The average life of a certian batary is 56.2 Hrs wth a standar deviation of 4.3hrs. Assume that it is normaly distribued. Bataries lower than 10% do not meet the minium requirements. What is the minimum reqiurement for a battery? Solution Normal distribution with = 56.2 and = 4.3 Using the z-table, find the z-score that has a p-value close to .1000 -----> z = -1.28 Now, use the z-formula to solve for x: z = (x - ) / ----> -1.28 = (x - 56.2) / 4.3 -----> x = 50.696 hours.

1. The average life of a certian batary is 56.2 Hrs wth a standar deviation of 4.3hrs. Assume that it
is normaly distribued. Bataries lower than 10% do not meet the minium requirements.
What is the minimum reqiurement for a battery?
Solution
Normal distribution with = 56.2 and = 4.3
Using the z-table, find the z-score that has a p-value close to .1000 -----> z = -1.28
Now, use the z-formula to solve for x:
z = (x - ) / ----> -1.28 = (x - 56.2) / 4.3 -----> x = 50.696 hours