Sampling Distribution of Mean 1. A manufacturer of batteries for a particular MP

Question

Sampling Distribution of Mean 1. A manufacturer of batteries for a particular MP3 player claims that its batteries, when fully charged, will last for 16 hours of play time before they need recharging. If this claim is true, if the standard deviation of play time is 1.5 hours, and play time has a normal distribution, what is the probability that a a. A randomly selected, fully charged battery will last less than 15 hours? b. Four randomly selected, fully charged batteries will last less than 15 hours on average? c. Sixteen randomly selected, fully charged batteries will last less than 15 hours on average? 2. To promote the use of the Ambassador Bridge, the management of the bridge claims that the time, during rush hour, between entering the bridge in Windsor and reaching a customs and immigration booth in Detroit is normally distributed with a mean time of only 19 minutes and a standard deviation of 4.5 minutes. If this claim is true, a. What is the probability that the actual time for a randomly selected automobile would exceed 22 minutes? b. What is the probability that the actual time for a randomly selected automobile would be less than 24 minutes? c, what is the time that 5% of the automobiles will exceed?

Explanation / Answer

1)a) P(X < 15)

   = P((X - mean)/sd < (15 - 16)/1.5)

   = P(z < -0.67)

   = 0.2514

b) P(X < 15)

= P((X - mean)/(sd/sqrt(n)) < (15 - 16)/(1.5/sqrt(4)))

= P(Z < -1.33)

= 0.0918

c) P(X < 15)

= P((X - mean)/(sd/sqrt(n)) < (15 - 16)/(1.5/sqrt(16)))

= P(Z < -2.67)

= 0.0038

2) a) P(X > 22)

   = P((X - mean)/sd > (22 - 19)/4.5)

= P(Z > 0.67)

= 1 - P(Z < 0.67)

= 1 - 0.7486

= 0.2514

b) P(X < 24)

   = P((X - mean)/sd > (24 - 19)/4.5)

= P(Z < 1.11)

= 0.8665

c) P(X > x) = 0.05

or, P(Z > (x - 19)/4.5) = 0.05

or, P(Z < (x - 19)/4.5) = 0.95

or, (x - 19)/4.5 = 1.645

or, x = 1.645 * 4.5 + 19

or, x = 26.4025

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