Calculate probability/ percentage a consumer advocacy group tested the "on- air" lifetime of a random study sample of 162 batteries . the mean lifetime was 2.9 hours with a standard deviation of 0.4 hours. estimate the percentage of batteries with lifetimes between 2.1 and 3.7 hours. The answer is 100% ~ 1 probability ??

Note that P(2.1 < x < 3.7) can be written as P(x=3.7) - P(x=2.1). Recall that the equation for a z-score given information about a single sample is:

z = (x – x̄) / s

Since we are given that the distribution is normal and that our sample mean is 2.9 hours we let our x̄ = 2.9 hours. Since our sample standard deviation is given we can let s = .4 With this information we can find the probabilities corresponding to the respective z-score values of P(x=3.7), P(x=2.1)