac{1}{9} = \frac{{119}}{9}{\text{ }}( = 13.2)$ (M1)A1

Note: If 120 is used instead of 119 award A0(M1)A0 for part (a) and apply follow through for parts (b)-(d). (b) is unaffected and in (c) the interval becomes $(234,{\text{ }}246)$ . In (d) the first 2 A1 marks are for $0.3633 \ldots$ and $0.0174 \ldots$ so the final answer will round to 0.017.

[3 marks]

Explain why a normal distribution can be used to give an approximate model for $X$ .

[2]

Verified

Solution

justified by the Central Limit Theorem R1

since $n$ is large A1

Note: Accept $n > 30$ .

[2 marks]

Use this model to find the values of $A$ and $B$ such that ${\text{P}}(A < X < B) = 0.9$ , where $A$ and $B$ are symmetrical about the mean of $X$ .

[7]

Verified

Solution

$X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)$

$Z = \frac{{X - 238}}{{\frac{{\sqrt {119} }}{3}}} \sim N(0,{\text{ }}1)$ (M1)(A1)

${\text{P}}(Z < q) = 0.95 \Rightarrow q = 1.644 \ldots$ (A1)

so ${\text{P}}( - 1.644 \ldots < Z < 1.644 \ldots ) = 0.9$ (R1)

${\text{P}}( - 1.644 \ldots < \frac{{X - 238}}{{\frac{{\sqrt {119} }}{3}}} < 1.644 \ldots ) = 0.9$ (M1)

interval is $232 < X < 244{\text{ }}({\text{3sf}}){\text{ }}(A = 232,{\text{ }}B = 244)$ A1A1

Notes: Accept the use of inverse normal applied to the distribution of $X$ .

Alternative is to use the GDC to find a pretend $Z$ confidence interval for a mean and then convert by multiplying by 119.

Either $A$ or $B$ correct implies the five implied marks.

Accept any numbers that round to these 3sf numbers.

[7 marks]

Calculate the probability that he makes a Type II error.

[5]

Verified

Solution

under ${{\text{H}}_1},{\text{ }}X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)$ (M1)

${\text{P}}(236 \leqslant X \leqslant 240) = 0.41769 \ldots$ (A1)

probability that all 4 values of $X$ lie in this interval is

${(0.41769 \ldots )^4} = 0.030439 \ldots$ (M1)(A1)

so probability of a Type II error is 0.0304 (3sf) A1

Note: Accept any answer that rounds to 0.030.

[5 marks]

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