Statistical inference — Question 18

QCAA 2024, Paper 2 · 6 marks

Q18 · 2024 · Technology-activeComplex unfamiliar6 marks

QUESTION 18 (6 marks)

A random variable

X

is normally distributed, with a known mean

\mu

and standard deviation

\sigma

. In figure 1, the shaded region between 4 and

\mu

represents 30% of the distribution of

X

Consider the distribution of

\bar X

based on repeated random sampling of

X

using a certain sample size. In figure 2, the shaded region between

\mu

and 6 represents 30% of the distribution of

\bar X

Given

P(4\le\bar X\le6)\approx0.77

, determine

P(4\le X\le6)

Figure 1: normal distribution with the region from 4 to the mean mu shaded.

Figure 2: sampling distribution of the sample mean with the region from mu to 6 shaded.

Question linkOriginal paper

WORKED SOLUTION

QCAA guide · typeset solution

6 marks

ANSWER

P(4\le X\le6)\approx0.45

Worked solution

From Figure 1, using a GDC,

\mu-4=0.8416\sigma.\tag{1}

Let the sample size be

n

. From Figure 2,

6-\mu=\frac{0.8416\sigma}{\sqrt n}.\tag{2}

Given

P(4\le\bar X\le6)\approx0.77

and

P(\mu\le\bar X\le6)=0.30

P(4\le\bar X\le\mu)\approx0.47,

\mu-4=\frac{1.8808\sigma}{\sqrt n}.\tag{3}

Dividing (2) by (3),

\frac{6-\mu}{\mu-4}=\frac{0.8416}{1.8808},

which gives

\mu\approx5.382

. From (1),

\sigma\approx1.642

. Therefore, using a GDC,

P(4\le X\le6)\approx0.45.

correctly determines an equation from Figure 1 in terms of

\mu

and

\sigma

[1 mark]

correctly determines an equation from Figure 2 in terms of

\mu

\sigma

and

n

[1 mark]

correctly determines an equation using

P(4\le\bar X\le6)\approx0.77

in terms of

\mu

\sigma

and

n

[1 mark]

determines

\mu

[1 mark]

determines

\sigma

[1 mark]

determines the required probability

[1 mark]

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Statistical inference — Question 18

QUESTION 18 (6 marks)

QCAA guide · typeset solution

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