Quick question about normal distributions

  • #1

Homework Statement



You purchase a chainsaw, and can buy one of two types of batteries to power it, namely Duxcell and Infinitycell. Batteries of each type have lifetimes before recharge that can be assumed independent and Normally distributed. The mean and standard deviation of the lifetimes of the Duxcell batteries are 10 and 2 minutes respectively, the mean and standard deviation for the Infinitycell batteries are 19 and 2 minutes respectively.



Part a) What is the probability that a Duxcell battery will last longer than an Infinitycell battery? Give your answer to two decimal places.

Part b) What is the probability that an Infinitycell battery will last more than twice as long as a Duxcell battery? Give your answer to two decimal places.

Part c) You are going to cut down a large tree and do not want to break off from the job to recharge your chainsaw battery. You buy two Duxcell batteries, and plan to use one until it runs out of power, after which you immediately replace it with the second battery. How long (in minutes) can the job last so that with probability 0.75 you can complete the job using the two Duxcell batteries in sequence?

Provide your answer to 1 decimal place.

Homework Equations


[/B]
z=(x-mean)/StdDev

The Attempt at a Solution



Just wanted to verify something quickly, for part b I had to use sqrt(20) as my standard deviation. I got this because I added the standard deviation of each battery:
sqrt(2^2 (for the infinitycell) + 4^2 (for the duxcell))

Meanwhile for part c), the total standard deviation I used was sqrt(2^2 +2^2), since it is 2 for each (duxcell).

My question is, in part b I could simply double the duxcell battery standard deviation, while in c I had to treat each battery standard deviation seperately, and first convert to variance before i could add them. Is this due to the difference between double the duration (part b) and two batteries in sequence (part c)?

Thanks
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



You purchase a chainsaw, and can buy one of two types of batteries to power it, namely Duxcell and Infinitycell. Batteries of each type have lifetimes before recharge that can be assumed independent and Normally distributed. The mean and standard deviation of the lifetimes of the Duxcell batteries are 10 and 2 minutes respectively, the mean and standard deviation for the Infinitycell batteries are 19 and 2 minutes respectively.



Part a) What is the probability that a Duxcell battery will last longer than an Infinitycell battery? Give your answer to two decimal places.

Part b) What is the probability that an Infinitycell battery will last more than twice as long as a Duxcell battery? Give your answer to two decimal places.

Part c) You are going to cut down a large tree and do not want to break off from the job to recharge your chainsaw battery. You buy two Duxcell batteries, and plan to use one until it runs out of power, after which you immediately replace it with the second battery. How long (in minutes) can the job last so that with probability 0.75 you can complete the job using the two Duxcell batteries in sequence?

Provide your answer to 1 decimal place.

Homework Equations


[/B]
z=(x-mean)/StdDev

The Attempt at a Solution



Just wanted to verify something quickly, for part b I had to use sqrt(20) as my standard deviation. I got this because I added the standard deviation of each battery:
sqrt(2^2 (for the infinitycell) + 4^2 (for the duxcell))

Meanwhile for part c), the total standard deviation I used was sqrt(2^2 +2^2), since it is 2 for each (duxcell).

My question is, in part b I could simply double the duxcell battery standard deviation, while in c I had to treat each battery standard deviation seperately, and first convert to variance before i could add them. Is this due to the difference between double the duration (part b) and two batteries in sequence (part c)?

Thanks

If ##X_d## is the Duxcell lifetime and ##X_i## is the Infinitycell lifetime, in part (b) you want to know ##P(X_i > 2 X_d)##, which is the same as ##P( X_i - 2X_d > 0)##. What can you say about the random variable ##Y = X_i - 2X_d##?
 

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