Question 3 (3 marks)
Quality checks are regularly conducted in a factory that produces rechargeable lithium-ion batteries. In a particular factory, the life span of a charged battery has a mean of 21 hours and standard deviation of 0.8 hours.
1
(a) A battery is chosen at random.
Find the probability that the battery has a life greater than 20.2 hours.
(b) Two batteries are chosen at random.
Use the section of the table showing the values of $P(Z< z)$ to find the probability that both batteries have a life of more than 22 hours.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$z$ & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09 \\
\hline 1.0 & 0.8413 & 0.8438 & 0.8461 & 0.8485 & 0.8508 & 0.8531 & 0.8554 & 0.8577 & 0.8599 & 0.8621 \\
\hline 1.1 & 0.8643 & 0.8665 & 0.8686 & 0.8708 & 0.8729 & 0.8749 & 0.8770 & 0.8790 & 0.8810 & 0.8830 \\
\hline 1.2 & 0.8849 & 0.8869 & 0.8888 & 0.8907 & 0.8925 & 0.8944 & 0.8962 & 0.8980 & 0.8997 & 0.9015 \\
\hline 1.3 & 0.9032 & 0.9049 & 0.9066 & 0.9082 & 0.9099 & 0.9115 & 0.9131 & 0.9147 & 0.9162 & 0.9177 \\
\hline 1.4 & 0.9192 & 0.9207 & 0.9222 & 0.9236 & 0.9251 & 0.9265 & 0.9279 & 0.9292 & 0.9306 & 0.9319 \\
\hline 1.5 & 0.9332 & 0.9345 & 0.9357 & 0.9370 & 0.9382 & 0.9394 & 0.9406 & 0.9418 & 0.9429 & 0.9441 \\
\hline 1.6 & 0.9452 & 0.9463 & 0.9474 & 0.9484 & 0.9495 & 0.9505 & 0.9515 & 0.9525 & 0.9535 & 0.9545 \\
\hline 1.7 & 0.9554 & 0.9564 & 0.9573 & 0.9582 & 0.9591 & 0.9599 & 0.9608 & 0.9616 & 0.9625 & 0.9633 \\
\hline 1.8 & 0.9641 & 0.9649 & 0.9656 & 0.9664 & 0.9671 & 0.9678 & 0.9686 & 0.9693 & 0.9699 & 0.9706 \\
\hline 1.9 & 0.9713 & 0.9719 & 0.9726 & 0.9732 & 0.9738 & 0.9744 & 0.9750 & 0.9756 & 0.9761 & 0.9767 \\
\hline
\end{tabular}
Answer
\(\boxed{\text{(b) The probability that two randomly chosen batteries have a life of more than 22 hours is approximately 0.0112.}}\)
Step 1 ：Calculate the z-scores for 20.2 hours and 22 hours using the formula: \(z = \frac{x - \text{mean}}{\text{standard deviation}}\)
Step 2 ：For 20.2 hours: \(z = \frac{20.2 - 21}{0.8} = -1.00\)
Step 3 ：For 22 hours: \(z = \frac{22 - 21}{0.8} = 1.25\)
Step 4 ：Use the Z-table to find the probabilities:
Step 5 ：For a life greater than 20.2 hours, find the complement probability: \(1 - P(Z < -1.00) = 1 - 0.1587 = 0.8413\)
Step 6 ：For two batteries with a life of more than 22 hours, find the complement probability and square it: \((1 - P(Z < 1.25))^2 = (1 - 0.8944)^2 = 0.0112\)
Step 7 ：\(\boxed{\text{(a) The probability that a randomly chosen battery has a life greater than 20.2 hours is approximately 0.8413.}}\)
Step 8 ：\(\boxed{\text{(b) The probability that two randomly chosen batteries have a life of more than 22 hours is approximately 0.0112.}}\)
