Question

Suppose that the tires of a certain brand have mean life 25,430 km and standard

deviation 1,800 km.

(a) If a tire is randomly selected, what is the probability that its life is longer than 25,000 km?

(b) Determine the minimal sample size needed so that the probability of the sample mean life longer than 25,000 km is at least 0.97. [15 marks] 

Answer 1

(a)

$$P(X>25000)=1-P(X\leq25000)$$

$$=1-P(Z\leq\dfrac{25000-25430}{1800})$$

$$\approx1-P(Z\leq-0.2389)\approx0.5944$$

(b)

$$P(\bar{X}>25000)=1-P(\bar{X}\leq25000)$$

$$=1-P(Z\leq\dfrac{25000-25430}{1800/\sqrt{n}})\geq0.97$$

$$P(Z\leq\dfrac{25000-25430}{1800/\sqrt{n}})\leq0.03$$

$$\dfrac{25000-25430}{1800/\sqrt{n}}\leq-1.880794$$

$$n\geq(\dfrac{1800(1.880794)}{25430-25000})^2$$

$$n\geq62$$

The minimal sample size is 62  tires.