Answer to Question #220490 in Statistics and Probability for Vickie

Question

Mean =12 hours ,a sample has 100 dry cells test to find the length of life , standard deviation ,= 3 hours assuming the data to be normally distributed, what percentage of battery is expected to have life more than 14 hrs , less than 18 hrs , between 10 and 14 hrs

Answer 1

$N=100 \\ \mu=12 \\ \sigma=3 \\ P(X>14) = 1 -P(X< 14) \\ = 1 -P(Z< \frac{14-12}{3}) \\ = 1 -P(Z< 0.6666) \\ = 1 - 0.7476 \\ = 0.2524 \\ = 25.24 \; \% \\ P(X<18) = P(Z< \frac{18-12}{3}) \\ = P(Z< 2) \\ = 0.9772 \\ = 97.72 \; \% \\ P(10<X<14) = P(X<14) -P(X<10) \\ = P(Z< \frac{14-12}{3}) -P(Z< \frac{10-12}{3}) \\ = P(Z< 0.6666) -P(Z< -0.6666) \\ = 0.7476 -0.2527 \\ = 0.49 \\ = 49 \%$