MATHHX B

MATHHX B

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12.4 Deskriptorer for grupperede observationer

Vi vil tage udgangspunkt i frekvenstabellen for højden på de 25 unge:

\(\begin {array}{|c|c|c|c|c|} \hline \text {Interval} & \text {intervalmidtpunkt} (m_i) & \text {Hyp.} (h_i) & \text {Frekv. } (f_i) & \text {Sum. frekv. } (F_i)\\ \hline ]150;160] & 155 & 1 & 0{,}04 & 0{,}04\\ \hline ]160;170] & 165 & 9 & 0{,}36 & 0{,}40\\ \hline ]170;180] & 175 & 12 & 0{,}48 & 0{,}88\\ \hline ]180;190] & 185 & 3 & 0{,}12 & 1\\ \hline \end {array}\)

Er man god til at regne deskriptorer for ugrupperede observationer er det nemt at tilpasse sig til de grupperede observationerne, når man blot husker på at bruge intervalmidtpunkterne som observationer.

Typeinterval

Typeintervallet er det interval som optræder flest gange. Det er intervallet \(]170;180]\), som har hyppigheden \(12\). Er der to eller flere intervaller som deler den højeste hyppighed, er der bare flere typeintervaller.

Fraktiler (\(x_p\))

Ligesom ved ugrupperede observationer er \(p\)-fraktilen den mindste observation, som har en summeret frekvens på mindst \(p\). Vi kan dog ikke aflæse den ud fra frekvenstabellen, da vi ikke har de enkelte observationer. I stedet må vi aflæse den på sumkurven som forklaret i afsnit 12.3.

Kvartiler, kvartilsæt og median

Bortset ændringen i den måde man finder fraktiler (og dermed også kvartiler) er der ingen ændringer i forhold til ugrupperede observationer. I sidste afsnit aflæste vi kvartilsættet (på sumkurven) til at være \((166,172,177)\).

Kvartilafstand

Igen ingen forskel i forhold til ugrupperede observationer, så kvartilafstanden bliver

\[177-166=11\]

Så spændet i højde hos den midterste halvdel er på \(11\) cm.

Gennemsnit

Vi regner gennemsnittet med formlen:

\[\overline {x}=m_1\cdot f_1+m_2\cdot f_2+\cdots +m_k\cdot f_k\]

Vi ser at det er samme formel som for ugrupperede observationer, bortset fra at intervalmidtpunktet erstatter observationerne. Hvis observationerne ligger jævnt fordelt i intervallet vil midtpunktet nemlig være en god repræsentant for observationerne. For vores observationer bliver det:

\begin{align*} \overline {x} = & m_1\cdot f_1+m_2\cdot f_2+\cdots +m_k\cdot f_k \\ = & 155\cdot 0{,}04+165\cdot 0{,}36+175 \cdot 0{,}48+185\cdot 0{,}12\\ = & 171{,}8 \end{align*}

Varians

Variansen betegnes \(\sigma ^2\) regnes med formlen:

\[\sigma ^2=(m_1-\overline {x})^2\cdot f_1+(m_2-\overline {x})^2\cdot f_2+\cdots +(m_k-\overline {x})^2\cdot f_k\]

Igen ser vi, at det er samme formel som for ugrupperede observationer bortset fra at intervalmidtpunktet erstatter observationerne. For vores observationer bliver det:

\begin{align*} \sigma ^2=&(m_1-\overline {x})^2\cdot f_1+(m_2-\overline {x})^2\cdot f_2+\cdots +(m_k-\overline {x})^2\cdot f_k\\ = & (155-171{,}8)^2\cdot 0{,}04+(165-171{,}8)^2\cdot 0{,}36\\ & +(175-171{,}8)^2\cdot 0{,}48+(185-171{,}8)^2\cdot 0{,}12\\ =& 53{,}76 \end{align*}

Standardafvigelse

Standardafvigelsen er stadig kvadratroden af variansen. Så for vores observationer bliver det:

\[\sigma =\sqrt {\sigma ^2}=\sqrt {53{,}76}=7{,}33\]

Så de unge har altså en gennemsnitshøjde på \(171{,}8\), men den typiske unge har altså en højde som afviger med ca. \(7\) cm fra dette.

Øvelse 12.4.1

Vi vender tilbage til observationssættet fra øvelse 12.3.1:

\begin{align*} & 11,11,12,13,14,\\ & 15,21,21,23,24,\\ & 25,25,27,28,30,\\ & 30,32,33,39,40 \end{align*}

  • a) Lav en frekvenstabel hvor du inddeler observationerne i intervallerne \(]10,20]\), \(]20,30]\) og \(]30,40]\). Du må godt genbruge den du lavet i øvelse 12.3.1.

  • b) Bestem typeintervallet

  • c) Bestem gennemsnittet.

  • d) Bestem variansen.

  • e) Bestem standardafvigelsen fortolk den.

Løsning 12.4.1

  • a)
    \(\begin {array}{|c|c|c|c|c|} \hline \text {Interval} & \text {intervalmidtpunkt} (m_i) & \text {Hyp.} (h_i) & \text {Frekv. } (f_i) & \text {Sum. frekv. } (F_i)\\ \hline ]10;20] & 15 & 6 & 0{,}3 & 0{,}3\\ \hline ]20;30] & 25 & 10 & 0{,}5 & 0{,}8\\ \hline ]30;40] & 35 & 4 & 0{,}2 & 1\\ \hline \end {array}\)

  • b) Typeintervallet er \(]20;30]\).

  • c) Gennemsnittet er \(24\).

  • d) \(\sigma ^2=49\).

  • e) Standardafvigelsen er på \(7\), hvilket kan fortolkes som at den typiske observation afviger med \(7\) fra gennemsnittet.