Question

DFor each question, construct a normal distribution curve and label the horizontal axis. Then answer eachquestion.10+11. The mean life of a tire is 30 000 km. The standard deviation is 2000 km.a) 68% of all tires will have a life betweenkm andkm.km.b) 95% of all tires will have a life betweenkm andc) What percent of the tires will have a life that exceeds 26 000 km?ad) If a company purchased 2000 tires, how many tires would you expect to last more than 28 000 km?

Answer 1

Solution:

The normal distribution curve is as shown;

Given:

$\begin{gathered} \mu=30,000km \\ \sigma=2000km \end{gathered}$

Question a:

If 68% of all tires will have a life when the Z-score is between -1 and 1.

Hence,

when Z = -1

$\begin{gathered} \\ Z=\frac{x-\mu}{\sigma} \\ -1=\frac{x-30000}{2000} \\ Cross\text{ multiplying;} \\ -1(2000)=x-30000 \\ -2000=x-30000 \\ -2000+30000=x \\ x=30000-2000 \\ x=28000km \end{gathered}$

when Z = 1

$\begin{gathered} \\ Z=\frac{x-\mu}{\sigma} \\ 1=\frac{x-30000}{2000} \\ Cross\text{ multiplying;} \\ 1(2000)=x-30000 \\ 2000=x-30000 \\ 2000+30000=x \\ x=30000+2000 \\ x=32000km \end{gathered}$

Therefore, 68% of all tires will have a life between 28,000km and 32,000km.

Question b:

If 95% of all tires will have a life when the Z-score is between -2 and 2.

Hence,

when Z = -2

$\begin{gathered} \\ Z=\frac{x-\mu}{\sigma} \\ -2=\frac{x-30000}{2000} \\ Cross\text{ multiplying;} \\ -2(2000)=x-30000 \\ -4000=x-30000 \\ -4000+30000=x \\ x=30000-4000 \\ x=26000km \end{gathered}$

when Z = 2

$\begin{gathered} \\ Z=\frac{x-\mu}{\sigma} \\ 2=\frac{x-30000}{2000} \\ Cross\text{ multiplying;} \\ 2(2000)=x-30000 \\ 4000=x-30000 \\ 4000+30000=x \\ x=30000+4000 \\ x=34000km \end{gathered}$

Therefore, 95% of all tires will have a life between 26,000km and 34,000km.

Question c:

The percent of tires that will have a life that exceeds 26,000km is;

$\begin{gathered} Z-score\text{ when x = 26000} \\ Z=\frac{x-\mu}{\sigma} \\ Z=\frac{26000-30000}{2000} \\ Z=-\frac{4000}{2000} \\ Z=-2 \end{gathered}$

From Z-score tables,

$\begin{gathered} P(x>Z)=0.97725 \\ As\text{ a percent,} \\ 0.97725\times100=97.725 \\ \approx97.73 \end{gathered}$

Therefore, the percent of tires that will have a life that exceeds 26,000 is 97.73%

Question d:

The probability of tires that will have a life that will last more than 28,000km is;

$\begin{gathered} Z-score\text{ when x = 28000} \\ Z=\frac{x-\mu}{\sigma} \\ Z=\frac{28000-30000}{2000} \\ Z=-\frac{2000}{2000} \\ Z=-1 \end{gathered}$

From Z-score tables,

$P(x>Z)=0.84134$

Hence, the number of tires that would last more than 28000km if the company purchased 2000 tires will be;

$\begin{gathered} E=n.P(x) \\ E=2000\times0.84134 \\ E=1682.68 \\ E\approx1683\text{ tires to the nearest whole number.} \end{gathered}$

Therefore, the number of tires that would last more than 28,000km if the company purchased 2000 tires will be approximately 1,683 tires to the nearest whole number.