Question

The lifetime of a certain type of battery is normally distributed with mean value 10 hours (a) If a pack of 4 batteries is purchased, what is the probability that the average lifetime of the (b) How many batteries must be purchased such that the probability that their average lifetime is at and standard deviation 1 hour batteries in the package is at least 9 hours? least 9.5 hours is .99?

Answer 1

lET X m a random variable, Given that mean (mu) = 10 hours standard deviation (sigma) = 1 hours (a) Lets Y is a random variable for 4 batters Now, mean for Y(mu_y) = 4 * mu = 4 * 10 mu_y = 40 and standard deviation (sigma_y) = squareroot mu_y = squareroot 40 = 6.324553203 TlideEqual 6.3245 therefore sigma_y = 6.3245 \r\n Now probability of the batteries in the package is at least a hours P(y greaterthanorequalto 9) = 1 - p(y lessthanorequalto 9) = 1 - (x - mu/sigma lessthanorequalto 9 - 40/6.3245) = 1 - (2 lessthanorequalto -31/6.3245) = 1 - (2 lessthanorequalto -4.9015732469) = 1 - 0.00001 from z scole calculation p(y greaterthanorequalto 9) = 0.99999 z - value for probability 0.99 is 2.326348 Now, x - mu/sigma = 2.326348 mu = 9.5, sigma = 1 hour x - 9.5/1 = 2.336348 x = 2.326348 + 9.5 = 11.826348 TlideEqual 12 therefore x = 12 batteries are required