Several states, including California, have passed legislation requiring automakers to sell a certain percentage of zero-emissions cars within their borders. One current alternative is battery-powered cars. The major problem with battery-operated cars is the limited time they can be driven before the batteries must be recharged. Longlife Battery, a start-up company, has developed a battery pack it claims will power a car at a sustained speed of 45 miles per hour for an average of 8 hours. But of course there will be variations: Some battery packs will last longer and some shorter than 8 hours. Current data indicate that the standard deviation of battery operation time before a charge is needed is 0.4 hours. Data show a normal distribution of uptime on these battery packs. Automakers are concerned that batteries may run short. For example, drivers might find an “8-hour” battery that lasts 7.5 hours or less unacceptable.

a) What are the chances of this happening with the Longlife battery pack?

b) The automakers will accept no more than a 2% chance. Longlife Battery asks the question, what would the mean uptime have to be to meet the 2% requirement?

http://davidmlane.com/hyperstat/z_table.html

a) 0.1056

b) 6.676

To answer these questions, we need to use the normal distribution and standard deviation given in the problem.

a) First, let's find the probability that a battery pack will last 7.5 hours or less. We can use the standard deviation and the mean (8 hours) to calculate this probability.

Using the properties of the normal distribution, we can calculate the z-score for the value 7.5 hours:

z = (x - mean) / standard deviation
z = (7.5 - 8) / 0.4
z = -0.5 / 0.4
z = -1.25

Next, we need to find the probability associated with this z-score. This probability can be found using a standard normal distribution table or a statistical calculator. From the table, we find that the area to the left of -1.25 is approximately 0.1056.

Therefore, the chances of a Longlife battery pack lasting 7.5 hours or less is approximately 10.56%.

b) To meet the 2% requirement, we need to find the mean uptime that corresponds to a 2% probability or an area of 0.02 in the tail of the distribution.

Using the standard normal distribution table, we can find the z-score that corresponds to an area of 0.02. The z-score is approximately -2.05.

Then, we can solve for the mean (x) using the z-score formula:

z = (x - mean) / standard deviation
-2.05 = (x - 8) / 0.4

Solving for x, we get:

-2.05 * 0.4 = x - 8
-0.82 + 8 = x
x = 7.18

Therefore, the mean uptime would have to be approximately 7.18 hours to meet the 2% requirement.