(1 point) The song-length of tunes in the Big Hair playlist of a
certain Statistics professors mp3-player vary from song to song.
This variation can be modeled by the Normal distribution, with a
mean song-length of ?=4.1μ=4.1 minutes and a standard deviation of
?=0.6σ=0.6 minutes. Note that a song that has a length of 4.5
minutes is a song that lasts for 4 minutes and 30 seconds.
While listening to a song, the professor decides to shuffle the
playlist, which means the mp3-player is to randomly pick a song
within this particular playlist, and play this next.
If using/working with ?z-values, use three decimals.
(a) What is the probability that the next song to
be played is between 4 and 4.6 minutes long? Answer to four
decimals.
(b) What proportion of all the songs in this
playlist are longer than 5 minutes? Use four decimals in your
answer.
(c) 15% of all the songs in this playlist are at
most how long, in minutes? Enter your answer to two decimals, and
keep your answer consistent with how the song length has been
expressed in this problem.
minutes
(d) There are 225 songs in the Big Hair playlist.
How many of these would you expect to be longer than 5 minutes in
length? Use two decimals in your answer.
songs
(e) From the time he set his mp3-player to
shuffle, there has been 17 songs randomly chosen and played in
succession. What is the chance that the 17-th song played is the
8-th to be longer than 4.1 minutes? Enter your answer to four
decimals.
for normal distribution z score =(X-μ)/σx | |
mean μ= | 4.1 |
standard deviation σ= | 0.60 |
a)
probability =P(4<X<4.6)=P((4-4.1)/0.6)<Z<(4.6-4.1)/0.6)=P(-0.17<Z<0.83)=0.7977-0.4338=0.3639 |
b)
probability =P(X>5)=P(Z>(5-4.1)/0.6)=P(Z>1.5)=1-P(Z<1.5)=1-0.9332=0.0668 |
c)
for 15th percentile critical value of z=-1.036 | ||
therefore corresponding value=mean+z*std deviation=3.5 |
d)
expected to be longer than 5 minutes =np=225*0.0668 =15.03
e)
here this follows negative binomial distribution with parameter r =8 and p=0.5 |
P(X=17)= | (x-1Cr-1) pr(1−p)(x-r) = | 0.0873 |
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