TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2022W/HW03.3

Tire company

A tire manufacturer believes that the tread life of its snow tires can be distributed by a Normal model with a mean of 32 000 miles and a standard deviation of 2 500 miles.

Use the values of the cumulative distribution function $\Phi (z)=P(Z\leq z)$ of the standard normal distribution $Z$ given in Table 1 to answer the following questions.

(a) Approximately what fraction of these tires can be expected to last between 30000 and 35000 miles?

(b) Calculate the interquartile range of this distribution.

Recall, the interquartile range is the diference between the upper $x_{0.75}$ and the lower quartile $x_{0.25}$, i.e., ${\text{IQR}}=x_{0.75}-x_{0.25}$.

(c) In a marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer does not want to take too big risk. If the dealer is willing to give refunds to no more than one of every 15 customers, for what milage can he guarantee these tires to last?

Dieses Beispiel hat einen unbekannten Lösungsstatus. Bitte editiere diese Seite und schreibe den dir bekannten Status ins Beispiel. Die möglichen Werte sind hier: Vorlage:Beispiel dokumentiert. Führe folgende Änderung durch:

{{Beispiel|1=
Angabetext
}}

oder

{{Beispiel|
Angabetext
}}

zu (im Falle einer korrekten, unverifizierten Lösung "solved". Auch möglich "unsolved", "wrong", "verified_by_tutor". Alle möglichen Werte sind hier: Vorlage:Beispiel dokumentiert.)

{{Beispiel|status=solved|1=
Angabetext
}}

Hilfreiche Tipps

Ähnliches Beispiel: Übungen 2019W - HW 3.6

Lösungsvorschlag von Simplex

(a)

$\mu =32000,\sigma =2500:X\sim {\mathcal {N}}(\mu ,\sigma ^{2})$ Standardized: $z={\frac {x-\mu }{\sigma }}\sim {\mathcal {N}}(0,1)$

$${\begin{aligned}P(30000\leq X\leq 35000)&=P(35000)-P(30000)\\P(30000)&={\frac {30000-32000}{2500}}=-0.8\Rightarrow 0.2119{\text{ (from table)}}\\P(35000)&={\frac {35000-32000}{2500}}=1.2\Rightarrow 0.8849{\text{ (from table)}}\\\end{aligned}}$$

$$P(30000\leq X\leq 35000)=0.673$$

(b)

$$0.25\approx -0.67\Rightarrow {\frac {x-32000}{2500}}=-0.67\Rightarrow x_{0.25}=30325$$

$$0.75\approx 0.67\Rightarrow {\frac {x-32000}{2500}}=0.67\Rightarrow x_{0.75}=30325$$

$${\text{IQR}}:x_{0.75}-x_{0.25}=3350$$

(c)

$${\frac {1}{15}}\approx 0.0{\overline {6}}\Rightarrow -1.5{\text{ (from table)}}$$

$$-1.5={\frac {X-32000}{2500}}\Rightarrow X=28250$$

--Simplex 19:51, 2. Feb. 2023 (CET)

Deine Lösung...