A sample of 100 dry battery cells tested to find the length of life have12x hr,3 hr. Assume date is normallydistributed, what percentage of battery cells are expected to have life (i) more that 15 hr (ii) less than 6hr(iii) between 10and 14 hr

To solve this problem, we need to use the properties of the normal distribution. The mean (μ) is 12 hours and the standard deviation (σ) is 3 hours.

(i) To find the percentage of battery cells that last more than 15 hours, we first need to calculate the z-score for 15 hours. The z-score is calculated as (X - μ) / σ. So, for 15 hours, the z-score is (15 - 12) / 3 = 1.

We then look up this z-score in a standard normal distribution table or use a calculator with a normal distribution function to find the area to the right of this z-score (since we want the percentage that last more than 15 hours). The area to the right of z = 1 is approximately 0.1587, or 15.87%.

(ii) To find the percentage of battery cells that last less than 6 hours, we calculate the z-score for 6 hours. The z-score is (6 - 12) / 3 = -2.

We then find the area to the left of this z-score (since we want the percentage that last less than 6 hours). The area to the left of z = -2 is approximately 0.0228, or 2.28%.

(iii) To find the percentage of battery cells that last between 10 and 14 hours, we calculate the z-scores for 10 and 14 hours. The z-score for 10 hours is (10 - 12) / 3 = -0.67 and the z-score for 14 hours is (14 - 12) / 3 = 0.67.

We then find the area between these two z-scores. The area between z = -0.67 and z = 0.67 is approximately 0.4967 - (-0.4967) = 0.9934, or 99.34%.

So, we expect approximately 15.87% of battery cells to last more than 15 hours, 2.28% to last less than 6 hours, and 99.34% to last between 10 and 14 hours.