# 7. (Sec. 5.4) The lifetime of a certain type of battery is normally distributed with mean value 10 hours and standard deviation 1 hour.(a) If a pack of 4 batteries is purchased, what is the probability that the average lifetime of the batteries in the package is at least 9 hours?(b) How many batteries must be purchased such that the probability that their average lifetime is at least 9.5 hours is .99?

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7. (Sec. 5.4) The lifetime of a certain type of battery is normally distributed with mean value 10 hours and standard deviation 1 hour.

(a) If a pack of 4 batteries is purchased, what is the probability that the average lifetime of the batteries in the package is at least 9 hours?

(b) How many batteries must be purchased such that the probability that their average lifetime is at least 9.5 hours is .99?

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Step 1

Part a)

To calculate the probability that the average lifetime of the batteries in the package is at least 9 hours ie P(X>=9) , we proceed to find the mean and the standard deviation of the pack. help_outlineImage TranscriptioncloseNormal variable X N(10, 1) for a certain type of batteries, Given 10 and o 1 N(10,0.5) A sample of 4 will follow normal distribution too but Y N(u, Y- P(Y2 9)-P(Z 2 9 10 -P(Z 2 0.5 P(Z 2-2) =1 - P(Z <-2) .using Excel function -NORM.S.DIST(-2,TRUE) =1-0.2275 0.97725 fullscreen
Step 2

Part b)

We have to find n  such... help_outlineImage TranscriptioncloseN(10,(1/n) A sample of n will follow normal distribution too but Y ~ N(/u, п 9.5 10 Y-A = P(Z2 -0.5n) =1 - P(Z < -0.5v/n) P(Y 9.5) P(Z 2 = P(Z 2 = 0.99 1-P(Z < -0.5 n) P(Z < -0.5n) = 0.01 Z value corresponding to probability of 0.01 is 2.32635 This can be computed using Excel function =NORM. S. INV(0.01) -2.32635 0.5n n 21.647 n 22 (rounding off) fullscreen

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