Home / IBDP Maths AI: Topic: AHL 4.15: A linear combination of n independent: IB style Questions HL Paper 2

IBDP Maths AI: Topic: AHL 4.15: A linear combination of n independent: IB style Questions HL Paper 2

Question

A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. However the weights vary with a standard deviation of \(0.54\) kg. A random sample of \(24\) bags is taken to check that the mean weight is \(28\) kg.

A.a.Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} {\rm{  .}}\][3]

A.b.(i)      Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .

(ii)     Hence find an approximate value for \({\rm{P}}( – 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .[6]

B.a.State and justify an appropriate test procedure giving the null and alternate hypotheses.[5]

B.b.What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?[7]

B.c.If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?[2]

▶️Answer/Explanation

Markscheme

\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)

\({{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = 1 + \left( { – \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} +  \ldots \)    M1A1

\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} – \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\)     A1

[3 marks]

A.a.

(i)     \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 – \frac{{{t^2}}}{2}}  + \frac{{{t^4}}}{8} – \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\)     M1

\( = \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\)     A1

\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} –  \ldots } \right)\)     R1A1

(ii)     \({\rm{P}}( – 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 – \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} – \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} –  \ldots } \right)\)     M1

\( = 0.38292 = 0.383\)     A1

[6 marks]

A.b.

this is a two tailed test of the sample mean \(\overline x \)

we use the central limit theorem to justify assuming that     R1

\(\overline X  \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1A1

\({{\rm{H}}_0}:\mu  = 28\)     A1

\({{\rm{H}}_1}:\mu  \ne 28\)     A1

[5 marks]

B.a.

since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\)     (M1)A1

and (\(\overline x  \le 28 – 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x  \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) )     (M1)(A1)(A1)

\(\overline x  \le 27.7676\) or \(\overline x  \ge 28.2324\)

so \(\overline x  \le 27.8\) or \(\overline x  \ge 28.2\)     A1A1

[7 marks]

B.b.

if \(\mu  = 28.1\)

\(\overline X  \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1

\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x  < 28.2324)\)

\( = 0.884\)     A1

Note: Depending on the degree of accuracy used for the critical region the answer  for part (c) can be anywhere from \(0.8146\) to \(0.879\).

[2 marks]

B.c.

Question

The random variable \(X\) has probability density function given by

\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{x{{\text{e}}^{ – x}},}&{{\text{for }}x \geqslant 0,} \\
{0,}&{{\text{otherwise}}}
\end{array}} \right..\]

A sample of size 50 is taken from the distribution of \(X\).

a.Use l’Hôpital’s rule to show that \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{{\text{e}}^x}}} = 0\).[3]

b.(i)     Find \({\text{E}}({X^2})\).

(ii)     Show that \({\text{Var}}(X) = 2\).[10]

c.State the central limit theorem.[2]

d.Find the probability that the sample mean is less than 2.3.[2]

▶️Answer/Explanation

Markscheme

attempt to apply l’Hôpital’s rule     M1

\(\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{{{\text{e}}^x}}}\)    A1

then \(\mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{{{\text{e}}^x}}}\)

then \(\mathop {\lim }\limits_{x \to \infty } \frac{6}{{{{\text{e}}^x}}}\)     A1

\( = 0\)    AG

[3 marks]

a.

(i)     \({\text{E}}({X^2}) = \mathop {\lim }\limits_{R \to \infty } \int\limits_0^R {{x^3}{{\text{e}}^{ – x}}{\text{d}}x} \)       M1

attempt at integration by parts      M1

the integral \( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + \mathop \smallint \limits_0^R 3{x^2}{{\text{e}}^{ – x}}{\text{d}}x\)       A1A1

\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + \int\limits_0^R {6x{{\text{e}}^{ – x}}{\text{d}}x} \)     M1

\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + [ – 6x{{\text{e}}^{ – x}}]_0^R + \int\limits_0^R {6{{\text{e}}^{ – x}}{\text{d}}x} \)     A1

\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + [ – 6x{{\text{e}}^{ – x}}]_0^R + [ – 6{{\text{e}}^{ – x}}]_0^R\)      A1

\( = 6\) when \(R \to \infty \)       R1

(ii)     \({\text{E}}(X) = 2\)      A1

\({\text{Var}}(X) = {\text{E}}({X^2}) – {\left( {{\text{E}}(X)} \right)^2} = 6 – {2^2}\)    M1

\( = 2\)    AG

[10 marks]

b.

if a random sample of size \(n\) is taken from any distribution \(X\), with \({\text{E}}(X) = \mu \) and \({\text{Var}}(X) = {\sigma ^2}\), then, for large n,     A1

the sample mean \(\bar X\) has approximate distribution \({\text{N}}\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}} \right)\)     A1

[2 marks]

c.

\(\bar X \sim {\text{N}}\left( {2,{\text{ }}\frac{2}{{50}} = {{(0.2)}^2}} \right)\)    (A1)

\({\text{P}}(\bar X < 2.3) = \left( {{\text{P}}(Z < 1.5)} \right) = 0.933\)    A1

[2 marks]

d.

Question

In a large population of sheep, their weights are normally distributed with mean \(\mu \) kg and standard deviation \(\sigma \) kg. A random sample of \(100\) sheep is taken from the population.

The mean weight of the sample is \(\bar X\) kg.

a.State the distribution of \(\bar X\) , giving its mean and standard deviation.[2]

b.The sample values are summarized as \(\sum {x = 3782} \) and \(\sum {{x^2} = 155341} \) where \(x\) kg is the weight of a sheep.

(i)     Find unbiased estimates for \(\mu \) and \({\sigma ^2}\).

(ii)     Find a \(95\%\) confidence interval for \(\mu \).[6]

c.Test, at the \(1\%\) level of significance, the null hypothesis \(\mu  = 35\) against the alternative hypothesis that \(\mu  > 35\).[5]

▶️Answer/Explanation

Markscheme

\(\bar X \sim N\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{{100}}} \right)\)     A1A1

Note: Award A1 for \(N\), A1 for the parameters.

a.

(i)     \(\bar x = \frac{{\sum x }}{n} = \frac{{3782}}{{100}} = 37.8\)     A1

\(s_{n – 1}^2 = \frac{{155341}}{{99}} – \frac{{{{3782}^2}}}{{9900}} = 124\)     M1A1

(ii)     \(95\% CI = 37.82 \pm 1.98\sqrt {\frac{{124.3006}}{{100}}} \)     (M1)(A1)

\( = (35.6,{\text{ }}40.0)\)     A1

b.

METHOD 1

one tailed t-test     A1

testing \(37.82\)     A1

\(99\) degrees of freedom

reject if \(t > 2.36\)     A1

t-value being tested is \(2.5294\)     A1

since \(2.5294 > 2.36\) we reject the null hypothesis and accept the alternative hypothesis     R1

METHOD 2

one tailed t-test     (A1)

\(p = 0.00650\)     A3

since \(p{\text{ – value}} < 0.01\) we reject the null hypothesis and accept the alternative hypothesis     R1

c.
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