Question

The first two images are the information to answer the 3rd page. The third page is all on question E which is what I need help with. I did answer the questions, but I was asked to give more information about them. For question e, i. I have be wrong standard deviation, but correct Z-score but I’m not sure how to change that. For ii. I have to two 2 z-scores and the pertenctage between them which I’m not sure about. And for iii. How many songs would my answer give me? For question C, I couldn’t find the range or standard deviation so if I could get help for that, if not all good!!

e is an Irish band that has been popular for over three decades. The lengths of 8o of their songs were selected at random and are listed in the table below. -399 Lengths of Randomly Selected U2 Songs (seconds) 382 222 214 38 559 334 262 219 384 331 418 195 338 255 22543 215 252 272 228 331 179 230 186 341 204 253 317 303337 70 286205 350 444 294 260 267 280 66186 308 240 25 279 276, 264 2 Use 50-second intervals to make a frequency distribution table for the data. a. Song Length(s) Frequency 50-99 l00-149 4 50-199 9 Хо-349 a50-a99 300- 349 35G 399 27 400- 449 0 156-49 500-59 5S0- 59 0

Answer 1

Solution :

The length (in seconds) of 80 of U2 (an Irish band) songs were selected at random.

From the data given , we have the following information about different measures (C).

$Standard\ Deviation\ (s)=\sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\overline{x})^2}=77.96433............(Ans)$

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(e) We assume that the given data is normally distributed and that the band's entire collection of songs has a mean and standard deviation equal to those calculated above.

Let X be the length (in seconds) of the songs of U2

$Thus,\ clearly\ X\sim N(\mu=267.2375\ ,\ \sigma^2=(77.96433)^2=6078.437)$

$And,\ Z-Score=\frac{X-\mu}{\sigma}\ \ ;\ \ where,\ Z\sim N(0,1)$

We have to find out the percentage of U2 songs that are expected to be the following:

(i) Over 180 seconds : Thus , we have to find P (X > 180)

$\therefore \ ,\ P(X>180)=P(\frac{X-\mu}{\sigma}>\frac{180-\mu}{\sigma})\ \ \ [\mu=267.2375\ ;\ \sigma=77.96433]$

  $=P(Z>\frac{180-267.2375}{77.96433})=P(Z>-1.11894)$

  $=1-P(Z<-1.11894)=1-(1-P(Z<1.11894))$

  $=1-1+P(Z<1.11894)=P(Z<1.11894)$

  $=\Phi(1.11894)\ \ \ \ \ [\Phi(*)\ is\ the\ Dist\overline{n}\ Func.\ of\ N(0,1)]$

  

Thus, the percentage of U2 songs that are expected to be over 180 seconds = 86.842 % ... (Ans)

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(ii) Between 210 seconds and 300 seconds : Thus , we have to find P (210 < X < 300)

$\therefore \ ,\ P(210<X<300)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{300-\mu}{\sigma})$

  $=P(\frac{210-267.2375}{77.96433}<Z<\frac{300-267.2375}{77.96433})$

  $=P(-0.73415<Z<0.42022)$

  $=P(Z<-0.73415)-P(Z<0.42022))$

  $=\Phi(0.42022)-\Phi(-0.73415)$

  

Thus, the percentage of U2 songs that are expected to be between 210 seconds and 300 seconds = 43.141 % ................... (Ans)

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(f) To find the length of time (in seconds) under which 90% of the U2 songs are expected to be :

Let the length of time (in seconds) of the U2 songs be .

Let the length of time (in seconds) under which there are 90% of the U2 songs be ""

Thus , clearly we have to find "" such that , P ( X < x ) = 0.90.

$\therefore ,\ P(X<x)=0.90\ \Rightarrow\ P(\frac{X-\mu}{\sigma}<\frac{x-\mu}{\sigma})=0.90$

267.2375 77.96433 P (Z < 0.90

$\Rightarrow\ \Phi(\frac{x-267.2375}{77.96433})=0.90=\Phi(1.28155)$

$\Rightarrow\ \frac{x-267.2375}{77.96433}=1.28155$

$\Rightarrow\ x=267.2375+1.28155*77.96433$

$\Rightarrow\ x=367.1527\ seconds..$

The length of time under which 90% of the U2 songs are expected to be = 367.1527 secs ...(Ans)

...Values of Range and Standard Deviation are given at the first ...

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