You purchase a chainsaw, and can buy one of two types of batteries to power it,

Question

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 18 and 3 minutes respectively. What is the probability that a Duxcell battery will last longer than an Infinity cell battery? Give your answer to two decimal places. 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. 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.

Explanation / Answer

Mean and standard deviation of duxcell = 10 and 2

Mean and sd of infinity cell = 18 and 3

distribution of infinity cell minus duxcell cell

mean difference = 18-10 = 8

standard deviation = sqrt(4+9) = sqrt(13) = 3.606

Probability that dux cell battery will lasts longer than an infinity cell is = P(duxcell mean-infinity cell >0)

= P[z> (0+8)/3.606]= P(z>2.2188) = 1- P(Z<=2.2188) = 0.01

b) Probability that infinity cell battery will last more than twice as long as Dux cell battery is

Let X is the mean of infinity cell and Y is Dux cell

then mean X-2Y = 18-20 =-2

variance (X-2Y ) =9+ 4*8 = 41; standard deviation =sqrt(41) = 6.403

= P[X-2Y>0 ) = P[z>(0+2)/6.403]= 1-P(z<=0.3123) = 0.38

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