A large playlist consists of songs with times which have mea

A large playlist consists of songs with times which have mean 2 minutes and standard deviation 10 seconds.

a.) A random sample of 36 songs is selected. What is the mean of the total length of the 36 songs, in minutes?

b.) If as above 36 songs were randomly selected, what is the standard deviation of the total length of the songs, in minutes?

(Give decimal answer to two places past decimal.)

c.) If 36 songs are selected at random, what is the probability than the total length of the songs is less than 70 minutes?

Give your answer to 3 places after the decimal, with no leading zero.

d.) What is the probability than more than 36 randomly chosen songs are required to fill a program which is 75 minutes long?

Give your answer to 3 places past the decimal, with no leading zero.

Solution

a) Mean = 2 x 36 = 72 mins

b) Standard deviation of total songs = 10 * sqrt(36) = 60 seconds = 1 min

c) P(length < 70) = P(Z < (70-72)/1) = 0.0228

d) In this case you need to find P(length < 75) = P(Z < (75-72)/1) = 0.9987. So there is a 1 - 0.9987 probability that you need more than 36 songs = 0.0013