A company produces batteries whose lifetimes are normally distributed with a mean of 100 hours. It is known that 90% of the batteries last at least 40 hours.
(i). Estimate the standard deviation lifetime. (3 marks)
(ii). What percentage of batteries will not last 70 hours?
To estimate the standard deviation of the battery lifetimes, we can use the concept of the cumulative standard normal distribution.
(i) Estimating the standard deviation lifetime:
1. Start by finding the z-score that corresponds to the desired cumulative probability, which in this case is 90%.
2. The z-score can be obtained from a standard normal distribution table or by using a calculator or software. The z-score corresponding to a cumulative probability of 90% is approximately 1.28.
3. Using the formula for the z-score: z = (x - μ) / σ, where z is the z-score, x is the observed value (40 hours), μ is the mean (100 hours), and σ is the standard deviation (unknown), we can rearrange the formula to solve for σ.
Rearranging the formula gives us: σ = (x - μ) / z
4. Plugging in the values, we have: σ = (40 - 100) / 1.28 = -60 / 1.28 ≈ -46.88 (rounded to the nearest hundredth). However, since standard deviation cannot be negative, we take the absolute value: σ ≈ 46.88.
Therefore, the estimated standard deviation of the battery lifetimes is approximately 46.88 hours.
(ii) Percentage of batteries that will not last 70 hours:
1. Using the z-score formula, we can calculate the z-score for the value of 70 hours, assuming a mean of 100 hours and the standard deviation obtained in part (i).
Plugging in the values: z = (70 - 100) / 46.88 ≈ -0.64.
2. Find the area under the standard normal curve from negative infinity to the calculated z-score of -0.64. This gives us the cumulative probability associated with the z-score.
3. Subtract the cumulative probability obtained from 1 to find the percentage of batteries that will not last 70 hours.
Using a standard normal distribution table or calculator, the cumulative probability associated with a z-score of -0.64 is approximately 0.2632.
Subtracting this from 1, we get: 1 - 0.2632 = 0.7368, or 73.68%.
Therefore, approximately 73.68% of the batteries will not last 70 hours.