Answer to Question #115631 in Math for Wale

Question

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 between ________ km and ________ km.
b) 95% of all tires will have a life between ________ km and ________ km.
c) What percent of the tires will have a life that exceeds 26,000 km?
d) If a company purchased 2000 tires, how many tires would you expect to last more than 28 000 km?

Answer 1

Let the random variable X denotes the life of a tire and it is normally distributed: $X\sim N(\mu, \sigma^2).$

Given $\mu=30000\ km, \sigma=2000 \ km.$

$$X\sim N(30000,2000^2)$$

(a) According to Empirical rule, the area covered between mean minus one standard deviation and mean plus one standard deviation of a normal distribution is 68%.

$$P(\mu-\sigma<X<\mu+\sigma)=68\%$$

$$P(30000-2000<X<30000+2000)=68\%$$

$$P(28000<X<32000)=68\%$$

68% of all tires will have a life between 28000 km and 32000 km. 

(b) According to Empirical rule, the area covered between mean minus two standard deviations and mean plus two standard deviations of a normal distribution is 95%.

$$P(\mu-2\sigma<X<\mu+2\sigma)=95\%$$

$$P(30000-2(2000)<X<30000+2(2000))=95\%$$

$$P(26000<X<34000)=95\%$$

95% of all tires will have a life between 26000 km and 34000 km. 

(c)

$$P(X>26000)=1-P(X\leq26000)=$$

$$=1-{1-P(26000<X<34000)\over 2}=1-{1-0.95\over 2}=0.975$$

$97.5\%$

(d)

$$P(X>28000)=1-P(X\leq28000)=$$

$$=1-{1-P(28000<X<32000)\over 2}=1-{1-0.68\over 2}=0.84$$

$$0.84(2000)=1680$$

1680 tires.