Question

The battery life of Westlight Electric's new battery is normally distributed with population mean of 600 hours and population standard deviation of 75 hours. A simple random sample of 16 batteries is planned to be taken to check adherence to the standard of 600 hours. Suggestion: Assign the appropriate symbol to the numbers in this problem. What is the probability that the sample mean will be less than 570 hours? Round your answer to 4 decimal places. Answer = What is the probability that the sample mean will be greater than 620 hours? Round your answer to 4 decimal places. Answer = What is the probability that the sample mean will be between 560 and 600 hours? Round your answer to 4 decimal places. Answer = What is the minimum sample size needed if the desired Margin of Error of a 99% confidence interval is 30 hours? Make sure your answer is a whole number. Answer =

Answer 1

Let the life of battery be denoted by X. It is given that, X~N(600,752). We have taken a sample of 16 batteries. So, by the property of Normal distribution, the sample mean will also follow Normal distribution.

XnN(600 752 sample size n=16. So XC76 @ P(X2570 - .. we know if Xn N (4,02 > N(MI) 72=X-4 ~ N(0,1) = P(ZC-16) = 0.0548 (from 2-tabletten 6 P(X>620) 0 107 = P(z> 1.07) = 1- P(z=1.07) -1-0.8577 (From 2 -table) = 0. 1423 © P(560X2000) -P1560-600 cх-u < 600-600 - PC Ta 75/ = P(-2.13 <zco) = P(220) - P(Z2-213) = 0.5-0.0166 (From = 0.4834 2 -table 213 0

③ gg.%. confidence interval for the mean of Normal distribution is given by, od X-Zool* 5 = 4 = x+20.0, * 1 So, Margin of emir is, 20.01 * 75 - 30 > 2.58 X75 - 30 7 30 Vño 2.58 X75 n=6.45 n= (6.45² = 41.6025 ~ 42
Note: for (d) part, the minimum sample size to get Margin of error is 30 hours is 42.